Question 1

An equilateral triangle ABC is inscribed in a circle of radius 20√3 cm. What is the length (in cm) of each side of the equilateral triangle?

Accepted Answer

Introduction / Context: This question checks your understanding of the relation between the side length of an equilateral triangle and the radius of the circumcircle in which it is inscribed. It is a standard geometry result that appears frequently in aptitude and competitive exams. Given Data / Assumptions: Triangle ABC is equilateral. Triangle ABC is inscribed in a circle, so all three vertices lie on the circle. The radius R of the circle (circumradius) is 20√3 cm. We need to find the side length a of the equilateral triangle. We use standard Euclidean geometry and exact values (no rounding until the end if needed). Concept / Approach: For an equilateral triangle of side a, the relationship between the side length and circumradius R is a well known formula: R = a / √3 This can be derived from splitting the equilateral triangle into two 30 60 90 right triangles by drawing a median. Once we know this relation, we can solve for a in terms of R: a = R * √3 Substituting R = 20√3 cm will give the required side length. Step-by-Step Solution: Step 1: Let a be the side length of the equilateral triangle ABC. Step 2: Use the formula for circumradius of an equilateral triangle: R = a / √3. Step 3: Rearrange the formula to express a: a = R * √3. Step 4: Substitute R = 20√3 cm. Step 5: So a = 20√3 * √3 = 20 * 3 = 60 cm. Step 6: Therefore, each side of triangle ABC is 60 cm long. Verification / Alternative check: We can confirm by checking the reverse relation. If a = 60 cm, then the circumradius R should be: R = a / √3 = 60 / √3 = (60√3) / 3 = 20√3 cm. This matches the given radius, so the answer 60 cm is fully consistent. Why Other Options Are Wrong: 30 cm: If a = 30 cm, then R = 30 / √3 = 10√3 cm, which is half of the given radius. 40 cm: For a = 40 cm, R = 40 / √3, which is not equal to 20√3 cm. 50 cm: For a = 50 cm, R = 50 / √3, which again does not match 20√3 cm. 20√3 cm: This option confuses side length with radius. 20√3 cm is the radius, not the side length. Common Pitfalls: Many students mistakenly use formulas for inradius instead of circumradius, or they forget the special relation of equilateral triangles and try to apply general triangle formulas. Another common error is mishandling square roots, such as writing (√3) * (√3) as √6 instead of 3. Keeping the formula R = a / √3 clearly in mind helps to avoid these mistakes. Final Answer: The side length of the equilateral triangle is 60 cm.

Question 2

An equilateral triangle ABC is inscribed in a circle of radius 20√3 cm. The centroid of triangle ABC lies at a distance d from vertex A. What is the value of d (in cm)?

Accepted Answer

Introduction / Context: This geometry question connects several important centers of an equilateral triangle: centroid, circumcenter and incenter. In an equilateral triangle, all these centers coincide at the same point. The problem asks for the distance from a vertex to the centroid, given the radius of the circumcircle. Given Data / Assumptions: Triangle ABC is equilateral. Triangle ABC is inscribed in a circle (circumcircle) of radius 20√3 cm. The centroid G of triangle ABC is the same point as the circumcenter and incenter, because the triangle is equilateral. We must find the distance d from vertex A to point G. Concept / Approach: In any triangle, the centroid is the intersection of medians. In an equilateral triangle, there is a special symmetry: centroid, circumcenter, orthocenter and incenter all coincide at the same point. The distance from a vertex to the circumcenter is exactly the circumradius R. Therefore, once we know that G is also the circumcenter, the distance AG is simply R. Step-by-Step Solution: Step 1: Recognize that in an equilateral triangle, centroid, circumcenter and incenter coincide. Step 2: The circle of radius 20√3 cm is the circumcircle of triangle ABC. Step 3: The circumcenter is the center of this circle and is at distance R from each vertex. Step 4: Since centroid and circumcenter are the same point G, the distance from vertex A to G is equal to the circumradius R. Step 5: Therefore, d = AG = 20√3 cm. Verification / Alternative check: We can also use the formula for the distance from a vertex to centroid in an equilateral triangle. If a is the side length and h is the height, then the centroid divides each median in the ratio 2:1 from the vertex. The height is h = (√3 / 2) * a, and the centroid lies at distance 2h / 3 from a vertex. For an equilateral triangle with circumradius R, we also have R = a / √3. Combining these gives distance from vertex to centroid as a / √3 = R. This confirms our reasoning that AG equals R, which is 20√3 cm. Why Other Options Are Wrong: 15 cm and 20 cm: Both are much smaller than the given radius and do not match the relation between centroid and circumcenter for an equilateral triangle. 30√3 cm: This is larger than the circumradius and would place the centroid outside the circle, which is impossible. 10√3 cm: This is half the radius, which would be the distance for some different construction, but not for centroid in an equilateral triangle. Common Pitfalls: A frequent mistake is to confuse centroid with the center of the incircle or to think that the centroid lies two thirds of the way from the vertex to the circumcenter, instead of along the median. In an equilateral triangle this confusion disappears if you remember that all the major centers coincide. Keeping this symmetry in mind avoids unnecessary calculations. Final Answer: The distance from vertex A to the centroid is 20√3 cm.

Question 3

The sum of the length, breadth and height of a cuboid is 22 cm, and the length of its space diagonal is 14 cm. What is the total surface area (in square cm) of the cuboid?

Accepted Answer

Introduction / Context: This problem uses algebraic identities related to a cuboid. You are given the sum of its three dimensions and the length of its space diagonal. From this, you must deduce the total surface area. It checks your ability to connect geometric formulas with algebraic identities. Given Data / Assumptions: Let the length, breadth and height of the cuboid be l, b and h (in cm). l + b + h = 22 cm. The space diagonal d satisfies d^2 = l^2 + b^2 + h^2 and we are told d = 14 cm. So l^2 + b^2 + h^2 = 14^2 = 196. We assume a standard right angled cuboid (rectangular box). We want the total surface area S = 2(lb + bh + hl). Concept / Approach: The key algebraic identity is: (l + b + h)^2 = l^2 + b^2 + h^2 + 2(lb + bh + hl) We already know the left side and l^2 + b^2 + h^2, so we can solve for lb + bh + hl. Once that sum is known, we multiply by 2 to get the surface area of the cuboid. Step-by-Step Solution: Step 1: Square the sum of dimensions: (l + b + h)^2 = 22^2 = 484. Step 2: Use the identity (l + b + h)^2 = l^2 + b^2 + h^2 + 2(lb + bh + hl). Step 3: Substitute l^2 + b^2 + h^2 = 196 into the identity. Step 4: So 484 = 196 + 2(lb + bh + hl). Step 5: Rearrange to find 2(lb + bh + hl) = 484 − 196 = 288. Step 6: The total surface area S of the cuboid is S = 2(lb + bh + hl) = 288 sq cm. Verification / Alternative check: Note that we did not need individual values of l, b and h. The combination of sum of squares and sum of dimensions is sufficient to determine 2(lb + bh + hl) uniquely. This shows that the surface area is fixed by the given data. No contradiction appears in the calculations, and all steps smoothly follow from standard identities, so 288 sq cm is reliable. Why Other Options Are Wrong: 216 sq cm and 144 sq cm: These would correspond to smaller values of lb + bh + hl, which contradict the computed value 144 (since 2 times that gives 288). Cannot be determined from the given information: This is incorrect because the identity involving the sum of dimensions and sum of squares is enough to fix the surface area uniquely. 200 sq cm: This is not supported by any consistent algebraic manipulation of the given data. Common Pitfalls: Students sometimes try to assume specific integer values for l, b and h, which is unnecessary and may lead to trial and error. Another error is forgetting the algebraic identity and thinking that surface area cannot be determined. Miscomputing 22^2 or 14^2, or making sign errors while rearranging, can also produce wrong answers. Keeping the identity clearly in mind avoids these mistakes. Final Answer: The total surface area of the cuboid is 288 sq cm.

