Question 1

In right triangle ABC, the right angle is at B (∠B = 90°). BD is drawn from B perpendicular to the hypotenuse AC, so BD is an altitude. The segments on AC satisfy AD = 9 cm and DC = 16 cm. What is the length (in cm) of the altitude BD?

Accepted Answer

Introduction: This problem uses the special relationships in a right triangle between the hypotenuse and the altitude drawn from the right angle to the hypotenuse. These relationships connect the altitude length with the segments into which it divides the hypotenuse. Given Data / Assumptions: Triangle ABC is right angled at B. BD is the altitude from B to the hypotenuse AC, so BD ⟂ AC. On the hypotenuse, AD = 9 cm and DC = 16 cm. We must find BD. Concept / Approach: In a right triangle, if an altitude is drawn from the right angle to the hypotenuse, then the square of the altitude equals the product of the segments into which it divides the hypotenuse. That is, BD² = AD * DC. This is a standard result from similar triangles or geometric mean properties and provides a direct way to compute the altitude without finding all side lengths. Step-by-Step Solution: We know AD = 9 cm and DC = 16 cm. Using the relation BD² = AD * DC, we get BD² = 9 * 16. Compute 9 * 16 = 144. So BD² = 144. Take the positive square root (since length is positive): BD = √144 = 12 cm. Verification / Alternative check: Alternatively, one could find the full hypotenuse AC = AD + DC = 25 cm and then show that triangles formed are similar and that the altitude is the geometric mean of the segments. This again leads to BD² = 9 * 16 = 144 and BD = 12 cm, confirming the result. Why Other Options Are Wrong: Values 6, 18, and 21 cm do not satisfy BD² = 144. For example, 6² = 36, 18² = 324, 21² = 441. Only 12² equals 144, so only 12 cm is consistent with the given segment lengths on the hypotenuse. Common Pitfalls: A common error is to assume BD is the arithmetic mean of AD and DC or to apply the Pythagorean theorem incorrectly to non right angles. Remember that the geometric mean relationship BD² = AD * DC specifically holds for the altitude drawn from the right angle to the hypotenuse. Final Answer: The length of BD is 12 cm.

Question 2

In right triangle DEF, the right angle is at E (∠E = 90°) and the measure of angle D is 30°. If the side opposite angle D is EF and EF = 6√3 cm, what is the length (in cm) of side DE?

Accepted Answer

Introduction: This problem involves a standard 30°–60°–90° right triangle. Such triangles have fixed side length ratios, so knowing one side lets you compute the others quickly. Recognizing which side corresponds to which angle is the key step. Given Data / Assumptions: Triangle DEF is right angled at E, so ∠E = 90°. ∠D = 30°, therefore ∠F = 60°. Side EF = 6√3 cm and EF is opposite angle D (30°). We must find DE, one of the other legs of the triangle. Concept / Approach: In a 30°–60°–90° triangle, if the side opposite 30° is a, then: Side opposite 60° is a√3. Hypotenuse (opposite 90°) is 2a. By identifying EF as the side opposite 30°, we can set EF = a and use these ratios to find DE, which is the side opposite 60° in this configuration. Step-by-Step Solution: Angle D = 30°, so side opposite D is EF. Given EF = 6√3 cm, and in a 30°–60°–90° triangle that side equals a. So let a = 6√3. Side opposite 60° (which is DE) has length a√3. Thus DE = a√3 = 6√3 * √3. Compute √3 * √3 = 3, so DE = 6 * 3 = 18 cm. Verification / Alternative check: We can also find the hypotenuse DF = 2a = 2 * 6√3 = 12√3 cm. Check Pythagoras: DE² + EF² = 18² + (6√3)² = 324 + 108 = 432. Hypotenuse squared: (12√3)² = 144 * 3 = 432. Both match, so the triangle is consistent. Why Other Options Are Wrong: Options 12 and 12√3 do not align with the fixed ratios of sides in a 30°–60°–90° triangle given EF = 6√3. The option 18√3 is much too large; squaring it would violate the Pythagorean equality. Only 18 cm satisfies the standard triangle ratios and Pythagorean theorem simultaneously. Common Pitfalls: Students sometimes swap which side is opposite 30° and 60°, or incorrectly think the hypotenuse is a√3 instead of 2a. Keeping a clear picture of the angle side relationships prevents these mistakes. Final Answer: The length of DE is 18 cm.

Question 3

A rhombus has diagonals of lengths 12 cm and 14 cm. Using the standard area formula for a rhombus in terms of its diagonals, what is its area (in cm²)?

