Question 1

The cost of levelling a circular field at a rate of 50 paise per square metre is Rs 7700. Using π = 22/7, what will be the cost (in rupees) of putting up a fence all around the boundary of this field at Rs 1.20 per metre?

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Introduction / Context: This question combines the concepts of area and circumference of a circle with unit cost calculations. First, we use the total cost and rate per square metre to determine the area of the circular field. From the area, we determine the radius of the circle using the formula for the area of a circle. Once we know the radius, we calculate the circumference, which represents the length of fencing required. Finally, we multiply this length by the cost per metre to obtain the total fencing cost. Given Data / Assumptions: Cost of levelling = Rs 7700. Rate of levelling = 50 paise per square metre = Rs 0.50 per square metre. The field is circular. Use π = 22/7. Cost of fencing = Rs 1.20 per metre. No wastage or additional costs are considered. Concept / Approach: The area of the circular field is given by area = cost of levelling / rate per square metre. Once we have the area, we use the formula area = π * r^2 to solve for r, the radius. After finding the radius, we find the circumference using C = 2 * π * r. The total cost of fencing is then the circumference multiplied by the rate per metre of fencing. All steps require careful handling of units and substitution of π as 22/7. Step-by-Step Solution: Rate of levelling = Rs 0.50 per m^2.Total cost = Rs 7700.Area of the circular field = 7700 / 0.50 = 15400 m^2.Use area formula: π * r^2 = 15400.With π = 22/7, we have (22 / 7) * r^2 = 15400.Multiply both sides by 7: 22 * r^2 = 15400 * 7.Compute 15400 / 22 = 700 (since 22 * 700 = 15400).Thus r^2 = 700 * 7 = 4900, so r = √4900 = 70 m.Circumference of the field = 2 * π * r = 2 * (22 / 7) * 70.Compute: (22 / 7) * 70 = 22 * 10 = 220, so circumference = 2 * 220 = 440 m.Fencing cost per metre = Rs 1.20.Total fencing cost = 440 * 1.20 = Rs 528. Verification / Alternative check: We can quickly recompute the intermediate values. Area: 7700 at Rs 0.50 per m^2 gives 7700 / 0.5 = 15400 m^2, correct. Radius: area = (22 / 7) * r^2 and r^2 = 15400 * 7 / 22 = 700 * 7 = 4900, hence r = 70 m. Circumference: 2 * (22 / 7) * 70 = 440 m, and 440 * 1.20 = 440 * (6 / 5) = 528, confirming the arithmetic. Every step checks out and supports the final answer. Why Other Options Are Wrong: Rs 132 and Rs 264 correspond to using a much smaller radius or miscalculating the circumference. Rs 1056 is exactly double the correct answer, which might come from mistakenly doubling the circumference or cost. Rs 880 would require a circumference or cost rate inconsistent with the given data. Only Rs 528 matches the correct calculation based on the actual radius and fencing rate. Common Pitfalls: Common errors include forgetting to convert 50 paise into Rs 0.50, miscomputing r^2 from the area, or using an approximate value for π instead of the specified 22/7. Some candidates also mistakenly treat 7700 as the area instead of converting it through the rate. Careful handling of units and stepwise substitution avoids these errors and leads directly to the correct fencing cost. Final Answer: The cost of putting up a fence all around the field is Rs 528.

Question 2

From each of the four corners of a rectangular sheet of cardboard of dimensions 25 cm × 20 cm, a square of side 2 cm is cut off. The remaining piece is then folded up along the cut edges to form an open rectangular box (without a lid). What is the v…

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Introduction / Context: This is a standard problem involving converting a flat rectangular sheet into an open box by cutting equal squares from each corner and folding up the sides. The cut-out squares determine the height of the box, and the remaining central rectangle determines the base dimensions. To find the volume of the resulting box, we calculate the length, breadth, and height of the box and then apply the volume formula for a cuboid (rectangular box). Given Data / Assumptions: Original sheet dimensions: length = 25 cm, breadth = 20 cm. From each corner, a square of side 2 cm is removed. The sheet is folded along the cut edges to form an open box (no lid). The height of the box equals the side of the square cut, that is 2 cm. We must find the volume of the resulting box in cubic centimetres. Concept / Approach: When squares are cut from each corner and the sides are folded up, the length and breadth of the base reduce by twice the side of the square cut, because we remove 2 cm at each end of the dimension. The height of the resulting box is the same as the side of the square. Once we know the effective length, breadth, and height, we use volume = length * breadth * height for the box. The problem therefore reduces to simple subtraction and multiplication. Step-by-Step Solution: Original length = 25 cm, original breadth = 20 cm.Side of each cut-out square = 2 cm.When we cut and fold, the height of the box = 2 cm.New length of the box base = 25 − 2 − 2 = 21 cm.New breadth of the box base = 20 − 2 − 2 = 16 cm.Height of the box = 2 cm.Volume of the box = length * breadth * height.So volume = 21 * 16 * 2.First 21 * 16 = 21 * (8 * 2) = (21 * 8) * 2 = 168 * 2 = 336.Now multiply by height 2: 336 * 2 = 672 cm³. Verification / Alternative check: We can multiply systematically: 25 − 4 = 21 and 20 − 4 = 16. Then 21 * 16 = (20 + 1) * 16 = 320 + 16 = 336, and times 2 gives 672. The calculations are simple and consistent. Also, since the original rectangle is 25 * 20 = 500 cm², the base 21 * 16 = 336 cm² is smaller due to the removed strips of width 2 cm along each side, which matches the folding logic. This confirms that the dimensions used are realistic and the volume is correctly computed. Why Other Options Are Wrong: The volume 828 cm³ would require at least one dimension to be larger than what is available after cutting. The value 500 cm³ would correspond to using the original area as volume with a height of 1 cm, which is incorrect. The value 1000 cm³ is far too large for these dimensions and height. The value 420 cm³ would come from incorrect length or breadth after cutting. Only 672 cm³ matches the exact calculation for the reduced base and given height. Common Pitfalls: Common mistakes include subtracting only 2 cm instead of 4 cm from each dimension, forgetting that the cut is taken from both ends. Others may misinterpret the height and think it is still the original thickness or some other value. Some also confuse area with volume and may incorrectly stop at 21 * 16 instead of multiplying by the height. Remembering that all three dimensions are needed for volume is crucial. Final Answer: The volume of the open box formed from the cardboard sheet is 672 cm³.

Question 3

The sum of the length and breadth of a rectangle is 6 cm. A square is constructed whose side is equal to the diagonal of this rectangle. If the ratio of the area of the square to the area of the rectangle is 5 : 2, what is the area (in cm²) of the s…

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Introduction / Context: This question links a rectangle and a square through the diagonal of the rectangle. The diagonal is used as the side length of the square. Additionally, a ratio of areas between the square and the rectangle is given. The problem involves using the relationships between the sides of a rectangle, its diagonal, and the area expression, along with the area ratio, to determine the area of the square. This tests algebraic manipulation and geometric understanding of rectangles and squares. Given Data / Assumptions: Let the rectangle have length l and breadth b. Sum of length and breadth: l + b = 6 cm. Diagonal of rectangle = d = √(l^2 + b^2). A square is formed with side equal to d, so area of square = d^2. Ratio of area of the square to the area of the rectangle is 5 : 2. Area of rectangle = l * b. Concept / Approach: The area of the square formed on the diagonal is d^2, and for the rectangle, the area is l * b. Given that the ratio of the areas is 5 : 2, we have d^2 / (l * b) = 5 / 2. But d^2 = l^2 + b^2. Additionally, (l + b)^2 = l^2 + b^2 + 2lb; and we know l + b = 6, which gives an equation linking l^2 + b^2 and lb. Using these two relationships, we solve for lb and then compute the area of the square using the ratio 5 : 2. Step-by-Step Solution: Let l and b be the sides of the rectangle.Given l + b = 6 cm.Area of rectangle = l * b.Diagonal d satisfies d^2 = l^2 + b^2.Given area ratio: (area of square) / (area of rectangle) = 5 / 2.Area of square = d^2 = l^2 + b^2.So (l^2 + b^2) / (l * b) = 5 / 2.This implies 2(l^2 + b^2) = 5 l b.Also, (l + b)^2 = l^2 + b^2 + 2 l b = 6^2 = 36.From 2(l^2 + b^2) = 5 l b, we have l^2 + b^2 = (5 / 2) l b.Substitute into (l + b)^2 equation: (5 / 2) l b + 2 l b = 36.Combine coefficients: (5 / 2 + 2) l b = (5 / 2 + 4 / 2) l b = (9 / 2) l b.Thus (9 / 2) l b = 36, so l b = 36 * (2 / 9) = 8.Area of rectangle = l b = 8 cm².Area of square : area of rectangle = 5 : 2, so area of square = (5 / 2) * 8 = 20 cm². Verification / Alternative check: We can check if there exist l and b satisfying l + b = 6 and l b = 8. These are roots of the quadratic t^2 − 6t + 8 = 0, so t = 2 or 4. Thus, l = 4 and b = 2 (or vice versa). Diagonal d^2 = 4^2 + 2^2 = 16 + 4 = 20, so area of square on the diagonal is indeed 20 cm². Area of rectangle is 4 * 2 = 8 cm², and 20 : 8 simplifies to 5 : 2. This confirms that the result is consistent with the given ratio and conditions. Why Other Options Are Wrong: The value 10 cm² would correspond to a ratio closer to 5 : 4, not 5 : 2. The value 4√5 is not even in square units and appears to be a misinterpretation of the diagonal length. The value 10√2 would also be a misinterpretation of the diagonal rather than its squared area. The value 25 cm² does not satisfy the ratio 5 : 2 when compared with a consistent rectangle area. Only 20 cm² fits both the algebraic derivation and the given ratio exactly. Common Pitfalls: Some students mistakenly use the diagonal directly as the answer instead of its square when asked for the area of the square. Others might misapply the relationship between (l + b)^2 and l^2 + b^2 + 2 l b or mishandle the algebra when solving for l b. Forgetting that the ratio refers to areas and not side lengths is another common source of error. Keeping track of each step and systematically using the algebraic identities avoids these mistakes. Final Answer: The area of the square constructed on the diagonal of the rectangle is 20 cm².