Question 4

The sum of the length, breadth and height of a cuboid is 22 cm and the length of its space diagonal is 14 cm. If S is the sum of the cubes of its three dimensions and V is its volume, then what is the value of (S − 3V) in cubic cm?

Accepted Answer

Introduction / Context: This question extends the previous cuboid problem and tests your knowledge of algebraic identities, specifically the identity for the sum of cubes in terms of the sum of numbers and their product. It mixes geometry (cuboid dimensions and diagonal) with algebra (sum of cubes and volume). Given Data / Assumptions: Dimensions of the cuboid are l, b and h (in cm). l + b + h = 22. Space diagonal d = 14 cm, so l^2 + b^2 + h^2 = 14^2 = 196. S is defined as S = l^3 + b^3 + h^3. Volume V is V = l * b * h. We need to find S − 3V. Concept / Approach: There is a powerful identity for three numbers a, b and c: a^3 + b^3 + c^3 − 3abc = (a + b + c)(a^2 + b^2 + c^2 − ab − bc − ca) In our case, a, b and c are l, b and h. Here S − 3V is exactly l^3 + b^3 + h^3 − 3 l b h, so we can directly use this identity. We already know l + b + h and l^2 + b^2 + h^2. To use the identity we also need the sum of pairwise products ab + bc + ca, which can be computed from another identity: (l + b + h)^2 = l^2 + b^2 + h^2 + 2(ab + bc + ca) Step-by-Step Solution: Step 1: Compute (l + b + h)^2 = 22^2 = 484. Step 2: Use the identity (l + b + h)^2 = l^2 + b^2 + h^2 + 2(ab + bc + ca). Step 3: Substitute l^2 + b^2 + h^2 = 196. Step 4: 484 = 196 + 2(ab + bc + ca). Step 5: So 2(ab + bc + ca) = 484 − 196 = 288, hence ab + bc + ca = 144. Step 6: Now apply the sum of cubes identity: S − 3V = (l + b + h)(l^2 + b^2 + h^2 − ab − bc − ca). Step 7: Substitute known values: l + b + h = 22, l^2 + b^2 + h^2 = 196, and ab + bc + ca = 144. Step 8: Compute the bracket: 196 − 144 = 52. Step 9: Therefore S − 3V = 22 * 52 = 1144 cubic cm. Verification / Alternative check: Although we do not know the individual dimensions l, b and h, the algebraic identities guarantee that S − 3V depends only on the combinations we already computed. We can even factor 1144 as 8 * 143 to see it is a reasonable moderate sized number, not extremely large. No contradictions arise, so 1144 cubic cm is consistent. Why Other Options Are Wrong: 572 cubic cm and 728 cubic cm: These are roughly half or smaller than the correct value and would correspond to a smaller value of the product (l + b + h)(l^2 + b^2 + h^2 − ab − bc − ca), which contradicts the exact computation. None of the above: This is wrong because 1144 cubic cm exactly matches the algebraic result. 900 cubic cm: This is a plausible looking round number but is not supported by any correct substitution into the identities. Common Pitfalls: Some learners forget the correct form of the sum of cubes identity or confuse it with (a + b + c)^3. Another common issue is not computing ab + bc + ca correctly and stopping at the sum of squares. Carefully applying both identities in sequence avoids these mistakes. It is also important to remember that S − 3V is exactly the expression produced by the identity, rather than S + 3V or any other combination. Final Answer: The value of S − 3V is 1144 cubic cm.

Question 5

The corners of a square of side a are cut away equally so that the remaining figure is a regular octagon. What is the length of each side of the regular octagon, in terms of a?

Accepted Answer

Introduction / Context: This problem is a classic geometry construction question. A regular octagon is formed by cutting congruent right isosceles triangles from the corners of a square. The goal is to relate the side length of the original square to the side length of the resulting regular octagon. It tests understanding of symmetry and basic coordinate or segment geometry. Given Data / Assumptions: We have a square with side length a. At each corner of the square, a congruent isosceles right triangle is cut away. The remaining figure is a regular octagon, so all eight sides of the octagon are equal. The cuts along each side of the square are of equal length from the corners. We need the side length of the regular octagon in terms of a. Concept / Approach: Let x be the length along each side of the square that is cut off from each corner. After cutting, each side of the octagon arises in two ways: One type is the middle segment on each side of the original square, which has length a − 2x. The other type is the hypotenuse of each cut off right isosceles triangle, which has length x√2. For a regular octagon, all sides must be equal, so these two expressions must match: a − 2x = x√2 Solving this equation for x and then for the common side gives the required formula. Step-by-Step Solution: Step 1: Let the square side be a and the length cut from each end of a side be x. Step 2: The central segment on each side of the square after cutting has length a − 2x. Step 3: Each cut off corner is an isosceles right triangle with legs x and x, so its hypotenuse (which becomes a side of the octagon) has length x√2. Step 4: For a regular octagon, these must be equal: a − 2x = x√2. Step 5: Rearrange: a = 2x + x√2 = x(2 + √2), so x = a / (2 + √2). Step 6: The side length s of the octagon is x√2. Step 7: Substitute x: s = (a / (2 + √2)) * √2. Step 8: Simplify by rationalizing: multiply numerator and denominator by (2 − √2). Step 9: s = a√2(2 − √2) / (4 − 2) = a√2(2 − √2) / 2. Step 10: Expand numerator: √2 * 2 = 2√2 and √2 * √2 = 2, so we get (2√2 − 2) / 2 = √2 − 1. Step 11: Therefore, s = a(√2 − 1). Verification / Alternative check: We can test with a convenient numerical value, for example a = 1. Then s = (√2 − 1). Computing x from a = x(2 + √2), we get x = 1 / (2 + √2). The central segment a − 2x and hypotenuse x√2 both evaluate numerically to the same value (√2 − 1), which confirms the derivation. Why Other Options Are Wrong: a(√3 − 1): This does not arise from any geometric relation in this construction and is numerically too large. a/(√2 + 2): This is actually the length x of the cut segments along the square sides, not the octagon side length. a/3 and a/2: These are simple fractions of the square side with no connection to the angle of 45 degrees in the corner triangles and do not satisfy the equality a − 2x = x√2 with x chosen accordingly. Common Pitfalls: Students often confuse x (the cut length) with the octagon side, or forget that the corner triangles are right isosceles, leading to incorrect hypo formulas. Another error is failing to enforce that the octagon has all sides equal, which is the defining property. Carefully labeling segments and writing the equality condition avoids these mistakes. Final Answer: The side length of the regular octagon is a(√2 − 1).