Accepted Answer

Introduction: This question checks whether you remember the area formula for a rhombus expressed in terms of its diagonals. A rhombus can be thought of as a special kite where the diagonals are perpendicular bisectors of each other. This gives a convenient formula for its area. Given Data / Assumptions: We have a rhombus. Lengths of the diagonals are d₁ = 12 cm and d₂ = 14 cm. We must find the area of this rhombus in square centimetres. Concept / Approach: For any rhombus (and in fact for any kite with perpendicular diagonals), the area can be calculated using the formula: Area = (1/2) * d₁ * d₂. This comes from the fact that the diagonals divide the rhombus into four right triangles of equal area, and their combined area equals half the product of the diagonals. Step-by-Step Solution: Given d₁ = 12 cm and d₂ = 14 cm. Use the formula: Area = (1/2) * d₁ * d₂. Substitute the values: Area = (1/2) * 12 * 14. Compute 12 * 14 = 168. Now multiply by 1/2: Area = 168 / 2 = 84 cm². Verification / Alternative check: If you draw the rhombus and its diagonals, you get four congruent right triangles, each with legs 6 cm and 7 cm (half of each diagonal). Area of one triangle is (1/2) * 6 * 7 = 21 cm². Multiply by 4 to get the rhombus area: 4 * 21 = 84 cm², confirming the formula based calculation. Why Other Options Are Wrong: The value 168 cm² is simply the product of the diagonals, without the necessary factor of 1/2. The values 42 cm² and 63 cm² are partial products that do not correspond to any standard area computation for rhombuses with these diagonals. Common Pitfalls: A typical mistake is to forget dividing by 2 and use d₁ * d₂ directly. Another is to mix formulas for rectangles or parallelograms with those of rhombuses. Remember that for rhombuses, diagonals and their perpendicular property are a powerful shortcut. Final Answer: The area of the rhombus is 84 cm².

Question 4

What is the area (in cm²) of an equilateral triangle whose side length is 14 cm? Use the standard area formula for equilateral triangles in terms of their side.

Accepted Answer

Introduction: Equilateral triangles are special triangles whose three sides and three angles are all equal. There is a direct formula for their area in terms of the side length, which avoids computing the height separately. This problem checks your ability to apply that formula correctly. Given Data / Assumptions: The triangle is equilateral. Each side has length 14 cm. We must compute its area in square centimetres. Concept / Approach: For an equilateral triangle with side length a, the area A is given by: A = (√3 / 4) * a². This comes from deriving the height using Pythagoras in a 30°–60°–90° triangle inside the equilateral triangle, then substituting into the general area formula (1/2 * base * height). Step-by-Step Solution: Given side length a = 14 cm. Use the formula A = (√3 / 4) * a². Compute a² = 14² = 196. Substitute: A = (√3 / 4) * 196. Simplify 196 / 4 = 49. Thus A = 49√3 cm². Verification / Alternative check: Alternatively, compute the height h first. For an equilateral triangle, h = (√3 / 2) * a = (√3 / 2) * 14 = 7√3. Then area A = (1/2) * base * height = (1/2) * 14 * 7√3 = 7 * 7√3 = 49√3 cm², which matches the formula based result. Why Other Options Are Wrong: Options (49/2)√3, (49/4)√3, and 98√3 result from misplacing the factor of 1/4 or incorrectly handling the square of 14. The area must scale with the square of the side, so doubling the side more than quadruples the area. Only 49√3 cm² is consistent with the correct formula and computations. Common Pitfalls: Learners sometimes forget to square the side or misapply the coefficient, using √3/2 instead of √3/4 in the area formula. Others confuse height and side. Keeping the formula A = (√3 / 4) * a² clearly in mind prevents such errors. Final Answer: The area of the equilateral triangle is 49√3 cm².

Question 5

Point P is the midpoint of segment AB. The coordinates of P are (3, 1), and the coordinates of point B are (5, −4). Using the midpoint formula in coordinate geometry, what are the coordinates of point A?

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Introduction: This question uses the midpoint formula from coordinate geometry to find the unknown endpoint of a line segment when one endpoint and the midpoint are known. Mastering this formula is essential for many geometry and analytic geometry problems. Given Data / Assumptions: P is the midpoint of segment AB. Coordinates of P are (3, 1). Coordinates of B are (5, −4). Coordinates of A are unknown and must be found. Concept / Approach: If A has coordinates (x₁, y₁) and B has coordinates (x₂, y₂), then the midpoint M of AB has coordinates: M = ( (x₁ + x₂) / 2 , (y₁ + y₂) / 2 ). We know M and B, so we can set up equations and solve for x₁ and y₁, the coordinates of A. Step-by-Step Solution: Let A = (x, y) and B = (5, −4). Midpoint P = (3, 1) is given by ((x + 5) / 2, (y − 4) / 2). Equate x coordinate of midpoint: (x + 5) / 2 = 3. Multiply both sides by 2: x + 5 = 6, so x = 1. Equate y coordinate of midpoint: (y − 4) / 2 = 1. Multiply both sides by 2: y − 4 = 2, so y = 6. Therefore, A = (1, 6). Verification / Alternative check: Check that the midpoint of A(1, 6) and B(5, −4) is indeed (3, 1). Average of x coordinates: (1 + 5) / 2 = 6 / 2 = 3. Average of y coordinates: (6 + (−4)) / 2 = 2 / 2 = 1. This matches P exactly, so the answer is correct. Why Other Options Are Wrong: Each alternative pair of coordinates, such as (−1, 7) or (1, −7), fails either the x coordinate or y coordinate midpoint condition, or both. Only (1, 6) produces midpoint (3, 1) when paired with B(5, −4). Common Pitfalls: A typical mistake is to subtract instead of adding coordinates, or to forget to divide by 2. Another error is mixing up which point is A and which is B, though the midpoint formula is symmetric so this usually does not change the result. Writing the formula explicitly helps prevent errors. Final Answer: The coordinates of point A are (1, 6).