Question 4

An equilateral triangle has side length 8 cm. A circumcircle is drawn around the triangle, and an incircle is drawn inside it, touching all three sides. Using π = 22/7, what is the area (in sq.cm) of the region lying between the circumcircle and the…

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Introduction / Context: This problem involves an equilateral triangle and the two special circles associated with it: the circumcircle (passing through all vertices) and the incircle (tangent to all three sides). The question asks for the area of the ring-shaped region between these two circles. To solve this, we use known formulas for the circumradius and inradius of an equilateral triangle in terms of its side, compute the areas of both circles, and subtract the incircle area from the circumcircle area. Given Data / Assumptions: Side length of the equilateral triangle, a = 8 cm. Use π = 22/7. Circumcircle radius is denoted by R. Incircle radius is denoted by r. We need area of the region between the two circles, that is πR^2 − πr^2. Concept / Approach: For an equilateral triangle of side a, the circumradius R and inradius r have standard formulas: R = a / √3 and r = a / (2√3). Once we compute R and r, the area of the circumcircle is πR^2 and the area of the incircle is πr^2. The required area is the difference π(R^2 − r^2). After simplifying R^2 − r^2 symbolically, we substitute the side length and then use π = 22/7 to get a numerical mixed fraction. Step-by-Step Solution: Given side a = 8 cm for the equilateral triangle.Circumradius R of an equilateral triangle: R = a / √3.Thus R = 8 / √3.Inradius r of an equilateral triangle: r = a / (2√3).Thus r = 8 / (2√3) = 4 / √3.Compute R^2: R^2 = (8 / √3)^2 = 64 / 3.Compute r^2: r^2 = (4 / √3)^2 = 16 / 3.Difference R^2 − r^2 = 64 / 3 − 16 / 3 = 48 / 3 = 16.Area between circumcircle and incircle = π(R^2 − r^2) = π * 16.Using π = 22 / 7, area = 16 * (22 / 7) = 352 / 7.Convert 352 / 7 to a mixed fraction: 7 * 50 = 350, remainder 2, so 352 / 7 = 50 2/7 sq.cm. Verification / Alternative check: We can also compute the areas directly as a check. Circumcircle area = πR^2 = (22 / 7) * (64 / 3) and incircle area = πr^2 = (22 / 7) * (16 / 3). Their difference is (22 / 7) * (64 / 3 − 16 / 3) = (22 / 7) * (48 / 3) = (22 / 7) * 16 = 352 / 7. This reconfirms the simplified result, showing that the intermediate steps are consistent and produce the same final area for the ring-shaped region. Why Other Options Are Wrong: The value 50 1/7 sq.cm would correspond to an area of 351 / 7, which is slightly less than the correct 352 / 7. Values 75 1/7 sq.cm and 75 2/7 sq.cm are substantially larger and do not match the derived difference of 16π. The value 40 sq.cm ignores the π factor and is not compatible with π = 22 / 7. Only 50 2/7 sq.cm matches the exact computation based on the given triangle side and circle formulas. Common Pitfalls: Typical errors include using incorrect formulas for R or r, such as mixing them up or forgetting the √3 in the denominators. Some students may subtract the radii instead of subtracting the squared radii in the area difference formula. Others may approximate π incorrectly or mishandle the conversion of an improper fraction to a mixed fraction. Carefully applying the correct radius formulas and performing arithmetic step by step avoids these mistakes. Final Answer: The area of the region between the circumcircle and the incircle of the equilateral triangle is 50 2/7 sq.cm.

Question 5

A solid sphere of radius 3 cm is melted and recast into a hollow right circular cylindrical tube of length 4 cm and external radius 5 cm. Assuming no loss of material and uniform thickness throughout, what is the thickness (in cm) of the cylindrical…

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Introduction / Context: This question combines volume conservation with the geometry of a hollow cylinder. A solid sphere is melted and reshaped into a cylindrical tube with a given external radius and length, but unknown internal radius. The difference between the outer and inner radii represents the thickness of the tube. Since no material is lost in the process, the volume of the sphere must equal the volume of the hollow cylinder. We use volume formulas to find the internal radius and then compute the thickness. Given Data / Assumptions: Radius of the sphere, R = 3 cm. The sphere is melted and formed into a hollow cylinder (tube). Length (height) of the cylindrical tube, h = 4 cm. External radius of the tube, R_outer = 5 cm. Internal radius of the tube, R_inner = r (unknown). No loss of volume during melting and recasting. Concept / Approach: The volume of the material in the hollow cylinder equals the volume of the sphere. The volume of the sphere is (4 / 3) * π * R^3. The volume of a hollow cylinder is π * (R_outer^2 − R_inner^2) * h. By equating these two volumes, we can solve for R_inner^2, then find R_inner. Finally, the thickness of the tube is the difference between the outer and inner radii: thickness = R_outer − R_inner. Step-by-Step Solution: Volume of the sphere = (4 / 3) * π * R^3, with R = 3 cm.Compute R^3: 3^3 = 27.So volume of sphere = (4 / 3) * π * 27 = 4 * 9 * π = 36π cubic centimetres.Volume of hollow cylinder = π * (R_outer^2 − R_inner^2) * h.Here R_outer = 5 cm, h = 4 cm, R_inner = r.So volume of hollow cylinder = π * (25 − r^2) * 4 = 4π(25 − r^2).Equate the volumes: 36π = 4π(25 − r^2).Cancel π from both sides to get 36 = 4(25 − r^2).Divide both sides by 4: 9 = 25 − r^2.So r^2 = 25 − 9 = 16.Thus r = 4 cm (internal radius).Thickness of the tube = external radius − internal radius = 5 − 4 = 1 cm. Verification / Alternative check: Substitute r = 4 cm back into hollow cylinder volume to check. Volume = 4π(25 − 16) = 4π * 9 = 36π cubic centimetres, which matches the volume of the sphere. This confirms that the inner radius is correct and that the derived thickness of 1 cm is consistent with the conservation of volume. Why Other Options Are Wrong: A thickness of 9 cm or 1.5 cm would imply an internal radius that is negative or unrealistically small, and the resulting hollow cylinder would not have the same volume as the sphere. A thickness of 0.6 cm or 2 cm similarly does not satisfy the volume equality and would give either too little or too much volume in the cylindrical shell. Only a thickness of 1 cm correctly balances the volumes and matches all the geometric constraints. Common Pitfalls: Common errors include forgetting to subtract the inner cylinder volume from the outer cylinder volume, miscomputing 25 − r^2, or confusing the sphere radius with the cylinder radius. Some students also forget to cancel π, which can complicate calculations unnecessarily. Being systematic about equating volumes and carefully handling squares and subtractions prevents these mistakes. Final Answer: The thickness of the hollow cylindrical tube is 1 cm.

Question 6

Two equal circles intersect in such a way that their centres and their two points of intersection form the vertices of a square of side 1 cm. What is the area (in sq.cm) of the region that is common to both circles?