Question 6

Three consecutive integers form the side lengths of a right angled triangle. How many distinct sets of such three consecutive positive integers are possible?

Accepted Answer

Introduction / Context: This question combines number theory with the Pythagoras theorem. It asks how many triples of three consecutive positive integers can appear as the side lengths of a right angled triangle. Such triples, if they exist, must satisfy the Pythagorean condition for some ordering of the three numbers. Given Data / Assumptions: The three side lengths are consecutive positive integers: n, n + 1 and n + 2 for some positive integer n. The triangle is right angled, so one side is the hypotenuse and the other two are legs. Because the numbers are consecutive, the largest side is n + 2. We are dealing only with positive integers as side lengths. Concept / Approach: For a right angled triangle with integer sides (a Pythagorean triple), we must have: (leg1)^2 + (leg2)^2 = (hypotenuse)^2 Given the sides are n, n + 1 and n + 2 with n + 2 as the largest, we need to test whether: n^2 + (n + 1)^2 = (n + 2)^2 Other arrangements where n or n + 1 is the hypotenuse are not possible, because the hypotenuse must be the largest side. Solving this equation will show how many integer values of n work. Step-by-Step Solution: Step 1: Let the three consecutive integers be n, n + 1 and n + 2. Step 2: Since n + 2 is the largest, consider it as the hypotenuse. Step 3: Apply Pythagoras theorem: n^2 + (n + 1)^2 = (n + 2)^2. Step 4: Expand the left side: n^2 + (n + 1)^2 = n^2 + (n^2 + 2n + 1) = 2n^2 + 2n + 1. Step 5: Expand the right side: (n + 2)^2 = n^2 + 4n + 4. Step 6: Set them equal: 2n^2 + 2n + 1 = n^2 + 4n + 4. Step 7: Rearrange: 2n^2 + 2n + 1 − n^2 − 4n − 4 = 0, which simplifies to n^2 − 2n − 3 = 0. Step 8: Factor: n^2 − 2n − 3 = (n − 3)(n + 1) = 0. Step 9: So n = 3 or n = −1. Only positive integer n is allowed, so n = 3. Step 10: The corresponding triple is (3, 4, 5), which is the famous Pythagorean triple. Step 11: Therefore, only one set of three consecutive positive integers works: 3, 4 and 5. Verification / Alternative check: Check the triple (3, 4, 5): 3^2 + 4^2 = 9 + 16 = 25 and 5^2 = 25, which satisfies the Pythagoras theorem exactly. For any larger n, the equation reduces to the same quadratic, which has no other positive integer solution. So no other consecutive positive integer triple can form a right angled triangle. Why Other Options Are Wrong: Only two or only three: These would require additional integer solutions to the quadratic equation, which do not exist. Infinitely many: This is impossible because the equation reduces to a fixed quadratic with only two roots, of which only one is positive. None: This would be true only if even 3, 4, 5 failed the Pythagorean condition, which is not the case. Common Pitfalls: A common mistake is to attempt guessing different triples without solving the quadratic systematically, which may lead to missing the fact that there is a single solution. Another error is considering negative values of n as valid side lengths. Always remember that triangle side lengths must be positive. Careful algebraic manipulation and factoring prevent these issues. Final Answer: There is only one such triple of consecutive positive integers, namely 3, 4 and 5.

Question 7

Two concentric circles (same centre) are drawn. The circumference of the smaller circle is 44 cm, and the circumference of the larger circle is double that of the smaller circle. What is the area (in square cm) of the region lying between these two …

Accepted Answer

Introduction / Context: This question is about concentric circles and the area of the ring shaped region (annulus) between them. It tests understanding of the relationship between circumference and radius, as well as the formula for the area of a circle. Given Data / Assumptions: The circumference of the smaller circle is 44 cm. The circumference of the larger circle is 88 cm, which is double 44 cm. Both circles share the same centre (they are concentric). We can use π = 22 / 7 for calculations. We need the area of the region between the two circles, that is π(R^2 − r^2), where R is the larger radius and r is the smaller radius. Concept / Approach: For any circle, the circumference C is related to its radius r by: C = 2πr From each given circumference we can compute the corresponding radius. Then we find the area of each circle using: Area = πr^2 Finally, the area between the two circles (the annulus) is: Area between = π(R^2 − r^2) Step-by-Step Solution: Step 1: For the smaller circle, C1 = 44 cm. Step 2: Use 44 = 2πr1. With π = 22 / 7, we have 2π = 44 / 7. Step 3: So r1 = 44 / (44 / 7) = 7 cm. Step 4: For the larger circle, C2 = 88 cm. Step 5: Again use C2 = 2πR, so R = 88 / (44 / 7) = 14 cm. Step 6: Now compute R^2 − r^2 = 14^2 − 7^2 = 196 − 49 = 147. Step 7: Area between the circles = π(R^2 − r^2) = π * 147. Step 8: Substitute π = 22 / 7 to get area = (22 / 7) * 147 = 22 * 21 = 462 square cm. Verification / Alternative check: We can also compute the full areas separately. Area of larger circle = πR^2 = (22 / 7) * 196 = 22 * 28 = 616 square cm. Area of smaller circle = πr^2 = (22 / 7) * 49 = 22 * 7 = 154 square cm. Difference = 616 − 154 = 462 square cm, which matches our direct computation. Why Other Options Are Wrong: 154 square cm: This is the area of the smaller circle alone, not the annular region. 308 square cm: This is neither the difference nor the sum of the two areas and does not match any correct calculation. 616 square cm: This is the area of the larger circle alone. 210 square cm: This value is not derived from any consistent use of circumference and area formulas. Common Pitfalls: A frequent mistake is to average the radii or circumferences instead of computing areas using the correct formulas. Some students also forget to subtract the smaller area from the larger one. Another common error is mishandling the constant π or using an inconsistent approximation. Always follow the sequence: circumference to radius, then radius to area, and finally area difference. Final Answer: The area of the region between the two circles is 462 square cm.

Question 8

A rectangular red carpet measures 6 ft by 12 ft. It has a dark red border of uniform width 6 inches all around the edges. What is the area (in square feet) of this dark red border?