Question 6

A rectangle has a diagonal of length 25 cm and one of its sides has length 24 cm. Using the Pythagorean theorem, what is the area (in cm²) of the rectangle?

Accepted Answer

Introduction: Every rectangle can be viewed as made up of right triangles when a diagonal is drawn. The diagonal acts as the hypotenuse of a right triangle whose legs are the sides of the rectangle. This allows us to apply the Pythagorean theorem to find the missing side and then compute the area. Given Data / Assumptions: Rectangle with diagonal d = 25 cm. One side (say length) is 24 cm. We must find the area, which is length * breadth. Concept / Approach: For a rectangle with sides a and b and diagonal d, the relationship by the Pythagorean theorem is: d² = a² + b². Once the unknown side is found, area = a * b. Here, we know d and one side, so we solve for the other side and multiply. Step-by-Step Solution: Let the given side be a = 24 cm, the unknown side be b, and diagonal d = 25 cm. Apply Pythagorean theorem: d² = a² + b². So 25² = 24² + b². Compute 25² = 625 and 24² = 576. Thus 625 = 576 + b². So b² = 625 − 576 = 49. Therefore b = √49 = 7 cm. Area = a * b = 24 * 7 = 168 cm². Verification / Alternative check: The triple (7, 24, 25) is a known Pythagorean triple. This confirms that a rectangle with sides 7 and 24 has diagonal 25. The area then is 7 * 24 = 168 cm², consistent with the calculation. Why Other Options Are Wrong: Options 132, 144, and 186 correspond to either incorrect side lengths or arbitrary multiplications that do not satisfy the Pythagorean relationship with a 25 cm diagonal. Only 168 arises from the correct side pair (24, 7) that forms a 7–24–25 right triangle. Common Pitfalls: Students sometimes incorrectly take the diagonal as the sum of the sides or misapply the theorem by adding the diagonal to a side. Always remember that for right triangles, the hypotenuse squared equals the sum of the squares of the legs, not the sum of the legs themselves. Final Answer: The area of the rectangle is 168 cm².

Question 7

The total area of the four walls of a rectangular room (excluding floor and ceiling) is 57 m². The room has length 5.5 m and height 3 m. Using the formula for the lateral surface area of a cuboid, what is the breadth (in m) of the room?

Accepted Answer

Introduction: This question tests the application of surface area formulas for a cuboid, specifically the lateral (four walls) surface area. Given the area of the four walls, length, and height, we can determine the missing breadth of the room. Given Data / Assumptions: Room is a rectangular cuboid. Length l = 5.5 m. Height h = 3 m. Area of four walls (lateral surface area) = 57 m². We must find breadth b (in metres). Concept / Approach: For a cuboid with length l, breadth b, and height h, the total area of the four vertical walls (lateral surface area) is: Lateral area = 2h(l + b). This is because there are two walls of dimension l by h and two walls of dimension b by h. With lateral area known, we can solve for b. Step-by-Step Solution: Given lateral area = 57 m². Use formula: 2h(l + b) = 57. Substitute h = 3, l = 5.5: 2 * 3 * (5.5 + b) = 57. This simplifies to 6(5.5 + b) = 57. Divide both sides by 6: 5.5 + b = 57 / 6 = 9.5. So b = 9.5 − 5.5 = 4 m. Verification / Alternative check: Check the result by recomputing the lateral area: l + b = 5.5 + 4 = 9.5. Then 2h(l + b) = 2 * 3 * 9.5 = 6 * 9.5 = 57 m², which matches the given lateral surface area exactly. Why Other Options Are Wrong: If breadth were 3 m, 3.5 m, or 4.5 m, the lateral area would be different from 57 m². For example, with b = 3, we get 2 * 3 * (5.5 + 3) = 49.5 m², not 57 m². Only b = 4 m yields the correct lateral area. Common Pitfalls: Students often confuse total surface area (including floor and ceiling) with lateral area. Others mistakenly use formula 2(lb + bh + lh). Here the question clearly specifies four walls only, so we must use 2h(l + b). Clarifying what surfaces are included avoids such confusion. Final Answer: The breadth of the room is 4 m.