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Introduction / Context: This problem concerns the intersection area (lens-shaped region) of two equal circles. The configuration is special: the centres of the circles and the two intersection points together form the four vertices of a square of side 1 cm. We must find the area common to both circles using this geometric information. This involves understanding the relationship between the distance between the centres, the radius of each circle, and the area of overlap of two equal circles. Given Data / Assumptions: Two circles have equal radii (say radius R). The centres of the circles and the two intersection points form a square of side 1 cm. All sides of this square measure 1 cm. The distance between the centres of the circles equals the length of one side of the square or its diagonal, depending on which vertices correspond to the centres. We must find the area common to both circles in square centimetres. Concept / Approach: In the described configuration, it turns out that the centres of the circles lie at opposite corners of the square and the intersection points lie at the other two corners. This means that the distance between the centres equals the diagonal of the square, which is side * √2 = 1 * √2 = √2. For two equal circles of radius R that intersect, the distance between centres is usually denoted by d. Here d = √2. Geometry of the square implies that the radius of each circle is R = 1. Once R and d are known, we use the standard formula for the overlap area of two equal circles of radius R with centre distance d. Step-by-Step Solution: Let R be the radius of each circle and d be the distance between the centres.Since the centres and intersection points form a square of side 1 cm, the distance between opposite vertices (a diagonal) is 1 * √2 = √2.The centres occupy opposite corners, so d = √2.In such a configuration, geometry shows that the radius R equals 1 cm (so that the intersection points also lie at distance 1 from each centre).The area of intersection of two equal circles of radius R with centre distance d is:Area = 2R^2 * cos^−1(d / (2R)) − (d / 2) * √(4R^2 − d^2).Here R = 1 and d = √2.Compute d / (2R) = √2 / 2 = 1 / √2.cos^−1(1 / √2) = 45° or π / 4 radians.Thus 2R^2 * cos^−1(d / (2R)) = 2 * 1^2 * (π / 4) = π / 2.Now compute the second term: (d / 2) * √(4R^2 − d^2).4R^2 − d^2 = 4 * 1^2 − (√2)^2 = 4 − 2 = 2.So √(4R^2 − d^2) = √2.Therefore (d / 2) * √(4R^2 − d^2) = (√2 / 2) * √2 = (√2 / 2) * √2 = 1.Hence the common area = (π / 2) − 1 square centimetres. Verification / Alternative check: We can reason qualitatively as well. Since R = 1, each circle has area π. The overlap region is less than half of either circle, and π / 2 − 1 is indeed less than π / 2. Numerically, π / 2 ≈ 1.57, so π / 2 − 1 ≈ 0.57 square centimetres, which seems reasonable for a modest overlapping lens from unit circles with centre distance √2. This rough magnitude check supports the correctness of the detailed formula-based calculation. Why Other Options Are Wrong: π/4 is roughly 0.785, which is larger than the derived overlap and does not match the standard overlap formula for circles at this distance. π/5 is even smaller and does not follow from any standard expression here. The value √2 − 1 is approximately 0.414, which is unrelated to the circle overlap formula. The option π − 1 would give an area close to 2.14, which is much larger than half of a unit circle and impossible for the overlap of two such circles. Only π/2 − 1 fits the exact expression derived from the known formula for the intersection of two equal circles with centre distance √2 and radius 1. Common Pitfalls: Many students are unfamiliar with the exact formula for the area of overlap of two circles and may try to approximate or guess the answer. Others might misinterpret the square configuration and incorrectly set the distance between centres equal to 1 instead of √2. Mistakes also arise from forgetting that the radius is 1, not 1/2 or another value. Careful geometric visualization of the square and consistent use of the standard formula are essential to solve this correctly. Final Answer: The area of the region common to both circles is π/2 − 1 square centimetres.

Question 7

PQRA is a rectangle with AP = 22 cm and PQ = 8 cm. A triangle ABC is drawn inside the rectangle so that its vertices lie on the sides of PQRA and BQ = 2 cm, QC = 16 cm along the sides meeting at Q. What is the length (in cm) of the line segment join…

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Introduction / Context: This problem places a triangle inside a rectangle with specific distances along the sides, then asks for the length of the line segment joining the midpoints of two sides of the triangle. It tests understanding of coordinate geometry, segment midpoints, and distance calculation. By assigning coordinates to the rectangle and triangle vertices, we can find the midpoints of AB and BC and then compute the distance between these two midpoints using the distance formula. Given Data / Assumptions: PQRA is a rectangle. AP = 22 cm and PQ = 8 cm. Triangle ABC is drawn with vertices on the sides of the rectangle. At vertex Q of the rectangle, BQ = 2 cm along one side and QC = 16 cm along the other adjacent side. We must find the length of the line segment joining the midpoints of AB and BC. Concept / Approach: We can set up a coordinate system to simplify calculations. Place the rectangle PQRA so that its vertices have convenient coordinates, then locate points B and C based on the given distances from Q along the sides. Choose A as the fourth corner of the rectangle that forms triangle ABC, so A, B, and C are three vertices of the triangle. Then we determine the midpoints of AB and BC and use the distance formula to find the length of the segment joining them. This coordinate approach avoids complicated geometry and leads to a straightforward algebraic solution. Step-by-Step Solution: Let us place the rectangle so that A = (0, 0), P = (8, 0), Q = (8, 22), and R = (0, 22).This makes AP = 22 cm (vertical) and PQ = 8 cm (horizontal), consistent with the given data.At vertex Q = (8, 22), side QP goes downward and side QR goes leftward.Given BQ = 2 cm along QR and QC = 16 cm along QP (or vice versa), a consistent placement is: B lies on QR and C lies on QP.Let B be 2 cm left of Q on QR: B = (8 − 2, 22) = (6, 22).Let C be 16 cm down from Q on QP: C = (8, 22 − 16) = (8, 6).Take A = (0, 0) as the rectangle corner opposite Q, so triangle ABC has vertices A(0, 0), B(6, 22), and C(8, 6).Find midpoint of AB: M = ((0 + 6) / 2, (0 + 22) / 2) = (3, 11).Find midpoint of BC: N = ((6 + 8) / 2, (22 + 6) / 2) = (7, 14).Distance MN = √[(7 − 3)^2 + (14 − 11)^2] = √(4^2 + 3^2) = √(16 + 9) = √25 = 5 cm. Verification / Alternative check: The coordinates of M and N are carefully derived as midpoints, and the distance calculation is straightforward. The vector from M to N is (7 − 3, 14 − 11) = (4, 3). The length of this vector is √(4^2 + 3^2) = √25 = 5, which is a familiar 3–4–5 right triangle relationship. This simple right triangle structure is another indication that the chosen coordinates and distances are correct and that the final answer is reliable. Why Other Options Are Wrong: 4√2 cm (approximately 5.66 cm) and 3√5 cm (approximately 6.71 cm) are larger than the correct value and would arise from miscalculating the differences in coordinates or using incorrect midpoints. Values 6 cm and 10 cm are rough guesses that do not correspond to the precise distance between the midpoints given the actual coordinates. Only 5 cm exactly matches the computed distance between M and N. Common Pitfalls: Some students may misinterpret the placement of B and C along the sides from Q and assign incorrect coordinates. Others might confuse which corner is A when forming triangle ABC, leading to a different and incorrect triangle. Mistakes in computing midpoints or in applying the distance formula (especially squaring and adding) are also common. A clear diagram, consistent coordinate assignment, and careful arithmetic prevent these errors. Final Answer: The length of the segment joining the midpoints of AB and BC is 5 cm.

Question 8

In triangle ABC, points D and E lie on sides AB and AC respectively such that DE is parallel to BC and AD : DB = 4 : 5. Segments CD and BE intersect at F. What is the ratio of the area of triangle DEF to the area of triangle CBF?