Accepted Answer

Introduction / Context: This aptitude question involves finding the area of a border around a rectangle. It tests your comfort with unit conversion (inches to feet) and with the idea that the area of a border equals the difference between the outer and inner rectangles. Given Data / Assumptions: Outer rectangle (the entire carpet) has length 12 ft and width 6 ft. The border all around is 6 inches wide. 1 foot = 12 inches, so 6 inches = 0.5 ft. The inner rectangle (central part without border) is smaller by this border width from all sides. We seek the area of the border region in square feet. Concept / Approach: The border forms a frame around a central rectangle. The area of the border is: Area of border = Area of outer rectangle − Area of inner rectangle We first compute the outer area from the original dimensions. Then we subtract twice the border width from each dimension to find the inner dimensions, compute the inner area and finally subtract. Step-by-Step Solution: Step 1: Convert the border width to feet. Given 6 inches, and 12 inches = 1 ft, we get 6 inches = 0.5 ft. Step 2: Outer rectangle dimensions: length L = 12 ft, width W = 6 ft. Step 3: Outer area A_outer = L * W = 12 * 6 = 72 square feet. Step 4: The border is 0.5 ft wide on each side, so total reduction in length is 2 * 0.5 = 1 ft, and total reduction in width is also 1 ft. Step 5: Inner rectangle dimensions: length = 12 − 1 = 11 ft, width = 6 − 1 = 5 ft. Step 6: Inner area A_inner = 11 * 5 = 55 square feet. Step 7: Area of dark red border = A_outer − A_inner = 72 − 55 = 17 square feet. Verification / Alternative check: We can think of the border as a union of four rectangles around the central area, but computing their individual areas is more tedious and will still sum to 17 square feet. Since the difference of areas approach is straightforward and produces a positive, reasonable result, it confirms that the answer is correct. Why Other Options Are Wrong: 9 square feet and 15 square feet: These are too small given that the border runs all around a fairly large carpet. They come from incorrect inner dimensions or missing part of the border area. 18 square feet: This is close but still off by 1 square foot and arises from mishandling one of the dimensions or forgetting that both sides shrink by twice the border width. 12 square feet: This would correspond to shrinking only in one direction or other incomplete reasoning. Common Pitfalls: Many students forget to convert inches to feet and attempt to mix units, which leads to wrong areas. Another common issue is subtracting the border width only once instead of twice (left and right, or top and bottom). Always remember that both dimensions are reduced on two sides when there is a uniform border. Final Answer: The area of the dark red border is 17 square feet.

Question 9

Consider the following statements about special quadrilaterals: (1) Every isosceles trapezium is cyclic. (2) Any parallelogram that is cyclic must be a rectangle. Which of the above statements is or are correct?

Accepted Answer

Introduction / Context: This is a conceptual geometry question about cyclic quadrilaterals and special quadrilaterals like trapeziums and parallelograms. It asks you to judge the truth of two statements involving isosceles trapeziums and cyclic parallelograms. Understanding properties of cyclic quadrilaterals and angle relationships is essential here. Given Data / Assumptions: Statement 1: Every isosceles trapezium is cyclic. Statement 2: Any parallelogram that is cyclic must be a rectangle. A trapezium (trapezoid) has exactly one pair of parallel sides. An isosceles trapezium is a trapezium whose non parallel sides (legs) are equal in length. A quadrilateral is cyclic if all its vertices lie on a single circle. For a cyclic quadrilateral, opposite angles are supplementary (their sum is 180 degrees). Concept / Approach: We analyse each statement separately using known theorems: For a trapezium to be cyclic, a necessary and sufficient condition is that its non parallel sides are equal, that is, the trapezium is isosceles. For a parallelogram to be cyclic, opposite angles must be supplementary. However, in a parallelogram, opposite angles are already equal. The only way for an angle to be equal to its supplement is for it to be 90 degrees, which makes the parallelogram a rectangle. These facts directly address each statement. Step-by-Step Solution: Step 1: Consider Statement 1: Every isosceles trapezium is cyclic. Step 2: In an isosceles trapezium, the legs are equal and the base angles at each base are equal. The sum of a pair of adjacent interior angles between a leg and a base is 180 degrees. Step 3: If a quadrilateral has one pair of opposite angles supplementary, and the quadrilateral is a trapezium with equal legs, it can be inscribed in a circle. In fact, a trapezium is cyclic if and only if it is isosceles. Thus every isosceles trapezium is cyclic, so Statement 1 is correct. Step 4: Now consider Statement 2: Any cyclic parallelogram is a rectangle. Step 5: In a parallelogram, opposite angles are equal. In a cyclic quadrilateral, opposite angles are supplementary (sum is 180 degrees). Step 6: Combine these: if a parallelogram is cyclic, then each pair of opposite equal angles must also be supplementary. This is only possible if each of those angles is 90 degrees. Step 7: A parallelogram with all right angles is by definition a rectangle. Hence any cyclic parallelogram must be a rectangle, so Statement 2 is also correct. Step 8: Since both statements are correct, the correct choice is that both 1 and 2 are true. Verification / Alternative check: You can draw an isosceles trapezium and see that its vertices lie on a circle by constructing equal arcs and using base angle equality. For the parallelogram case, try drawing a non right angled parallelogram and attempt to inscribe it in a circle: the sum of opposite angles will exceed or fall short of 180 degrees, making a circumcircle impossible. Only a rectangle satisfies both parallelogram and cyclic conditions at once. Why Other Options Are Wrong: 1 only or 2 only: Each would assert that exactly one statement is true. We have shown that both statements 1 and 2 are correct, so these options are invalid. Neither 1 nor 2: This would mean both statements are false, which directly contradicts established theorems. Cannot be determined: The truth of these statements comes from standard Euclidean geometry and does not require additional numerical data, so it can be determined. Common Pitfalls: One common misconception is to think that any trapezium can be cyclic, which is not correct. Only isosceles trapeziums guarantee cyclicity. Another mistake is to forget that a cyclic quadrilateral has supplementary opposite angles and not equal opposite angles in general. Confusing these properties may lead to wrong conclusions about parallelograms. Remembering that cyclicity and parallelogram properties together force right angles helps avoid error. Final Answer: Both statements are correct, so the right choice is Both 1 and 2.

Question 10

Consider the following statements about circles and segments: (1) The number of circles that can be drawn through three non collinear points is infinity. (2) The angle formed in the minor segment of a circle is acute. Which of the above statements i…

Accepted Answer

Introduction / Context: This conceptual question is about basic properties of circles: how many circles can pass through given points, and the nature of angles in segments of a circle. It tests whether you correctly recall some fundamental uniqueness and angle properties in circle geometry. Given Data / Assumptions: Statement 1: The number of circles that can be drawn through three non collinear points is infinity. Statement 2: The angle formed in the minor segment of a circle is acute. Non collinear points means the three points do not lie on a straight line. A segment of a circle is the region bounded by a chord and the corresponding arc. The smaller region is the minor segment, and the larger is the major segment. Concept / Approach: We analyse each statement using well known results: Through any three non collinear points, there exists exactly one circle, called the circumcircle of the triangle formed by those points. Angles subtended by a chord at points on the same arc have a fixed relation. The inscribed angle subtended by a chord on the same side as the minor arc is obtuse, while on the side of the major arc it is acute. Thus, the angle in the minor segment is typically obtuse, not acute. Step-by-Step Solution: Step 1: Consider the three non collinear points as vertices of a triangle. Step 2: In geometry, it is a standard theorem that there is a unique circle passing through the three vertices of a triangle. This circle is called the circumcircle of the triangle. Step 3: Therefore, Statement 1, which says there are infinitely many circles through three non collinear points, is false. There is exactly one. Step 4: Now consider Statement 2 about angles in a minor segment. Step 5: Take a chord AB of a circle. The minor segment is the smaller region bounded by chord AB and the minor arc AB. Any angle with vertex on the minor arc and intercepting chord AB actually subtends the major arc. Step 6: Since the angle at the circumference equals half the measure of the intercepted arc, an angle intercepting a major arc (greater than 180 degrees) is obtuse. Step 7: Hence, angles formed in the minor segment of the circle are typically obtuse, not acute. Step 8: Therefore, Statement 2, which claims such an angle is acute, is also false. Step 9: Since both statements are false, the correct option is that neither 1 nor 2 is correct. Verification / Alternative check: You can verify Statement 1 by constructing different sets of three non collinear points and trying to draw more than one circle through them; all attempts will coincide with a single unique circle. For Statement 2, a careful diagram of a chord and its arcs will show that the angle in the major segment is acute, while the angle in the minor segment is obtuse, confirming the reasoning above. Why Other Options Are Wrong: 1 only or 2 only: These would require one of the statements to be true, but we have seen that both contradict standard theorems about circles and segments. Both 1 and 2: This is impossible since both are individually false. Cannot be determined: The correctness of the statements follows from well established geometric facts and does not require any numerical data, so it can be determined. Common Pitfalls: Learners often confuse the case of two points (infinitely many circles through a fixed chord) with three non collinear points (only one circle). Another common error is mixing up the angle in the minor segment with the angle in the major segment. Keeping a clear mental or drawn diagram of chords, arcs and segments helps avoid these misunderstandings. Final Answer: Both statements are false, so the correct choice is Neither 1 nor 2.