Question 8

The height of an equilateral triangle is 18 cm. Using properties of equilateral triangles and the relationship between side and height, what is its area (in cm²)?

Accepted Answer

Introduction: Here we are given the height of an equilateral triangle, not its side, and asked to find the area. This uses the geometric relationship between the side length and the height of an equilateral triangle, combined with the basic triangle area formula. Given Data / Assumptions: The triangle is equilateral. Height h = 18 cm. We must find its area in square centimetres. Concept / Approach: For an equilateral triangle of side a, the height h is related by: h = (√3 / 2) * a. We can solve this for a in terms of h, then compute area using either: A = (1/2) * base * height, or A = (√3 / 4) * a². Both methods give the same answer when used correctly. Step-by-Step Solution: Given h = 18 cm and h = (√3 / 2) * a. Solve for a: a = (2h) / √3 = (2 * 18) / √3 = 36 / √3. Rationalize: 36 / √3 = (36√3) / 3 = 12√3 cm. So side a = 12√3 cm. Now use area formula A = (1/2) * base * height. Base = a = 12√3, height = 18. So A = (1/2) * 12√3 * 18 = 6√3 * 18. Compute 6 * 18 = 108, so A = 108√3 cm². Verification / Alternative check: Alternatively, use A = (√3 / 4) * a² with a = 12√3. Then a² = (12√3)² = 144 * 3 = 432. So A = (√3 / 4) * 432 = (432√3) / 4 = 108√3 cm². This matches the previous result. Why Other Options Are Wrong: Options with square metres are incorrect units. The value 108 sq. cm. ignores the √3 factor, which is essential due to the geometry of equilateral triangles. Only 108√3 sq. cm. fits both the correct formula and the given height. Common Pitfalls: Errors often include forgetting the √3 in the height formula, mixing units, or failing to square the side correctly. Another mistake is using the height directly as a side in the equilateral triangle area formula. Carefully distinguishing between side and height avoids confusion. Final Answer: The area of the equilateral triangle is 108√3 sq. cm.

Question 9

The sides of a triangle are in the ratio 1/2 : 1/3 : 1/4. If the perimeter of the triangle is 104 cm, what is the length (in cm) of the longest side?

Accepted Answer

Introduction: This aptitude question uses proportional reasoning. The sides of a triangle are given in a ratio involving fractions, and the total perimeter is known. By converting the fractional ratio to a simpler whole number ratio, we can find the actual side lengths. Given Data / Assumptions: Side lengths are in ratio 1/2 : 1/3 : 1/4. Perimeter of the triangle is 104 cm. We must determine the longest side. Concept / Approach: Ratios involving fractions can be simplified by multiplying through by the least common multiple (LCM) of the denominators, which converts the ratio into whole numbers. Once we have a:b:c in simplest whole number form, we can write side lengths as ka, kb, kc, use the perimeter to find k, and then deduce each length. The largest of these is the longest side. Step-by-Step Solution: Original ratio: 1/2 : 1/3 : 1/4. LCM of denominators 2, 3, 4 is 12. Multiply each term by 12: (1/2)*12 = 6, (1/3)*12 = 4, (1/4)*12 = 3. So the side ratio becomes 6 : 4 : 3. Let the sides be 6k, 4k, and 3k. Perimeter = 6k + 4k + 3k = 13k = 104 cm. Solve for k: k = 104 / 13 = 8. Thus the sides are 6k = 48 cm, 4k = 32 cm, 3k = 24 cm. The longest side is 48 cm. Verification / Alternative check: Check that the perimeter is indeed 48 + 32 + 24 = 104 cm. Also, the largest ratio term was 6, and 6k gives 48 cm. This is consistent and respects the original fractional ratio when scaled back down. Why Other Options Are Wrong: Values 52, 32, and 26 do not preserve the 6 : 4 : 3 ratio when matched with the other sides under total perimeter 104 cm. For instance, 52 would require a different k and give a perimeter larger than 104. Only 48 cm fits all conditions simultaneously. Common Pitfalls: Mistakes include using 1/2, 1/3, 1/4 directly as side lengths or failing to find the correct LCM of denominators. Others sum the fractions incorrectly. Converting to simple integer ratios first is the safest method. Final Answer: The length of the longest side is 48 cm.