Accepted Answer

Introduction / Context: This geometry problem involves a triangle with a segment parallel to one side, creating similar triangles, and then concurrent segments intersecting at a point inside the triangle. The question asks for the ratio of areas of two smaller triangles formed by these constructions. It tests understanding of similarity, area ratios in similar triangles, and the effect of cevians (segments like BE and CD) on area distribution when they intersect inside a triangle. Given Data / Assumptions: Triangle ABC is a general triangle. D is on AB and E is on AC such that DE ∥ BC. AD : DB = 4 : 5, so D divides side AB in that ratio. Segments CD and BE intersect at F. We must find area(ΔDEF) : area(ΔCBF). No specific side lengths are given, so we can assume convenient coordinates for the vertices. Concept / Approach: Since DE is parallel to BC, triangle ADE is similar to triangle ABC. The ratio AD : AB gives the linear scaling between these triangles. From similarity, the ratio of their areas equals the square of the ratio of corresponding sides. After finding the area of ΔADE in terms of area of ΔABC, we infer how much area is left in the trapezoidal region between DE and BC. Using a coordinate approach, we then locate D and E, find the intersection F of BE and CD, and compute the areas of ΔDEF and ΔCBF to get their ratio. Because the scaling is consistent, the final ratio is independent of the particular coordinates chosen. Step-by-Step Solution: Let us place triangle ABC in coordinates for convenience.Take A = (0, 0), B = (9, 0), and C = (0, 9), so AB = 9 and AC = 9, forming a right triangle for easier arithmetic.On AB, AD : DB = 4 : 5, so AB = 9 corresponds to 4k + 5k = 9k, giving k = 1.Thus AD = 4 and DB = 5, so D = (4, 0).DE ∥ BC. The line BC goes from B(9, 0) to C(0, 9) with direction vector (−9, 9).A line through D that is parallel to BC has the same direction, so starting from D(4, 0) and moving in direction (−9, 9) will meet AC.AC runs from A(0, 0) to C(0, 9), so it is the y-axis segment.The line through D has equation x = 4 − 9t, y = 0 + 9t.It meets AC where x = 0, so 4 − 9t = 0 gives t = 4 / 9, so y = 9 * (4 / 9) = 4.Hence E = (0, 4).Now, line BE connects B(9, 0) and E(0, 4), and line CD connects C(0, 9) and D(4, 0). Their intersection is F.By solving these line equations, we obtain F = (36 / 13, 36 / 13).We now compute area(ΔDEF) and area(ΔCBF).Using coordinate geometry or polygon area formulas, area(ΔDEF) = 40 / 13 and area(ΔCBF) = 405 / 26.The ratio area(ΔDEF) : area(ΔCBF) = (40 / 13) : (405 / 26) = (40 / 13) * (26 / 405).Simplify: (40 * 26) / (13 * 405) = (40 * 2) / 405 = 80 / 405 = 16 / 81.Thus area(ΔDEF) : area(ΔCBF) = 16 : 81. Verification / Alternative check: We can reason more conceptually. Since AD : AB = 4 : 9, triangle ADE is similar to triangle ABC with linear ratio 4 : 9. Therefore, area(ΔADE) : area(ΔABC) = 4^2 : 9^2 = 16 : 81. The region between DE and BC accounts for the remaining area, which is area(ΔABC) − area(ΔADE) = (81 − 16) / 81 of the total. It is consistent that ΔDEF is a small part of ΔADE and ΔCBF is a part of the lower larger region, giving a ratio of 16 : 81 overall. The coordinate computations align with this conceptual expectation. Why Other Options Are Wrong: The ratio 16 : 25 would arise from incorrectly using side length ratios rather than squared side ratios. The ratio 81 : 16 is simply the inverse of the correct answer. The ratios 4 : 9 and 25 : 16 come from misapplied similarity arguments or mixing linear and area ratios. Only 16 : 81 matches both the precise coordinate geometry calculation and the similarity-based reasoning. Common Pitfalls: Students often confuse linear similarity ratios with area ratios and may use 4 : 9 directly instead of 16 : 81. Others might misplace points D and E or incorrectly assume symmetric positions that do not match the ratio AD : DB = 4 : 5. Algebraic mistakes while solving for F or computing triangle areas can also occur. Working with clear coordinates and remembering that area scales with the square of the linear factor helps avoid these errors. Final Answer: The ratio of the areas of triangles DEF and CBF is 16 : 81.

Question 9

In trapezium ABCD, AB is parallel to CD and the diagonals AC and BD intersect at point O. The longer base AB is twice the shorter base CD (AB = 2 CD). What is the ratio of the area of triangle AOB to the area of triangle COD?

Accepted Answer

Introduction / Context: In this geometry question we work with a trapezium, a quadrilateral that has exactly one pair of parallel sides. The problem focuses on the relationship between the parallel sides of the trapezium and the areas of two triangles formed by its diagonals. Understanding how diagonals intersect inside a trapezium and how areas depend on base lengths and heights is essential for solving many coordinate and Euclidean geometry questions in competitive exams. Given Data / Assumptions: ABCD is a trapezium with AB parallel to CD. The diagonals AC and BD intersect at point O inside the trapezium. The longer base AB is twice the length of the shorter base CD, so AB = 2 * CD. We are asked to find the ratio of areas: area(ΔAOB) : area(ΔCOD). All points lie in a plane and the figure is a standard Euclidean trapezium. Concept / Approach: The key concept is that when diagonals of a trapezium intersect, triangles formed between one base and the intersection point share the same height but have different bases along the parallel sides. For triangles with the same altitude, the ratio of their areas equals the ratio of their corresponding bases. In this problem, the area of triangle AOB is related to base AB, and the area of triangle COD is related to base CD, after we analyze how the diagonals intersect and form similar or proportionate triangles inside the trapezium. Step-by-Step Solution: Step 1: Consider trapezium ABCD with AB parallel to CD and AB = 2 * CD. Step 2: Draw the diagonals AC and BD so that they intersect at point O inside the trapezium. Step 3: Use the known property of a trapezium: the diagonals intersect in such a way that the areas of triangles formed near the parallel sides are proportional to the squares of the corresponding bases, or equivalently, that the segments of diagonals relate to the bases. Step 4: A convenient method is to place the trapezium on a coordinate plane. Let AB be the longer base on the x-axis and CD be the shorter base above it. A coordinate model shows that area(ΔAOB) becomes proportional to AB * some height, while area(ΔCOD) becomes proportional to CD * a related height. Step 5: More directly, using the known result derived from coordinate geometry or vector methods, when AB is twice CD, the intersection of diagonals divides the figure so that area(ΔAOB) : area(ΔCOD) = (AB / CD)^2 = (2 / 1)^2 = 4 : 1. Step 6: Therefore the required ratio of areas of triangles AOB and COD is 4 : 1. Verification / Alternative check: An easy verification is to assume specific coordinates. For example, take AB from (0,0) to (2,0) and CD from (0,1) to (1,1), creating a simple trapezium with AB = 2 and CD = 1. Compute the intersection point of diagonals AC and BD, then use the coordinate formula for the area of a triangle. When you do this, you will find that area(ΔAOB) is four times area(ΔCOD), which confirms the theoretical ratio 4 : 1 found by the algebraic argument. Why Other Options Are Wrong: Option 1: 1:4 is the inverse of the correct ratio. It would suggest that the triangle near the shorter base has a larger area, which contradicts the effect of AB being longer than CD. Option 3: 2:1 ignores the fact that the relationship involves the square of the base ratio, not just the simple linear ratio of the bases. Option 4: 3:1 has no geometric justification from the properties of intersecting diagonals in a trapezium where AB = 2 * CD. Option 5: None of these is incorrect because one of the provided ratios, 4:1, matches the correct area ratio. Common Pitfalls: Students often confuse the ratio of areas with the ratio of sides directly, forgetting that in many configurations the area ratio involves squares or products of sides. Another common mistake is to assume that diagonals of a trapezium behave the same way as diagonals in a parallelogram, which is not true. Failing to distinguish between which triangles share a common height and which triangles share a common base can easily lead to an incorrect ratio. It is also easy to accidentally invert the ratio if you do not keep consistent order, such as writing area(ΔCOD) : area(ΔAOB) instead of the ratio requested. Final Answer: The ratio of the areas of triangles AOB and COD is 4:1.

Question 10

In triangle ABC, O is the orthocentre (the point where all three altitudes meet). If the measure of angle BOC is 100°, what is the measure of angle BAC?

Accepted Answer

Introduction / Context: This question tests a standard but very important property of the orthocentre of a triangle. The orthocentre is the common point of intersection of the three altitudes of a triangle. There is a special relationship between the angles at the orthocentre and the angles at the vertices of the triangle. Many geometry questions in exams use this relationship, so recognizing the formula will save time and reduce calculations. Given Data / Assumptions: ABC is a triangle. O is the orthocentre of triangle ABC. The measure of angle BOC is 100°. We are asked to find the measure of angle BAC. The triangle is assumed to be acute-angled so that the orthocentre lies inside the triangle, which is the usual context for this standard relation. Concept / Approach: For an acute triangle, there is a well known relation between the angle at a vertex and the angle at the orthocentre opposite that vertex. In particular, the angle between the lines OB and OC at the orthocentre is related to the angle at vertex A by the formula: ∠BOC = 180° − ∠A where ∠A is the same as angle BAC. Using this formula, we can directly compute the missing vertex angle once ∠BOC is known, by simply rearranging the equation. Step-by-Step Solution: Step 1: Recall the relation for an acute triangle: ∠BOC = 180° − ∠A, where ∠A = ∠BAC. Step 2: The question gives ∠BOC = 100°. Substitute this into the formula: 100° = 180° − ∠BAC Step 3: Rearrange to find angle BAC: ∠BAC = 180° − 100° = 80° Step 4: Therefore, the measure of angle BAC is 80°. Verification / Alternative check: A quick verification can be done by constructing an approximate acute triangle and measuring angles. If we choose a triangle with angle A = 80°, then the relation predicts ∠BOC = 180° − 80° = 100°, which matches the given value. This consistency confirms that our formula has been applied correctly and the computed angle is accurate. Why Other Options Are Wrong: Option 2: 50° does not satisfy the relation 180° − ∠A = 100°. If ∠A were 50°, ∠BOC would be 130°, not 100°. Option 3: 100° is simply the value of ∠BOC. Using it again as angle BAC ignores the known geometric relation between these angles. Option 4: 60° would give ∠BOC = 120°, which contradicts the given data. Option 5: None of these is incorrect because 80° is a valid option and satisfies the exact formula. Common Pitfalls: Students sometimes confuse the orthocentre relation with the circumcentre relation. For the circumcentre of a triangle, the angle at the centre opposite a side is 2 times the angle at the circumference. For the orthocentre, the relation involves 180° minus the vertex angle instead. Another common pitfall is to assume the triangle is right angled or obtuse without checking context. In a right triangle, the orthocentre lies at the right angled vertex, and different relations apply. Here the use of the standard formula for an acute triangle is appropriate and leads directly to the correct result. Final Answer: The measure of angle BAC is 80°.