Question 11

Consider the following inequalities for the sides of any triangle ABC: (1) AC − AB < BC (2) BC − AC < AB (3) AB − BC < AC. Which of the above inequalities are always true for a valid triangle?

Accepted Answer

Introduction / Context: This question is based on the triangle inequality, a fundamental property of any triangle. It tests whether you understand not only that each side is less than the sum of the other two, but also that the difference between any two sides is less than the third side. Given Data / Assumptions: We have a triangle ABC with side lengths AB, BC and AC. Three inequalities are given: (1) AC − AB < BC (2) BC − AC < AB (3) AB − BC < AC All side lengths are positive real numbers. We assume ABC is a valid triangle in Euclidean geometry. Concept / Approach: For any triangle with sides x, y and z, the triangle inequality states: x + y > z y + z > x z + x > y From these inequalities we can also derive that the absolute difference between any two sides is less than the third side: |x − y| < z and similarly for other pairs. Each of the given inequalities is essentially one form of such a difference inequality, but without absolute values they still hold because if the left side becomes negative, it is automatically less than a positive side length. Step-by-Step Solution: Step 1: Consider inequality (1): AC − AB < BC. Step 2: Starting from the triangle inequality AC < AB + BC, subtract AB from both sides: AC − AB < BC. So (1) is always true. Step 3: Consider inequality (2): BC − AC < AB. Step 4: From BC < AC + AB, subtract AC from both sides: BC − AC < AB. So (2) is always true. Step 5: Consider inequality (3): AB − BC < AC. Step 6: From AB < BC + AC, subtract BC from both sides: AB − BC < AC. So (3) is always true. Step 7: In cases where the difference is negative, such as AC − AB being negative, the inequality remains valid because a negative number is automatically less than any positive side length. Step 8: Therefore all three inequalities (1), (2) and (3) hold for every valid triangle ABC. Verification / Alternative check: You can test with a concrete example, such as a 3 4 5 triangle. Let AB = 3, BC = 4 and AC = 5. Then: AC − AB = 5 − 3 = 2 which is less than BC = 4, so (1) holds. BC − AC = 4 − 5 = −1 which is less than AB = 3, so (2) holds. AB − BC = 3 − 4 = −1 which is less than AC = 5, so (3) holds. This numerical check supports the algebraic reasoning. Why Other Options Are Wrong: 1 and 2 only, 2 and 3 only, 1 and 3 only: Each of these options wrongly excludes one of the inequalities, even though all are valid consequences of the triangle inequality. None of these: This would imply none of the combinations are correct, which contradicts the derivation showing that all three are true. Common Pitfalls: Learners sometimes believe that the triangle inequality only provides information about sums of sides, not differences. Another mistake is to worry about the sign of expressions like BC − AC and think a negative result invalidates the inequality. Remember that a negative value is always less than a positive side length, so the inequality still holds. Final Answer: All three inequalities are always true, so the correct choice is 1, 2 and 3.

Question 12

Triangle ABC is a right angled triangle with angle A = 90°, AB = 5 cm and AC = 12 cm. The triangle is rotated about its hypotenuse BC to form a double cone (two right circular cones with a common base). What is the approximate total volume (in cubic…

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Introduction / Context: This is an advanced geometry and mensuration problem that combines right triangles, rotation in three dimensions and the volume formula for cones. When a right angled triangle is rotated about its hypotenuse, it generates a solid known as a double cone (two right circular cones joined at a common base). The problem requires careful use of right triangle properties and cone volume formulas. Given Data / Assumptions: Triangle ABC is right angled at A. AB = 5 cm and AC = 12 cm are the legs of the right triangle. Hypotenuse BC is the axis of rotation. On rotation of triangle ABC around BC, a double cone is formed. We use π = 22 / 7 for approximate volume calculation. We must find the total volume of the two cones formed. Concept / Approach: First, we find the length of the hypotenuse BC using Pythagoras theorem. When the triangle rotates about BC, the right angle vertex A traces a circle whose radius equals the perpendicular distance from A to BC (the altitude from A to BC). The point where this altitude meets BC splits BC into two segments; each segment is the height of one cone, and the common base radius is this altitude. The total volume of the double cone is: V_total = (1 / 3) * π * r^2 * h1 + (1 / 3) * π * r^2 * h2 = (1 / 3) * π * r^2 * (h1 + h2) Here h1 + h2 equals the hypotenuse BC. The altitude r from the right angle to the hypotenuse can be expressed in terms of the legs AB and AC. Step-by-Step Solution: Step 1: Use Pythagoras theorem to find BC: BC^2 = AB^2 + AC^2 = 5^2 + 12^2 = 25 + 144 = 169, so BC = 13 cm. Step 2: The altitude from A to BC, call it h, can be found using the area of the triangle. Step 3: Area of triangle ABC = (1 / 2) * AB * AC = (1 / 2) * 5 * 12 = 30 square cm. Step 4: The same area is also (1 / 2) * BC * h, so 30 = (1 / 2) * 13 * h. Step 5: Solve for h: h = (30 * 2) / 13 = 60 / 13 cm. Step 6: When rotated about BC, point A traces a circle of radius r = h = 60 / 13 cm, which becomes the radius of the common circular base of the double cone. Step 7: The two cones have heights equal to the two segments into which BC is divided, but their sum is BC = 13 cm, so h1 + h2 = 13. Step 8: Therefore total volume is V_total = (1 / 3) * π * r^2 * (h1 + h2) = (1 / 3) * π * (60 / 13)^2 * 13. Step 9: Simplify r^2: (60 / 13)^2 = 3600 / 169. Step 10: So V_total = (1 / 3) * π * (3600 / 169) * 13 = (1 / 3) * π * (3600 / 13). Step 11: Thus V_total = (1200 / 13) * π cubic cm. Step 12: Using π ≈ 22 / 7, V_total ≈ (1200 / 13) * (22 / 7) cubic cm. Step 13: Compute approximately: 1200 * 22 = 26400, and 13 * 7 = 91, so V_total ≈ 26400 / 91 ≈ 290 cubic cm (approximately). Verification / Alternative check: We can also compute the lengths of the two segments of BC and volumes of each cone separately using the relations AB^2 = (segment1) * BC and AC^2 = (segment2) * BC. This leads back to the same total volume expression, confirming that the shortcut formula V_total = (1 / 3) * π * AB^2 * AC^2 / BC is correct and yields the same numerical result near 290 cubic cm. Why Other Options Are Wrong: 145 cubic cm: This is roughly half the correct value and would correspond to using only one cone volume instead of both. 435 cubic cm and 580 cubic cm: These values are significantly larger than the correct volume and would arise from miscalculations such as squaring the hypotenuse again or adding extra factors of π. 260 cubic cm: This underestimates the volume and does not correspond to any consistent application of the formula. Common Pitfalls: A key difficulty is visualizing the double cone and identifying the correct radius and heights. Another common mistake is to treat one of the legs as the radius instead of the altitude to the hypotenuse. Errors in using the area relation or mishandling the fractions also occur. Keeping the geometric picture clear and systematically using area and Pythagoras relations helps avoid these issues. Final Answer: The approximate total volume of the double cone is 290 cubic cm.