Question 10

A room has length 25 m, breadth 12 m, and height 10 m. The four walls and the ceiling (but not the floor) are to be painted. Painter A can paint 200 m² in 5 days, while Painter B can paint 250 m² in 2 days. If A and B work together at their respecti…

Accepted Answer

Introduction: This problem combines geometry (surface area of a cuboid) with work and time concepts. First we compute the total area that must be painted and then work out how long two painters, working together at different rates, will take to finish the job. Given Data / Assumptions: Room dimensions: length l = 25 m, breadth b = 12 m, height h = 10 m. Surfaces to be painted: four walls and ceiling (no floor). Painter A paints 200 m² in 5 days. Painter B paints 250 m² in 2 days. Both painters work at constant rates and can work together. Concept / Approach: Step one is to compute the total area to be painted. The four walls of a cuboid have area 2h(l + b). The ceiling has area l * b. Next, we find each painter's daily painting rate and add them to get a combined rate. Time required equals total area divided by total rate. Step-by-Step Solution: Area of four walls = 2h(l + b) = 2 * 10 * (25 + 12) = 20 * 37 = 740 m². Area of ceiling = l * b = 25 * 12 = 300 m². Total area to paint = 740 + 300 = 1040 m². Painter A's rate = 200 m² / 5 days = 40 m² per day. Painter B's rate = 250 m² / 2 days = 125 m² per day. Combined rate = 40 + 125 = 165 m² per day. Time required = total area / combined rate = 1040 / 165 days. Compute 1040 / 165: 165 * 6 = 990, remainder = 50. So time = 6 + 50/165 = 6 + 10/33 days. Verification / Alternative check: Check that in 6 10/33 days they cover 1040 m²: 165 * (6 10/33) = 165 * ( (6*33 + 10) / 33 ) = 165 * (208 / 33) = (165 / 33) * 208 = 5 * 208 = 1040 m², confirming the calculation. Why Other Options Are Wrong: If the time were 6, 7 10/33, or 8 days, multiplying by 165 m² per day would not yield 1040 m². For example, 6 days gives only 990 m². Therefore those options correspond to either underestimating or overestimating the work done. Common Pitfalls: A common mistake is to forget the ceiling area or to include the floor by accident. Others miscompute individual rates or incorrectly add them. Always handle area and rate calculations separately and carefully. Final Answer: Working together, the painters will complete the job in 6 10/33 days.

Question 11

A square and a circle are constructed such that the side length of the square is equal to the diameter of the circle. What is the ratio of the area of the square to the area of the circle (take π = 22/7)?

Accepted Answer

Introduction: This question compares the areas of two shapes, a square and a circle, when they share a common linear dimension: the side of the square equals the diameter of the circle. It tests your ability to manipulate formulas and form an exact ratio of areas. Given Data / Assumptions: Side of square = s. Diameter of circle = s (same as square side). Radius of circle r = s / 2. Use π = 22/7 for calculations. We must find area(square) : area(circle). Concept / Approach: Area of a square with side s is s². Area of a circle with radius r is πr². By substituting r = s/2 and simplifying, we can express both areas in terms of s and π. Then we can form their ratio and get a fraction independent of s. Step-by-Step Solution: Area of square Aₛ = s². Radius of circle r = s/2. Area of circle A𝚌 = πr² = π(s/2)² = π(s² / 4) = (π/4)s². We need Aₛ : A𝚌 = s² : (π/4)s². Cancel s² (since s ≠ 0): ratio = 1 : (π/4) = 4 : π. Now use π = 22/7, so ratio = 4 : (22/7). Multiply both terms by 7 to clear denominator: 4 * 7 : 22 = 28 : 22. Simplify 28 : 22 by dividing both parts by 2: 14 : 11. Verification / Alternative check: If you choose a specific value, say s = 14 units, then square area is 14² = 196. Circle radius r = 7, area = π * 7² = (22/7) * 49 = 154. Ratio 196 : 154 simplifies to 14 : 11, confirming the general derivation. Why Other Options Are Wrong: Ratios 28 : 11, 7 : 22, and 22 : 7 either ignore the step of simplification or invert the ratio. Only 14 : 11 maintains the exact proportional relationship consistent with both the formulas and the numerical check. Common Pitfalls: A common error is to forget that the circle's radius is half the diameter, leading to area proportional to s² instead of (s/2)². Others may accidentally invert the ratio and give area(circle) : area(square). Being explicit with formulas avoids these mistakes. Final Answer: The ratio of the area of the square to the area of the circle is 14 : 11.

Question 12

A piece of wire 132 cm long is bent successively to form three different closed shapes: (1) an equilateral triangle, (2) a square, and (3) a circle. For which of these shapes will the enclosed area be greatest?