Question 11

A triangle has sides of lengths 6 cm, 8 cm and 10 cm and is inscribed in a circle. What is the area, in square centimetres, of the circumcircle of this triangle?

Accepted Answer

Introduction / Context: In this question we relate a right angled triangle to its circumcircle. When a triangle is right angled, the circumcircle has a very special property: the hypotenuse is the diameter of the circumcircle. This allows us to find the radius quickly and then compute the area of the circle. Problems of this type are common in aptitude tests and help reinforce the connection between triangle geometry and circles. Given Data / Assumptions: The sides of the triangle are 6 cm, 8 cm and 10 cm. The triangle is inscribed in a circle, so the circle is its circumcircle. We must find the area of this circumcircle in square centimetres. No value of π is explicitly given, so we can safely use π = 22/7 to match the answer forms expressed as a rational multiple of 1/7. Concept / Approach: First, notice that 6, 8 and 10 form a Pythagorean triple: 6^2 + 8^2 = 36 + 64 = 100 = 10^2. This means the triangle is right angled, with the side of length 10 cm as the hypotenuse. For a right angled triangle, the circumcircle has a diameter equal to the hypotenuse. Thus, the radius r of the circumcircle is half the hypotenuse. Once the radius is known, the area of the circle can be found using the formula: Area = π * r^2 Step-by-Step Solution: Step 1: Verify the triangle is right angled. Compute 6^2 + 8^2 = 36 + 64 = 100 and 10^2 = 100, so the triangle is right angled with hypotenuse 10 cm. Step 2: Use the property of the circumcircle of a right angled triangle: the hypotenuse is the diameter. Thus, diameter = 10 cm, so radius r = 10 / 2 = 5 cm. Step 3: Write the area formula for the circle: Area = π * r^2. Substitute r = 5 cm: Area = π * 5^2 = 25π. Step 4: Express area as a rational number using π = 22/7 to match the form of the options. Area = 25 * (22 / 7) = (25 * 22) / 7 = 550 / 7 square centimetres. Step 5: Compare with the options and choose 550/7 sq cm. Verification / Alternative check: We can quickly approximate 550/7. Since 7 * 78 = 546 and 7 * 79 = 553, 550/7 is a little more than 78.5 square centimetres. If we approximate π as 3.14, the area from 25π is 25 * 3.14 = 78.5 square centimetres, which matches the approximation from 550/7. This confirms that the conversion from 25π to 550/7 is correct and consistent with the approximate decimal value. Why Other Options Are Wrong: Option 1: 275/7 sq cm represents half of the correct area and would correspond to 12.5π, which is not the area of a circle with radius 5 cm. Option 3: 2200/7 sq cm is four times the correct answer and would correspond to 100π, as if the radius were 10 cm instead of 5 cm. Option 4: 1100/7 sq cm is twice the correct answer and would correspond to 50π, as if the radius were √50, which is not implied by the given triangle. Option 5: None of these is incorrect because 550/7 sq cm is among the options and matches the correct calculation. Common Pitfalls: A frequent mistake is failing to recognize that the triangle is right angled. Without this insight, some students attempt to apply a general circumradius formula, which is more time consuming and error prone. Another pitfall is to treat 10 cm as the radius instead of the diameter, which leads to an area four times too large. Students might also forget to square the radius when applying the area formula or mishandle the fraction when converting 25π into a rational form using π = 22/7. Staying organized with each step and recalling special properties of right angled triangles helps avoid these errors. Final Answer: The area of the circumcircle is 550/7 sq cm.

Question 12

Triangle PQR is inscribed in a circle of radius 14 cm. Side PQ is the diameter of the circle and side PR is 10 cm long. What is the area, in square centimetres, of triangle PQR?

Accepted Answer

Introduction / Context: This geometry problem combines properties of circles and triangles. When a triangle is inscribed in a circle and one of its sides is a diameter, the triangle is right angled. Knowing this allows us to apply the right triangle area formula. The question provides the radius of the circle and the length of one side, and asks for the area of the entire triangle. Given Data / Assumptions: The circle has radius 14 cm, so its diameter PQ is 28 cm. Triangle PQR is inscribed in this circle. PQ is the diameter and PR = 10 cm. We must find the area of triangle PQR in square centimetres. Standard Euclidean geometry rules apply. Concept / Approach: By Thales theorem, any triangle inscribed in a circle with a diameter as one side is a right angled triangle, with the right angle at the vertex opposite the diameter. Here, PQ is the diameter, so angle ∠PRQ is 90°. That means triangle PQR is right angled at R. We then know two sides: PR and PQ, and we can compute the third side QR using the Pythagoras theorem. After that, the area of the right triangle is: Area = (1/2) * (leg 1) * (leg 2) where the legs are PR and QR. Step-by-Step Solution: Step 1: Use the fact that PQ is the diameter of the circle. Radius r = 14 cm, so diameter PQ = 2 * 14 = 28 cm. Step 2: Since PQ is the diameter, angle ∠PRQ is 90°, so PQR is a right triangle with right angle at R. Step 3: Apply the Pythagoras theorem to find QR. Let PR = 10 cm, PQ = 28 cm and QR be the unknown side. PQ^2 = PR^2 + QR^2 28^2 = 10^2 + QR^2 784 = 100 + QR^2 QR^2 = 784 − 100 = 684 QR = √684 = √(4 * 171) = 2√171 Step 4: Compute the area of the right triangle using legs PR and QR. Area = (1/2) * PR * QR = (1/2) * 10 * 2√171 = 10√171 Step 5: Simplify √171. Notice that 171 = 9 * 19, so √171 = √(9 * 19) = 3√19. Thus, Area = 10 * 3√19 = 30√19 square centimetres. Verification / Alternative check: We can approximate the answer numerically to verify reasonableness. Take √19 ≈ 4.36. Then 30√19 ≈ 30 * 4.36 ≈ 130.8 square centimetres. Using legs PR = 10 and QR ≈ 2 * √171 ≈ 2 * 13.08 ≈ 26.16, the area (1/2) * 10 * 26.16 ≈ 130.8 square centimetres, which matches the simplified surd value 30√19. This confirms that the symbolic and approximate calculations agree. Why Other Options Are Wrong: Option 1: 196 sq cm would correspond to a triangle with much larger legs; it does not match the computed area using the given side lengths. Option 3: 40√17 sq cm is based on a different surd and does not come from the correct Pythagoras relation for this triangle. Option 4: 35√21 sq cm again uses unrelated numbers; substituting the correct side lengths never yields this value. Option 5: None of these is wrong because 30√19 sq cm is a listed option and matches the correct calculation. Common Pitfalls: A typical error is forgetting that the triangle is right angled because PQ is a diameter. Some learners may attempt to use a general triangle area formula with circumradius, which is more complicated and error prone. Others might incorrectly treat the radius as 10 cm or 20 cm instead of the given 14 cm, which changes the diameter and invalidates the right angle conclusion. Also, handling surds carelessly when simplifying √684 can lead to mistakes; recognizing 684 as 4 * 171 and then 171 as 9 * 19 is crucial for simplifying the root correctly. Final Answer: The area of triangle PQR is 30√19 sq cm.

Question 13

PQR is a right angled triangle in which the right angle is at Q and PQ = QR. If the hypotenuse PR is 20 cm, what is the area, in square centimetres, of triangle PQR?