Question 13

PQ is a diameter of a circle with centre O. RS is a chord parallel to PQ and subtends an angle of 40° at the centre O. If PR and QS are produced to meet at T, what is the measure (in degrees) of angle ∠PTQ?

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Introduction / Context: This geometry problem involves a circle, a diameter, a chord parallel to the diameter and the angle formed by intersecting secants. It tests your ability to combine properties of central angles, chords and the angle between two lines that intersect outside the circle. Given Data / Assumptions: PQ is a diameter of the circle with centre O. RS is a chord parallel to PQ. Chord RS subtends an angle of 40 degrees at the centre O, so angle ROS = 40 degrees. Segments PR and QS are extended to meet at point T outside the circle. We need to find angle PTQ. Concept / Approach: There is a known theorem about the angle between two secants that intersect outside a circle. If two secants from an external point T cut the circle at points P, R and Q, S, then: ∠PTQ = 1 / 2 (difference of measures of intercepted arcs) In this case, arcs intercepted by the two secants are major arc PQ and minor arc RS (or vice versa). Once we know the measures of these arcs in degrees, we can apply the secant angle formula. Step-by-Step Solution: Step 1: Since RS subtends a central angle of 40 degrees at O, the minor arc RS also has measure 40 degrees. Step 2: PQ is a diameter, so arc PQ is a semicircle with measure 180 degrees. Step 3: Because RS is a chord parallel to diameter PQ and the geometry is symmetrical, the secants PR and QS will intercept arc PQ and arc RS in a way that allows use of the exterior secant angle formula. Step 4: For two secants TPQ and TRS intersecting the circle at P, Q and R, S respectively, the angle at the external point T is given by: ∠PTQ = 1 / 2 (measure of far arc − measure of near arc). Step 5: Here, the far arc is the semicircular arc PQ with measure 180 degrees, and the near arc is the smaller arc RS with measure 40 degrees. Step 6: So ∠PTQ = 1 / 2 (180 − 40) = 1 / 2 (140) = 70 degrees. Verification / Alternative check: If drawn accurately, the configuration shows that T lies on the perpendicular bisector of segment OS, and symmetry makes the intersection angle at T acute and larger than 60 degrees but less than 90 degrees. A value of 70 degrees fits this visual intuition. Also, using coordinate geometry with a unit circle and appropriate points for P, Q, R and S confirms the computed angle as 70 degrees. Why Other Options Are Wrong: 55 and 60 degrees: These do not match the exact formula 1 / 2 (180 − 40) and would come from incorrect use of average or misinterpreting which arcs are involved. 90 degrees: This would require the difference of arc measures to be 180 degrees, which is not the case here. 50 degrees: This would correspond to using 1 / 2 (180 − 80) or another incorrect combination of arcs. Common Pitfalls: The main pitfall is confusing the sum and difference of arcs in the secant angle theorem, or using the wrong arcs (for example, using the major arc corresponding to RS instead of the minor arc). Another mistake is treating T as if it were on the circle rather than outside. Carefully identifying which arcs are intercepted by each secant and applying the correct formula avoids these errors. Final Answer: The measure of angle ∠PTQ is 70 degrees.

Question 14

The diagonal of a square is 10 cm long. What is the length (in cm) of each side of the square?

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Introduction / Context: This is a straightforward geometry question involving the relationship between the side and diagonal of a square. It tests your ability to use Pythagoras theorem in a simple but important special case of a right angled isosceles triangle formed by the sides of a square and its diagonal. Given Data / Assumptions: The figure is a square. The diagonal of the square is given as 10 cm. All sides of a square are equal, and each interior angle is 90 degrees. We must find the side length of the square. Concept / Approach: If the side of the square is s, then the diagonal is the hypotenuse of a right angled triangle whose legs are the two sides of the square. By Pythagoras theorem: Diagonal^2 = s^2 + s^2 = 2s^2 So: s^2 = (Diagonal^2) / 2 and s = Diagonal / √2 This formula is specific to a square and is a useful shortcut to remember. Step-by-Step Solution: Step 1: Let s be the side of the square (in cm). Step 2: The diagonal d is given as 10 cm. Step 3: Apply Pythagoras theorem to the right triangle formed by two sides and the diagonal: d^2 = s^2 + s^2 = 2s^2. Step 4: Substitute d = 10, so 10^2 = 2s^2, which gives 100 = 2s^2. Step 5: Divide by 2 to get s^2 = 50. Step 6: Therefore, s = √50 = √(25 * 2) = 5√2 cm. Verification / Alternative check: We can quickly verify by squaring the side: s = 5√2 gives s^2 = 25 * 2 = 50. Then the diagonal should be √(s^2 + s^2) = √(50 + 50) = √100 = 10 cm, which matches the given value. Why Other Options Are Wrong: 5 cm: This would give a diagonal of √(5^2 + 5^2) = √50, which is less than 10 cm. 10√2 cm: This is larger than required and would correspond to a diagonal of √2 times the actual one. 5/√2 cm: This is smaller than 5 cm and does not produce a diagonal of 10 cm. 10 cm: This would incorrectly assume that side equals diagonal, which is never true in a square. Common Pitfalls: Some students reverse the roles and multiply the diagonal by √2 instead of dividing by √2. Others forget to apply Pythagoras theorem and try to use perimeter or area formulas. Remember that the diagonal of a square is always side times √2, so the side is diagonal divided by √2. Final Answer: The side length of the square is 5√2 cm.

Question 15

Point A divides the line segment BC internally in the ratio 5 : 1. The coordinates of B are (6, −4) and the coordinates of C are (0, 8). What are the coordinates of point A?