Accepted Answer

Introduction: This is a conceptual geometry question about maximizing area for a given perimeter. We are given a fixed wire length and asked which shape encloses the largest possible area when the wire is bent into different forms. Understanding isoperimetric properties and regular shapes is crucial here. Given Data / Assumptions: Wire length (perimeter or circumference) = 132 cm. Shapes formed: equilateral triangle, square, circle. Perimeter or circumference of each shape equals 132 cm. We must decide which shape has the largest area. Concept / Approach: Among all plane figures with a given perimeter, the circle encloses the maximum area. Among polygons with a fixed perimeter, a regular polygon with more sides encloses more area than one with fewer sides. Therefore, we expect the circle to have the largest area. We can also confirm this by computing approximate areas of each shape using the common perimeter. Step-by-Step Solution: For the equilateral triangle: perimeter = 3a = 132 ⇒ a = 44 cm. Area triangle Aₜ = (√3 / 4) * a² = (√3 / 4) * 44² = (√3 / 4) * 1936 ≈ 484√3 ≈ 838.6 cm². For the square: perimeter = 4s = 132 ⇒ s = 33 cm. Area square Aₛ = s² = 33² = 1089 cm². For the circle: circumference = 2πr = 132 ⇒ r = 132 / (2π) = 66 / π. Using π ≈ 3.14, r ≈ 21.02 cm. Area circle A𝚌 = πr² ≈ 3.14 * (21.02)² ≈ 3.14 * 441.8 ≈ 1387.2 cm². Verification / Alternative check: We see numerically that A𝚌 ≈ 1387.2 cm², Aₛ = 1089 cm², and Aₜ ≈ 838.6 cm². Therefore, the circle clearly has the largest area. This matches the theoretical result that the circle is the most area efficient shape for a fixed perimeter. Why Other Options Are Wrong: The equilateral triangle and the square, although regular, each enclose less area than the circle for the same perimeter. The areas are not equal across shapes, so the option stating equality is also incorrect. Only the circle consistently maximizes area with fixed perimeter. Common Pitfalls: Some learners think more sides always mean maximum area and stop at the square, forgetting that the circle can be seen as a polygon with infinitely many sides. Others may not know or recall the general isoperimetric property. A quick area comparison confirms the theoretical expectation. Final Answer: The enclosed area is largest when the wire is bent into a circle.

Question 13

In an isosceles triangle, each of the two equal sides is twice as long as the third side. An equilateral triangle is constructed such that its perimeter is the same as that of the isosceles triangle. What is the ratio of the area of the isosceles tr…

Accepted Answer

Introduction: This geometry problem compares the areas of two different triangles that share the same perimeter: one is isosceles with a specific side relationship, and the other is equilateral. It tests knowledge of triangle area formulas, height computation, and careful ratio simplification. Given Data / Assumptions: Isosceles triangle: equal sides each have length 2a, base has length a. Perimeter of isosceles triangle = 2 * 2a + a = 5a. Equilateral triangle has the same perimeter, so each side is 5a / 3. We must find area(isosceles) : area(equilateral). Concept / Approach: For the isosceles triangle, we find its height using the Pythagorean theorem by dropping a perpendicular from the apex to the base, splitting the base into two segments of length a/2. For the equilateral triangle, we use the standard area formula in terms of side. After computing both areas in terms of a, we form a ratio and simplify. Step-by-Step Solution: Isosceles triangle sides: 2a, 2a, a. Drop altitude to base a, splitting it into a/2 and a/2. Height hᵢ = √[(2a)² − (a/2)²] = √[4a² − a²/4] = √[(16a² − a²)/4] = √(15a²/4) = (a/2)√15. Area isosceles Aᵢ = (1/2) * base * height = (1/2) * a * (a/2)√15 = (a²/4)√15. Equilateral triangle side s = 5a / 3. Area equilateral Aₑ = (√3 / 4) * s² = (√3 / 4) * (25a² / 9) = (25√3 a²) / 36. Form ratio Aᵢ : Aₑ = [(a²√15)/4] : [(25√3 a²)/36]. Cancel a² and simplify: = (√15 / 4) * (36 / 25√3) = (36 / 100) * (√15 / √3). Compute √15 / √3 = √(15/3) = √5. So ratio = (36 / 100)√5 = 36√5 : 100. Verification / Alternative check: You can choose a convenient value, for example a = 1, compute both areas numerically with that choice, and confirm that their ratio simplifies to 36√5 : 100. This matches the algebraic derivation and confirms the correctness of the simplification. Why Other Options Are Wrong: Ratios 30√5 : 100, 32√5 : 100, or 42√5 : 100 correspond to using incorrect heights or incorrect equilateral area formulas. Only 36√5 : 100 arises from the correct application of the Pythagorean theorem and standard area formulas. Common Pitfalls: Errors usually occur in computing the isosceles triangle height or in misusing the equilateral triangle area formula, such as omitting the √3 factor or mishandling fractions when simplifying. Being systematic with algebraic steps prevents these mistakes. Final Answer: The ratio of the areas is 36√5 : 100.

Question 14

In an equilateral triangle ABC, G is the centroid (intersection point of the medians). Each side of the triangle is 6 cm long. What is the length (in cm) of segment AG, measured from vertex A to the centroid?