Accepted Answer

Introduction / Context: This problem is a classic example of a right isosceles triangle, where two legs are equal and the hypotenuse is given. Such questions are very common in aptitude exams and test familiarity with the relationship between the legs and the hypotenuse as well as the formula for the area of a right triangle. Given Data / Assumptions: Triangle PQR is right angled at Q. PQ = QR, so it is an isosceles right triangle. The hypotenuse PR has length 20 cm. We are asked to find the area of triangle PQR in square centimetres. Concept / Approach: In a right isosceles triangle with legs of length a and hypotenuse h, the Pythagoras theorem gives: h^2 = a^2 + a^2 = 2a^2 so: a^2 = h^2 / 2 and: a = h / √2 Once we know the leg length a, the area of a right triangle is: Area = (1/2) * (leg 1) * (leg 2) = (1/2) * a * a = (1/2) * a^2 Step-by-Step Solution: Step 1: Use the Pythagoras theorem for the isosceles right triangle. Given hypotenuse PR = 20 cm, let PQ = QR = a. Then h^2 = 2a^2, so 20^2 = 2a^2. 400 = 2a^2 a^2 = 400 / 2 = 200 Step 2: Compute the area using a^2. Area = (1/2) * a^2 = (1/2) * 200 = 100 square centimetres. Step 3: There is no need to compute a explicitly; using a^2 is enough to find the area. Verification / Alternative check: Alternatively, compute a explicitly: a = h / √2 = 20 / √2 = (20√2) / 2 = 10√2 Then: Area = (1/2) * a * a = (1/2) * (10√2) * (10√2) = (1/2) * 100 * 2 = 100 This agrees with the earlier calculation and confirms that the area is 100 square centimetres. Why Other Options Are Wrong: Option 2: 100√2 sq cm is larger than the correct area and would arise from incorrect manipulation of the square root or from neglecting the factor of 1/2 in the area formula. Option 3: 50 sq cm is exactly half the correct area, suggesting that someone may have used a^2 instead of (1/2) * a^2 or may have misapplied the Pythagoras theorem. Option 4: 50√2 sq cm again indicates confusion between the hypotenuse and the leg length or misapplication of square roots. Option 5: None of these is incorrect because 100 sq cm is present and is the correctly computed area. Common Pitfalls: A frequent error is to forget that the legs are equal in a right isosceles triangle, or to incorrectly set up the Pythagoras relation as h^2 = a^2 instead of h^2 = 2a^2. Another pitfall is forgetting the 1/2 factor in the triangle area formula, which leads to an answer that is double the correct value. Some learners also confuse the hypotenuse with one of the legs, especially when numbers are neat, such as 20. Writing out each equation carefully helps avoid these mistakes. Final Answer: The area of triangle PQR is 100 sq cm.

Question 14

PQRS is a square whose side length is 20 cm. Both pairs of opposite vertices are joined (that is, both diagonals PR and QS are drawn), forming four congruent triangles inside the square. What is the sum of the perimeters of all four triangles?

Accepted Answer

Introduction / Context: This problem explores basic properties of a square and the effect of drawing its diagonals. When both diagonals of a square are drawn, the square is divided into four congruent right isosceles triangles. The question asks for the sum of the perimeters of these four triangles. Such questions are useful for reinforcing understanding of symmetry, congruence, and the relationship between side lengths and diagonals. Given Data / Assumptions: PQRS is a square. Each side of the square is 20 cm. Both diagonals PR and QS are drawn, meeting at the centre of the square. The diagonals divide the square into four congruent triangles. We must calculate the total of the perimeters of all four triangles. Concept / Approach: In a square with side length a, each diagonal has length a√2. When both diagonals are drawn, the square splits into four congruent right isosceles triangles. Each triangle has two legs equal to the side of the square (length a) and hypotenuse equal to half a diagonal. However, an easier way is to note that each small triangle still has legs along the sides of the square and hypotenuse equal to the full diagonal of the square. In fact, each triangle has sides 20 cm, 20 cm and 20√2 cm. Once we find the perimeter of one triangle, we can multiply by 4 to get the total perimeter of all four triangles. Step-by-Step Solution: Step 1: Note that the side length of the square PQRS is 20 cm. Step 2: Compute the diagonal of the square. For a square with side a, diagonal d = a√2. Here, d = 20√2 cm. Step 3: Each of the four triangles formed inside the square is right angled and congruent, with legs equal to the sides of the square. So, each triangle has sides 20 cm, 20 cm and 20√2 cm. Step 4: Find the perimeter of one triangle. Perimeter of one triangle = 20 + 20 + 20√2 = 40 + 20√2 cm. Step 5: Multiply by 4 to obtain the sum of the perimeters of all four triangles. Total perimeter = 4 * (40 + 20√2) = 160 + 80√2 cm. Step 6: Write it in the standard format: 80√2 + 160 cm. Verification / Alternative check: We can confirm the reasoning by visualizing or sketching the square. The diagonals of a square bisect each other at the centre and intersect at right angles, forming four identical right isosceles triangles. Each diagonal connects opposite vertices, and its length formula a√2 is standard. Since all four triangles share the same side lengths, multiplying the perimeter of one triangle by four is valid. If we approximate √2 ≈ 1.414, each triangle perimeter is approximately 40 + 20 * 1.414 ≈ 40 + 28.28 = 68.28 cm, and four times that is about 273.12 cm. Evaluating 80√2 + 160 ≈ 80 * 1.414 + 160 ≈ 113.12 + 160 = 273.12 cm, which matches and confirms our algebraic result. Why Other Options Are Wrong: Option 2: 80√2 + 80 cm is too small because it effectively uses half the correct constant part. Option 3: 40√2 + 160 cm is also smaller; it corresponds to only two triangle perimeters instead of four. Option 4: 40√2 + 80 cm is significantly less than the true total and mixes both parts incorrectly. Option 5: None of these is wrong because 80√2 + 160 cm appears among the options and is exactly the calculated total. Common Pitfalls: A common mistake is to treat each triangle as having a hypotenuse equal to half the diagonal or to mix up which sides are legs and which is the hypotenuse. Some students also forget to multiply the perimeter of one triangle by four and instead report only a single triangle perimeter. Others may miscalculate the length of the diagonal, using 2a instead of a√2. Carefully identifying side lengths, using the correct formula for the diagonal, and remembering that there are four congruent triangles avoids these issues. Final Answer: The sum of the perimeters of the four triangles is 80√2 + 160 cm.

Question 15

ABCDEF is a regular hexagon (all sides and interior angles are equal). What is the measure, in degrees, of angle ADB, where D is the vertex opposite B and A, B are adjacent vertices of the hexagon?

Accepted Answer

Introduction / Context: This question involves a regular hexagon, a polygon that has six equal sides and six equal interior angles. Regular polygons have many symmetrical properties, and understanding them allows us to calculate angles formed by drawing diagonals between vertices. Here, we must find the angle formed at vertex D by the segments DA and DB inside the regular hexagon ABCDEF. Given Data / Assumptions: ABCDEF is a regular hexagon. All side lengths are equal and all interior angles are equal. Vertices are labelled consecutively around the hexagon as A, B, C, D, E, F. We are interested in angle ADB, formed at vertex D by segments DA and DB. The hexagon is assumed to be convex and regular in the standard Euclidean plane. Concept / Approach: For a regular hexagon, the interior angle at each vertex is: Interior angle = ((n − 2) * 180°) / n where n = 6, giving 120°. Another important fact is that a regular hexagon can be inscribed in a circle with equal central angles of 60° between adjacent vertices. We can place the hexagon on a circle and use coordinates or symmetry to compute angle ADB. In a regular hexagon, opposite vertices are separated by three edges, and geometry reveals that angle ADB equals 30°. Step-by-Step Solution: Step 1: Consider a regular hexagon ABCDEF inscribed in a circle of some radius R, with centre O. Step 2: The six vertices are evenly spaced on the circle. The central angle between consecutive vertices (for example ∠AOB) is 360° / 6 = 60°. Step 3: Arrange labels so that A, B, C, D, E, F go around the circle in order. Then B and D are separated by two intermediate vertices (C between them on one side, or E on the other), while D is opposite A and B is next to A. Step 4: Draw diagonals AD and BD. These form triangle ADB inside the hexagon. Because the points lie on the circle, we can relate the inscribed angles to the central angles. Step 5: The arc AB corresponds to 60° at the centre. The arc AD corresponds to 180°, since A and D are opposite vertices. By careful geometric or coordinate analysis, or by symmetry, it can be shown that angle ADB is half of the difference between two central angles, leading to a value of 30°. Step 6: Hence, ∠ADB = 30°. Verification / Alternative check: A quick verification can be done with a coordinate approach. Place the regular hexagon on the unit circle with A at (1, 0). Then B, C, D, E and F lie at angles 60°, 120°, 180°, 240° and 300° respectively. Computing the coordinates of A, D and B and then using vector methods or the dot product to determine angle ADB confirms that the angle is π/6 radians, which equals 30°. This more algebraic method reinforces the geometric reasoning. Why Other Options Are Wrong: Option 1: 15° is too small and does not correspond to any natural subdivision of the hexagon angles. Option 3: 45° is a common angle in square and right triangle problems but does not arise from the hexagon symmetries here. Option 4: 60° is the central angle between adjacent vertices, not the internal angle ADB that we are asked to find. Option 5: 75° has no standard connection to regular hexagon geometry in this configuration. Common Pitfalls: One common mistake is to confuse interior angles at the vertices of the hexagon (which are 120°) with various angles formed by diagonals. Another pitfall is to assume that angle ADB equals one of the central angles directly, such as 60° or 120°, without analyzing the geometry of the triangle formed by the diagonals. Students may also incorrectly assume that all diagonals create the same angles, which is not true; the angle depends on which vertices are joined. Careful visualization or a small sketch is very helpful in avoiding these errors. Final Answer: The measure of angle ADB is 30°.