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Introduction / Context: This is a coordinate geometry question involving internal division of a line segment. It tests your knowledge of the section formula, which gives the coordinates of a point dividing a segment in a given ratio. Given Data / Assumptions: Segment BC has endpoints B(6, −4) and C(0, 8). Point A divides BC internally in the ratio 5 : 1. The ratio 5 : 1 means BA : AC = 5 : 1, so A is closer to C because AC is the smaller part. We must find the coordinates (x, y) of point A. Concept / Approach: For a segment joining points P(x1, y1) and Q(x2, y2), if a point R divides PQ internally in the ratio m : n, where PR : RQ = m : n, then the coordinates of R are given by the section formula: R = ((n x1 + m x2) / (m + n), (n y1 + m y2) / (m + n)) Here, P corresponds to B, Q corresponds to C, and R corresponds to A, with BA : AC = 5 : 1, so m = 5 and n = 1. Step-by-Step Solution: Step 1: Let B(x1, y1) = (6, −4) and C(x2, y2) = (0, 8). Step 2: The ratio BA : AC is 5 : 1, so m = 5 and n = 1 in the section formula. Step 3: Using the formula for internal division, coordinates of A are: A = ((n x1 + m x2) / (m + n), (n y1 + m y2) / (m + n)). Step 4: Substitute x1 = 6, y1 = −4, x2 = 0, y2 = 8, m = 5 and n = 1. Step 5: x coordinate of A: xA = (1 * 6 + 5 * 0) / (5 + 1) = 6 / 6 = 1. Step 6: y coordinate of A: yA = (1 * (−4) + 5 * 8) / 6 = (−4 + 40) / 6 = 36 / 6 = 6. Step 7: Therefore, A has coordinates (1, 6). Verification / Alternative check: We can confirm that A is closer to C than to B. The vertical difference from C(0, 8) to A(1, 6) is only 2 units in y and 1 unit in x, while B(6, −4) is much farther away. Also, the direction from B to C roughly moves from (6, −4) to (0, 8), and A(1, 6) lies along that line segment, which visually supports the calculation. Why Other Options Are Wrong: (-1, 6) and (-1, -6): These place A on the left side of both B and C and do not lie on the segment connecting B and C correctly for the given ratio. (1, -6): This point has the correct x coordinate from the calculation but a completely incorrect y coordinate and lies outside the segment. (2, 4): This point is on the line passing through B and C but corresponds to a different division ratio, not 5 : 1. Common Pitfalls: Students sometimes reverse the ratio in the section formula or mistakenly use m x1 + n x2 instead of n x1 + m x2. Another error is to treat the ratio as a fraction of total distance without using coordinates properly. Carefully labeling which part corresponds to which weight in the formula helps avoid these mistakes. Final Answer: The coordinates of point A are (1, 6).

Question 16

What is the area (in square units) of the rectangle enclosed by the lines x = −4, x = 2, y = −2 and y = 3 in the Cartesian plane?

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Introduction / Context: This question tests the ability to interpret straight line equations in the coordinate plane and compute the area of the rectangular region they enclose. It is a straightforward application of distance between parallel lines and the area formula for a rectangle. Given Data / Assumptions: Vertical boundary lines: x = −4 and x = 2. Horizontal boundary lines: y = −2 and y = 3. The region enclosed by these four lines is a rectangle. We must find its area in square units. Concept / Approach: A rectangle formed by two vertical and two horizontal lines has side lengths equal to the differences in the corresponding coordinates: Width (horizontal side) = difference in x values = x_right − x_left. Height (vertical side) = difference in y values = y_top − y_bottom. The area of a rectangle is then: Area = Width * Height Step-by-Step Solution: Step 1: Identify the left and right boundaries: x = −4 and x = 2. Step 2: Compute width = x_right − x_left = 2 − (−4) = 2 + 4 = 6 units. Step 3: Identify the bottom and top boundaries: y = −2 and y = 3. Step 4: Compute height = y_top − y_bottom = 3 − (−2) = 3 + 2 = 5 units. Step 5: Compute area = width * height = 6 * 5 = 30 square units. Verification / Alternative check: Plotting the points of intersection of the lines confirms a rectangle with vertices at (−4, −2), (2, −2), (2, 3) and (−4, 3). The horizontal distance from (−4, −2) to (2, −2) is 6 units, and the vertical distance from (2, −2) to (2, 3) is 5 units. Multiplying these side lengths again yields 30 square units, reinforcing the calculation. Why Other Options Are Wrong: 12 and 24: These correspond to smaller areas and would result from using only one dimension correctly or using incorrect differences between coordinates. 60: This is double the correct area and could arise if one mistakenly took both directions as 10 units (for example by misreading the lines) and then multiplied. 15: This might appear if someone added lengths and then halved incorrectly, but it does not correspond to the true area of the rectangle. Common Pitfalls: Students sometimes forget to subtract negative values correctly, leading to incorrect side lengths. Remember that subtracting a negative is equivalent to adding the absolute value. Another mistake is to confuse the coordinate values with distances directly without considering their difference. Final Answer: The area of the rectangle is 30 square units.

Question 17

AB is a chord of a circle with centre O. ON is perpendicular from the centre O to the chord AB. If the length of chord AB is 20 cm and ON = 2√11 cm, what is the radius (in cm) of the circle?

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Introduction / Context: This geometry question uses the relation between a chord, the perpendicular from the centre to the chord and the radius of a circle. It is a standard application of the Pythagoras theorem in a right triangle formed by radius, half chord and the perpendicular distance from the centre. Given Data / Assumptions: Circle has centre O. AB is a chord of the circle with length AB = 20 cm. ON is the perpendicular from O to chord AB, and ON = 2√11 cm. We need to find the radius r of the circle. Because ON is perpendicular to AB, it bisects the chord AB. Concept / Approach: When a perpendicular is drawn from the centre of a circle to a chord, it bisects the chord. Thus each half of AB is 10 cm. In the right triangle formed by the radius, half the chord and the perpendicular from the centre, we can apply Pythagoras theorem: r^2 = (half chord)^2 + (distance from centre)^2 So: r^2 = 10^2 + (2√11)^2 Step-by-Step Solution: Step 1: Since ON is perpendicular to AB at N, it bisects AB. Step 2: Therefore, AN = NB = AB / 2 = 20 / 2 = 10 cm. Step 3: Consider right triangle ONA, where OA is the radius r, ON = 2√11 cm and AN = 10 cm. Step 4: Apply Pythagoras theorem: OA^2 = ON^2 + AN^2. Step 5: Compute ON^2: (2√11)^2 = 4 * 11 = 44. Step 6: Compute AN^2: 10^2 = 100. Step 7: So r^2 = OA^2 = 44 + 100 = 144. Step 8: Therefore, r = √144 = 12 cm. Verification / Alternative check: We can check that a radius of 12 cm is reasonable. The distance from the centre to the chord is 2√11 ≈ 6.64 cm, and half the chord is 10 cm. Together with the radius 12 cm, these values form a valid right triangle since 6.64^2 + 10^2 ≈ 144, confirming the calculation. Why Other Options Are Wrong: 10 cm: This would imply the radius equals half the chord, which is only true if the chord is a diameter. Here, ON is not zero, so AB is not a diameter. 13 cm and 15 cm: These are larger than necessary and do not satisfy the equation r^2 = 44 + 100 exactly. 9 cm: This is too small; with such a radius, half chord 10 cm would exceed the radius, which is impossible in a circle. Common Pitfalls: A common error is forgetting that the perpendicular from the centre to the chord bisects the chord, leading to incorrect half chord length. Another mistake is miscomputing (2√11)^2 or misapplying Pythagoras theorem. Keeping the right triangle picture in mind and carefully squaring the radical avoids these mistakes. Final Answer: The radius of the circle is 12 cm.