Accepted Answer

Introduction: This question focuses on two well known properties of triangles: the length of the median in an equilateral triangle and the position of the centroid along that median. It uses basic geometry facts to compute the distance from a vertex to the centroid. Given Data / Assumptions: Triangle ABC is equilateral. Each side is 6 cm. G is the centroid of triangle ABC. We must find the length AG. Concept / Approach: In any triangle, the centroid divides each median in the ratio 2 : 1, counting from the vertex to the midpoint of the opposite side. That is, the vertex to centroid segment is two thirds of the median. In an equilateral triangle, each median is also an altitude, and its length can be found using the 30°–60°–90° triangle relations. Step-by-Step Solution: Side length a = 6 cm. In an equilateral triangle, the height (and also the median) from any vertex is h = (√3 / 2) * a. So median length from vertex A, call it AM, is h = (√3 / 2) * 6 = 3√3 cm. The centroid G divides median AM in the ratio 2 : 1 from vertex to base. Therefore, AG = (2/3) * AM. AG = (2/3) * 3√3 = 2√3 cm. Verification / Alternative check: You can also coordinate the triangle with A at the top and base on the x axis, compute the centroid as the average of vertices, and then distance from A to G. This will also produce AG = 2√3 cm for side length 6, confirming the median based reasoning. Why Other Options Are Wrong: Values 2√2 cm and 3√2 cm relate to right triangles with 45° angles, not to equilateral triangles. The value 3√3 cm is the entire median length, not the distance from vertex to centroid. According to centroid properties, that distance must be two thirds of the median, giving 2√3 cm instead. Common Pitfalls: Confusing centroid with incenter or circumcenter can lead to wrong assumptions about distances. Another common error is to forget the 2 : 1 division of medians and instead use half. Remember that the centroid lies closer to the side than to the vertex. Final Answer: The length of AG is 2√3 cm.

Question 15

In a circle of radius 5 cm, AB is a diameter and PQ is a chord that is not a diameter. The diameter AB and the chord PQ intersect at point X at right angles (AX ⟂ PQ). The segments on AB satisfy AX : BX = 3 : 2, where X lies between A and B. What is…

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Introduction: This geometry problem involves a circle with a diameter and a chord intersecting at right angles. Knowing the ratio in which the intersection point divides the diameter, and the radius of the circle, allows us to use coordinate geometry or chord length formulas to find the length of the chord. Given Data / Assumptions: Circle radius R = 5 cm. AB is a diameter, so AB = 2R = 10 cm. Chord PQ is perpendicular to AB at point X. X lies on AB with AX : XB = 3 : 2. We must find the length of chord PQ. Concept / Approach: We can place the circle in a coordinate system with center at the origin. Take AB as a horizontal diameter. Then we can locate point X using the given ratio along AB. Since PQ is perpendicular to AB at X, it will be a vertical line through X. Using the circle equation x² + y² = R², we can find the y coordinates of the intersection points and hence the chord length. Step-by-Step Solution: Let the center O be at (0, 0). Then radius R = 5, so circle equation is x² + y² = 25. Take diameter endpoints as A(−5, 0) and B(5, 0). AX : XB = 3 : 2 and AX + XB = 10 ⇒ 3k + 2k = 10 ⇒ 5k = 10 ⇒ k = 2. So AX = 6 and XB = 4. Starting from A(−5, 0), moving 6 units right gives X at (−5 + 6, 0) = (1, 0). Chord PQ is perpendicular to AB, so it lies on the vertical line x = 1. Substitute x = 1 into x² + y² = 25 ⇒ 1 + y² = 25 ⇒ y² = 24 ⇒ y = ±√24 = ±2√6. Thus P = (1, 2√6) and Q = (1, −2√6). Length of PQ is vertical distance: |2√6 − (−2√6)| = 4√6 cm. Verification / Alternative check: Use the formula for chord length when distance from center to chord is known. Distance OX = 1. Chord length L = 2√(R² − OX²) = 2√(25 − 1) = 2√24 = 4√6 cm. This matches the coordinate calculation. Why Other Options Are Wrong: Lengths 2√13 cm, 5√3 cm, and 6√5 cm correspond to different distances from the center or radius values and do not match the condition that OX = 1 and R = 5. Only 4√6 cm satisfies the geometry and the circle equation simultaneously. Common Pitfalls: Errors can occur when placing coordinates for A and B, or when translating the ratio AX : XB to actual segment lengths. Another error is miscalculating the distance from the center to the chord. Using either coordinate geometry or direct chord length formulas carefully avoids these issues. Final Answer: The length of chord PQ is 4√6 cm.

Question 16

Let x and y be two acute angles such that x + y < 90° and sin(2x − 20°) = cos(2y + 20°). What is the value of tan(x + y)?