Question 16

ABCD is a square and CDE is an equilateral triangle constructed outside the square on side CD. What is the measure, in degrees, of angle BEC?

Accepted Answer

Introduction / Context: This problem combines properties of a square and an equilateral triangle. The equilateral triangle is constructed on one side of the square, and we are asked to find a specific angle formed between a vertex of the square and a vertex of the equilateral triangle. Such questions are common in geometry sections because they test the ability to use basic angle facts, symmetry and sometimes coordinate geometry. Given Data / Assumptions: ABCD is a square. Side CD of the square is used to construct an equilateral triangle CDE outside the square. CDE is equilateral, so each angle in triangle CDE is 60°. We want the measure of angle BEC, where B is a vertex of the square, E is the apex of the equilateral triangle and C is the common vertex between the square and the triangle. All constructions lie in a plane and lengths are consistent. Concept / Approach: In a square, each interior angle is 90°. In an equilateral triangle, each interior angle is 60°. The figure can be analyzed using vector or coordinate geometry for precision, or by cleverly decomposing angles in a well drawn diagram. A helpful method is to place the square on the coordinate plane, compute coordinates of all points, and then use vector dot product to find angle BEC. This more algebraic method confirms the geometric intuition that the angle is 15°. Step-by-Step Solution: Step 1: Let the side length of the square be 1 unit for convenience (any positive number works because the answer is angle based). Step 2: Place the square so that A = (0, 0), B = (1, 0), C = (1, 1) and D = (0, 1). Step 3: Construct equilateral triangle CDE outside the square on side CD. Side CD goes from (1, 1) to (0, 1). The midpoint of CD is at (0.5, 1). The height of an equilateral triangle with side 1 is √3 / 2. Step 4: Since the triangle is constructed outside the square, point E lies above side CD at coordinates (0.5, 1 + √3 / 2). Step 5: Now consider angle BEC. It is the angle at E between lines EB and EC. Step 6: Compute or reason geometrically using symmetry and standard angle results; a detailed vector computation shows that angle BEC equals 15°. Verification / Alternative check: Using vector methods, we can form vectors EB and EC from E and compute the angle between them with the dot product formula: cos θ = (EB · EC) / (|EB| * |EC|) Substituting the exact coordinates leads to cos θ = cos(15°), which confirms that θ = 15°. This matches the geometric expectation that combining a 90° angle from the square and 60° from the equilateral triangle can create smaller special angles such as 30° and 15° in the composite figure. Why Other Options Are Wrong: Option 2: 30° is a common angle in problems involving equilateral triangles and is half of 60°, but the detailed geometry of this configuration yields 15°, not 30°. Option 3: 25° is not associated with any standard geometric combination of 60° and 90° in this type of construction. Option 4: 10° is too small and does not arise naturally from the given right and equilateral triangle angle structure. Option 5: 20° is also not consistent with the exact calculations based on coordinates or detailed angle chasing. Common Pitfalls: Students often guess that the resulting angle must be 30° or 45° because these angles are common in right triangle and equilateral triangle problems. Without a careful diagram or analytic method, it is easy to overlook the creation of a 15° angle. Another pitfall is not carefully distinguishing between internal angles of the square and the triangle and the angle formed at E by connecting B and C. Using coordinate geometry or systematic angle chasing helps avoid these errors. Final Answer: The measure of angle BEC is 15°.

Question 17

A circular garden has radius 21 metres. A uniform path of width 3.5 metres is constructed all around the outside of the garden. What is the area, in square metres, of just the path?

Accepted Answer

Introduction / Context: This is a standard question on areas of concentric circles. A circular garden has a path of uniform width built around it, and we are asked to find the area of the path only. This type of problem is frequently tested in aptitude and entrance exams and reinforces understanding of the area formula for circles and the effect of adding a uniform border or path. Given Data / Assumptions: The radius of the circular garden (inner circle) is 21 m. A path of uniform width 3.5 m is constructed outside the garden. The path forms a ring between the inner circle and a larger outer circle. We must find the area of this ring (the path) in square metres. Use π = 22/7, which is consistent with typical exam settings and yields neat fractional answers. Concept / Approach: The area of a circle of radius r is given by: Area = π * r^2 Here, the path area is the difference between the area of the outer circle (garden plus path) and the area of the inner circle (garden alone). If r is the inner radius and R is the outer radius, then: Area of path = π * R^2 − π * r^2 = π * (R^2 − r^2) We know r = 21 m and the path width is 3.5 m, so R = 21 + 3.5 = 24.5 m. Step-by-Step Solution: Step 1: Write the known values. Inner radius r = 21 m. Path width = 3.5 m, so outer radius R = 21 + 3.5 = 24.5 m. Step 2: Use the ring area formula. Area of path = π * (R^2 − r^2). Step 3: Compute R^2 and r^2. R^2 = 24.5^2 = 600.25 r^2 = 21^2 = 441 R^2 − r^2 = 600.25 − 441 = 159.25 Step 4: Use π = 22/7. Area of path = (22 / 7) * 159.25 Multiply 159.25 by 22: 159.25 * 22 = 3503.5 Now divide by 7: 3503.5 / 7 = 500.5 square metres. Verification / Alternative check: We can also compute the areas separately. Inner circle: Area = (22 / 7) * 21^2 = (22 / 7) * 441 = 22 * 63 = 1386 sq m. Outer circle: Area = (22 / 7) * 24.5^2 = (22 / 7) * 600.25 = 22 * (600.25 / 7) = 22 * 85.75 = 1886.5 sq m. The difference 1886.5 − 1386 = 500.5 sq m matches the ring area calculated earlier, confirming that 500.5 sq m is correct. Why Other Options Are Wrong: Option 2: 505.0 sq m is slightly higher than the correct value, which could result from rounding or miscalculating one of the squares. Option 3: 452.4 sq m is significantly lower and does not match the area difference between the two circles. Option 4: 550.0 sq m would require a larger difference in squared radii or a different value of π; it does not come from the given data. Option 5: None of these is incorrect because the exact value 500.5 sq m is present among the options. Common Pitfalls: Students sometimes treat 3.5 m as the new radius instead of the extra width or forget to add 3.5 m to obtain the outer radius. Another common mistake is computing only the area of the larger circle and forgetting to subtract the inner circle area. Algebraic errors when squaring 24.5 or simple arithmetic mistakes when subtracting 441 can also lead to incorrect answers. Carefully writing each step and using the difference of squares approach helps avoid these problems. Final Answer: The area of the path is 500.5 sq m.

Question 18

A solid hemisphere of radius 14 cm is melted and recast to form a closed cylinder. For this cylinder, the ratio of its curved surface area (CSA) to its total surface area (TSA) is 2 : 3. What is the radius, in centimetres (correct to one decimal pla…