Question 18

The largest chord of a circle has length 20.4 cm. What is the radius of the circle?

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Introduction / Context: This question checks a very simple but important fact about circles: the largest possible chord is the diameter. Recognizing this allows you to immediately find the radius from the length of the largest chord. Given Data / Assumptions: The largest chord of the circle has length 20.4 cm. We know that the largest chord in any circle is its diameter. We must find the radius of the circle. Concept / Approach: In any circle, the diameter is the longest chord. Every other chord has a length strictly less than the diameter. Therefore, if the problem states that the largest chord has length 20.4 cm, this length must equal the diameter of the circle. The radius is half the diameter: Radius = Diameter / 2 Step-by-Step Solution: Step 1: Identify that the largest chord of a circle is its diameter. Step 2: Given largest chord length is 20.4 cm, so Diameter D = 20.4 cm. Step 3: Radius r = D / 2. Step 4: Compute r = 20.4 / 2 = 10.2 cm. Verification / Alternative check: Any chord that is not a diameter lies strictly inside the circle and must be shorter than 20.4 cm. Therefore, there is no ambiguity: a circle with radius 10.2 cm has diameter 20.4 cm, and this chord is indeed the largest possible chord. Why Other Options Are Wrong: Greater than 10.2 cm or greater than or equal to 10.2 cm: These suggest the radius could be larger than 10.2 cm while still having a largest chord of 20.4 cm, which is impossible because then the diameter would exceed 20.4 cm, contradicting the statement that 20.4 cm is the largest chord. Less than 10.2 cm: This would give a diameter less than 20.4 cm, so 20.4 cm could not be a chord in that circle at all. Cannot be determined: The information given is sufficient, because largest chord directly identifies the diameter, so the radius is uniquely determined. Common Pitfalls: Sometimes, students overcomplicate such problems by thinking about other chords or by not recalling that the diameter is always the longest chord. Keeping this basic property in mind allows for a quick and accurate solution. Final Answer: The radius of the circle is 10.2 cm.

Question 19

In a rectangle, the length of the diagonal is 17 centimetres and the breadth is 8 centimetres. Find the perimeter of the rectangle in centimetres.

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Introduction / Context: This aptitude question tests your understanding of the relationship between the sides and the diagonal of a rectangle, and how to use that relationship to find the perimeter. Such questions are very common in competitive exams and help reinforce concepts from coordinate and Euclidean geometry. Given Data / Assumptions: The figure is a rectangle. The diagonal of the rectangle is 17 cm. The breadth (shorter side) of the rectangle is 8 cm. We must find the perimeter of the rectangle in centimetres. The rectangle has opposite sides equal and all angles are 90 degrees. Concept / Approach: In a rectangle, the diagonal, length, and breadth form a right angled triangle. Therefore, we can apply the Pythagoras theorem: length^2 + breadth^2 = diagonal^2. Once we find the length, we use the formula for the perimeter of a rectangle: Perimeter = 2 * (length + breadth). Step-by-Step Solution: Step 1: Let the length of the rectangle be L cm and breadth be B = 8 cm. Step 2: The diagonal D = 17 cm. By Pythagoras theorem: L^2 + B^2 = D^2. Step 3: Substitute values: L^2 + 8^2 = 17^2. Step 4: Compute squares: L^2 + 64 = 289. Step 5: Rearrange: L^2 = 289 - 64 = 225. Step 6: Take square root: L = 15 cm (length is positive). Step 7: Now compute perimeter: Perimeter = 2 * (L + B) = 2 * (15 + 8). Step 8: Perimeter = 2 * 23 = 46 cm. Verification / Alternative check: We can quickly verify by rechecking the diagonal: 15^2 + 8^2 = 225 + 64 = 289, and 17^2 = 289, so the diagonal value is consistent. Therefore the dimensions are correct, and the perimeter must be 46 cm. Why Other Options Are Wrong: 42 cm: This would correspond to 21 cm total for length plus breadth, which is not correct for the given diagonal. 50 cm: This implies a different combination of length and breadth that does not give a diagonal of 17 cm. 84 cm: Much too large and does not match the computed side lengths. 92 cm: Exactly double of 46 cm and clearly not appropriate for this rectangle. Common Pitfalls: Students sometimes confuse the formula for the perimeter with that for the area. Another common error is to forget to apply the Pythagoras theorem correctly or to make mistakes in squaring or square root computation. Some also mistakenly double only one side rather than doubling the sum of both sides when calculating the perimeter. Final Answer: The perimeter of the rectangle is 46 cm.

Question 20

If any two sides of a triangle are divided by a line in the same ratio, then the line must be __________ to the third side of the triangle.

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Introduction / Context: This conceptual geometry question is based on a very important theorem about triangles that explains when a line segment inside a triangle is parallel to one of its sides. Understanding this theorem helps in solving many problems involving similar triangles, proportional segments, and midpoints. Given Data / Assumptions: We are given a triangle. A line cuts any two sides of the triangle. The line divides these two sides in the same ratio. We are asked what relation this line has with the third side of the triangle. Standard Euclidean geometry is assumed. Concept / Approach: The relevant result is the Basic Proportionality Theorem, also called Thales theorem. It states: If a line divides any two sides of a triangle in the same ratio, then that line is parallel to the third side of the triangle. Conversely, if a line is drawn parallel to one side of a triangle, it will cut the other two sides in the same ratio. This strong two way relationship links proportional division and parallelism. Step-by-Step Solution: Step 1: Consider triangle ABC with points D on AB and E on AC. Step 2: Suppose line DE intersects sides AB and AC such that AD / DB = AE / EC. Step 3: The condition that the ratios are equal means the two sides are divided proportionally. Step 4: By the Basic Proportionality Theorem, if a line divides two sides of a triangle proportionally, the line must be parallel to the third side. Step 5: Therefore DE must be parallel to BC, the third side of the triangle. Verification / Alternative check: You can imagine drawing a line parallel to the third side first. Because of similarity of the smaller triangle and the original triangle, ratios of corresponding segments on the other two sides are equal. This confirms that proportional division and parallelism always go together in this setting. Why Other Options Are Wrong: Equal: A line cannot be equal to a side of a triangle; equality is not the correct geometric relation here. Perpendicular: Perpendicular lines form right angles, but that has nothing to do with equal ratios on two sides. Non parallel: This directly contradicts the theorem; with equal ratios, the line must be parallel, not non parallel. Collinear: Collinear refers to points on the same straight line, not the relationship between a line inside the triangle and a side of the triangle. Common Pitfalls: Students sometimes confuse this theorem with properties of medians or angle bisectors. Others think any line that cuts two sides might be parallel, without checking the proportionality condition. Always remember: equal ratios on two sides imply the line is parallel to the third side, and parallelism in turn guarantees proportional division. Final Answer: The line must be parallel to the third side of the triangle.

Volume and Surface Area Questions