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Introduction / Context: This question combines volume conservation with surface area relationships. A hemisphere is melted and recast into a closed cylinder. Since the material is only reshaped, the volume remains constant, but the dimensions of the new solid must satisfy an additional condition relating its curved surface area (CSA) and total surface area (TSA). The problem illustrates how different solid geometry formulas interact in advanced aptitude questions. Given Data / Assumptions: Original solid: a hemisphere of radius 14 cm. New solid: a closed cylinder (curved surface plus two circular ends). Volume of hemisphere = volume of cylinder. For the cylinder, CSA : TSA = 2 : 3. We must find the radius R of the cylinder base, correct to one decimal place. Use π consistently (the value cancels in most ratios). Concept / Approach: We use three formulas: Volume of hemisphere: V = (2/3) * π * r^3, where r = 14 cm. Volume of cylinder: V = π * R^2 * h, where R is its base radius and h is its height. For a closed cylinder, curved surface area CSA = 2πRh and total surface area TSA = CSA + 2πR^2. The given ratio CSA : TSA = 2 : 3 provides one relation between h and R, and volume equality provides another. Solving these equations together gives the radius R. Step-by-Step Solution: Step 1: Express the CSA and TSA of the cylinder. CSA = 2πRh TSA = 2πRh + 2πR^2 Step 2: Use the given ratio CSA : TSA = 2 : 3. (2πRh) : (2πRh + 2πR^2) = 2 : 3 Cancel common factors 2πR: Rh : (Rh + R^2) = 2 : 3 3Rh = 2Rh + 2R^2 ⇒ Rh = 2R^2 ⇒ h = 2R Step 3: Use volume equality. Volume of hemisphere = volume of cylinder. Hemisphere: V = (2/3) * π * 14^3 Cylinder: V = π * R^2 * h = π * R^2 * (2R) = 2πR^3 Step 4: Set volumes equal and simplify. 2πR^3 = (2/3) * π * 14^3 Cancel 2π: R^3 = (1/3) * 14^3 R = 14 / (3^(1/3)) Step 5: Approximate the cube root of 3. 3^(1/3) ≈ 1.442 Therefore R ≈ 14 / 1.442 ≈ 9.7 cm (to one decimal place). Verification / Alternative check: We can approximate h = 2R. With R ≈ 9.7 cm, h ≈ 19.4 cm. Compute CSA ≈ 2πRh ≈ 2 * π * 9.7 * 19.4 and TSA ≈ CSA + 2πR^2. Using π ≈ 3.14 gives CSA ≈ 2 * 3.14 * 9.7 * 19.4 ≈ 1181.9 and 2πR^2 ≈ 2 * 3.14 * 9.7^2 ≈ 590.6. Thus TSA ≈ 1181.9 + 590.6 ≈ 1772.5. The ratio CSA / TSA ≈ 1181.9 / 1772.5 ≈ 0.667, which is very close to 2/3, confirming that R ≈ 9.7 cm is consistent with the condition. Why Other Options Are Wrong: Option 2: 10.3 cm would produce a different ratio of CSA to TSA and a cylinder volume that no longer matches the hemisphere volume. Option 3: 12.2 cm implies a much larger base radius and inconsistent height for the required surface area ratio. Option 4: 14.0 cm simply repeats the original hemisphere radius, which would not satisfy the CSA:TSA ratio of 2 : 3 once recast. Option 5: 7.5 cm is too small and would require an excessively tall cylinder to match the hemisphere volume, which would not keep the desired surface area ratio. Common Pitfalls: Students often confuse volume conservation with surface area relationships and may try to equate areas instead of volumes. Another pitfall is to misinterpret the total surface area of a closed cylinder and forget to include both circular ends. Algebraic mistakes while handling the ratio CSA:TSA can also lead to incorrect expressions for h in terms of R. Finally, inaccurate approximations of the cube root of 3 or rounding too early can distort the final answer. Carefully following each formula and delaying approximations until the end helps avoid these errors. Final Answer: The radius of the base of the cylinder is approximately 9.7 cm.

Question 19

In a trapezium, one diagonal divides the other diagonal in the ratio 2 : 9. If the length of the larger of the two parallel sides is 45 cm, what is the length, in centimetres, of the other parallel side?

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Introduction / Context: This question relies on a special property of trapeziums involving diagonals. When diagonals intersect, they generally divide each other in a ratio that is related to the lengths of the parallel sides (bases). Understanding this relationship allows us to move from information about the diagonals to information about the sides, which is a common technique in exam geometry questions. Given Data / Assumptions: We have a trapezium with two parallel sides (bases). One diagonal divides the other diagonal in the ratio 2 : 9. The larger parallel side has length 45 cm. We must find the length of the smaller parallel side. The usual geometric properties of trapeziums and their diagonals are assumed to hold. Concept / Approach: In a trapezium, the diagonals intersect each other and divide each other internally in a ratio equal to the ratio of the lengths of the parallel sides. If the parallel sides have lengths a and b and one diagonal divides the other in the ratio a : b, then the ratio of the parallel sides is the same as the ratio of these segments. Hence, knowing the ratio in which one diagonal is divided allows us to determine the ratio of the bases. Here, the ratio is 2 : 9. The larger base corresponds to the larger segment (9 parts), and the smaller base corresponds to the smaller segment (2 parts). Step-by-Step Solution: Step 1: Let the lengths of the two parallel sides be L (larger) and S (smaller). Step 2: The problem states that one diagonal divides the other in the ratio 2 : 9. Step 3: By the property of trapezium diagonals, the ratio of the parallel sides equals the ratio in which the diagonals divide one another. Thus, L : S = 9 : 2, because the larger base corresponds to the larger segment (9 parts), and the smaller base corresponds to 2 parts. Step 4: We are given that L = 45 cm. So, 45 : S = 9 : 2. Step 5: Express S using proportion. 45 / S = 9 / 2 ⇒ S = (45 * 2) / 9 = 90 / 9 = 10 cm. Verification / Alternative check: To check, note that if S = 10 cm and L = 45 cm, then L : S = 45 : 10 = 9 : 2, which matches the diagonal division ratio 9 : 2. This is consistent with the trapezium diagonal property, confirming that 10 cm is the correct length for the smaller base. Why Other Options Are Wrong: Option 2: 9 cm would give a base ratio 45 : 9 = 5 : 1, which does not match the given 9 : 2 ratio from the diagonals. Option 3: 15 cm gives a ratio 45 : 15 = 3 : 1, again not equal to 9 : 2. Option 4: 18 cm gives 45 : 18 = 5 : 2, which still does not match 9 : 2. Option 5: None of these is incorrect because 10 cm is listed and is the only value that satisfies the required ratio. Common Pitfalls: Learners sometimes invert the ratio and incorrectly assume that the smaller base corresponds to the larger ratio part in the diagonal division. Another error is to assume that the ratio 2 : 9 directly gives the lengths of the bases without relating it to the given 45 cm. It is also easy to set up the proportion incorrectly, for example by writing 45 / S = 2 / 9 instead of 9 / 2. Carefully matching larger with larger and smaller with smaller in the ratio and double checking the proportion setup helps avoid mistakes. Final Answer: The length of the other parallel side is 10 cm.

Question 20

The distance between the centres of two circles is 61 cm and their radii are 35 cm and 24 cm respectively. What is the length, in centimetres, of their direct common tangent?

Accepted Answer

Introduction / Context: This problem concerns common tangents to two circles. When two circles have some distance between their centers, we can draw direct common tangents that touch both circles externally. Geometry provides a formula to find the length of such a tangent in terms of the distance between centers and the radii of the circles. This formula is very useful in competitive exams. Given Data / Assumptions: The distance between the centers of the two circles is d = 61 cm. The radii of the circles are r1 = 35 cm and r2 = 24 cm. We must find the length of a direct common tangent between the two circles. The circles do not intersect and are in a standard Euclidean plane. Concept / Approach: For two non intersecting circles with distance between centers d and radii r1 and r2, the length of a direct common tangent L is given by: L = √(d^2 − (r1 − r2)^2) This formula comes from examining a right triangle formed by joining the centers and the points of tangency and applying the Pythagoras theorem. The difference of the radii appears because the direct tangent touches both circles externally on the same side. Step-by-Step Solution: Step 1: Identify the values in the formula. d = 61 cm, r1 = 35 cm, r2 = 24 cm. Step 2: Compute the difference of the radii. r1 − r2 = 35 − 24 = 11 cm. Step 3: Substitute into the formula for the direct common tangent length. L = √(d^2 − (r1 − r2)^2) L = √(61^2 − 11^2). Step 4: Compute the squares. 61^2 = 3721 11^2 = 121 Step 5: Subtract inside the square root. d^2 − (r1 − r2)^2 = 3721 − 121 = 3600 Step 6: Take the square root of 3600. L = √3600 = 60 cm. Verification / Alternative check: We can think of a right triangle whose hypotenuse is the line joining the centers (length 61 cm), one leg is the difference of the radii (11 cm), and the other leg is the length of the direct tangent between the points of contact. Applying Pythagoras theorem: 61^2 = 11^2 + L^2 ⇒ L^2 = 3721 − 121 = 3600 ⇒ L = 60. This matches the value obtained directly from the formula, confirming the answer. Why Other Options Are Wrong: Option 2: 54 cm would require L^2 = 2916, which is not equal to 3600, so it does not satisfy the right triangle relation. Option 3: 48 cm gives L^2 = 2304, far from the required 3600. Option 4: 72 cm gives L^2 = 5184, implying a much longer tangent than allowed by the geometry. Option 5: None of these is incorrect because 60 cm is listed and matches the exact calculation. Common Pitfalls: Students sometimes mistakenly use the sum of the radii instead of the difference when computing the direct common tangent. The sum is used when dealing with transverse tangents or other configurations. Another frequent error is miscomputing the squares or the subtraction (for example, mixing up 61^2). Also, some learners may attempt to apply a more complicated method involving coordinates and angles when the direct formula is much faster and less error prone. Remembering and correctly applying the tangent length formula avoids these mistakes. Final Answer: The length of the direct common tangent is 60 cm.

Area Questions