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Introduction: This trigonometry question tests your understanding of angle relationships and basic trigonometric identities. You are given a relation between sine and cosine involving transformed angles and asked to find the tangent of a sum of angles, tan(x + y). Recognising how to convert cosine into sine and then solving the resulting equation is the core idea here. Given Data / Assumptions: x and y are acute angles, so 0° < x < 90° and 0° < y < 90°. x + y < 90°. sin(2x − 20°) = cos(2y + 20°). We are required to find tan(x + y). Concept / Approach: We use the identity cos A = sin(90° − A) to convert the cosine term to a sine term. This lets us equate arguments of sine by using the fact that sin θ = sin φ implies either θ = φ + 360°k or θ = 180° − φ + 360°k. Under the given acute angle conditions and the restriction x + y < 90°, we select the valid relation and compute x + y, then evaluate tan(x + y). Step-by-Step Solution: Rewrite cos(2y + 20°) as a sine function:cos(2y + 20°) = sin(90° − (2y + 20°)) = sin(70° − 2y).So the equation becomes sin(2x − 20°) = sin(70° − 2y).For principal values, one possibility is 2x − 20° = 70° − 2y.Then 2x + 2y = 90° ⇒ x + y = 45°.The alternative solution 2x − 20° = 180° − (70° − 2y) = 110° + 2y would lead to x − y = 65°, which combined with x and y acute gives x + y > 90°, contradicting the condition x + y < 90°. So we reject that.Thus x + y = 45°.Now tan(x + y) = tan 45° = 1. Verification / Alternative check: You can choose specific values satisfying x + y = 45° and the conditions, for example x = 30°, y = 15°. Then 2x − 20° = 40°, 2y + 20° = 50°. We have sin 40° and cos 50°, and since cos 50° = sin(40°), the equation holds. For these x and y, x + y = 45°, and tan 45° = 1, confirming our result. Why Other Options Are Wrong: √3 and 1/√3 correspond to 60° and 30° tangents, which would require x + y to be 60° or 30°, not 45°. The value 2 + √2 does not match any standard simple angle tangent. Zero would require x + y to be 0° or 180°, which is impossible here since x and y are acute and their sum is less than 90°. Common Pitfalls: Common errors include forgetting the identity cos A = sin(90° − A), mixing up the general solutions of sin θ = sin φ, or accepting the second sine solution without checking it against the given inequality x + y < 90°. Always verify which branch of the trigonometric equation is compatible with the constraints. Final Answer: The required value of tan(x + y) is 1.

Question 17

Two metal pipes have lengths 1.5 m and 1.2 m. They are to be cut into smaller pieces of equal length without leaving any waste. What is the greatest possible length (in metres) of each piece so that all pieces are of the same size?

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Introduction: This problem is a practical application of highest common factor (HCF) or greatest common divisor (GCD) in a geometry and measurement context. The idea is that if two lengths have to be cut into equal smaller pieces without leaving any leftover material, the maximum length of each small piece must exactly divide both original lengths. Given Data / Assumptions: First pipe length = 1.5 m. Second pipe length = 1.2 m. All small pieces must be of the same length. No extra length or waste should be left over from either pipe. We want the greatest possible length of each piece. Concept / Approach: The greatest possible equal length that divides both pipe lengths evenly is their HCF. Since the numbers are decimals, it is easier to convert them into centimetres and work with whole numbers. After finding the HCF of the two whole numbers, we convert the result back to metres to answer the question in the required units. Step-by-Step Solution: Convert to centimetres:1.5 m = 150 cm, 1.2 m = 120 cm.Find the HCF of 150 and 120.Prime factors: 150 = 2 × 3 × 5 × 5, and 120 = 2 × 2 × 2 × 3 × 5.Common factors: one 2, one 3, one 5 → HCF = 2 × 3 × 5 = 30.So the greatest possible piece length is 30 cm.Convert back to metres: 30 cm = 0.3 m. Verification / Alternative check: Check that 0.3 m divides both lengths exactly: 1.5 ÷ 0.3 = 5 pieces; 1.2 ÷ 0.3 = 4 pieces. Both are whole numbers, and using any longer piece (for example 0.4 m) will not divide both lengths exactly. Thus 0.3 m is the largest possible common piece length with no wastage. Why Other Options Are Wrong: 0.13 m and 0.41 m do not divide either length neatly. 0.4 m divides 1.2 m but not 1.5 m. 0.2 m divides both but is smaller than 0.3 m, so it is not the greatest possible length. Only 0.3 m satisfies all conditions. Common Pitfalls: Some learners attempt trial and error with decimals instead of converting to centimetres and using HCF, which can be error prone. Others may mistakenly take the least common multiple instead of the greatest common divisor, which would correspond to a larger length that does not fit exactly into both pipes. Final Answer: The greatest possible length of each equal pipe piece is 0.3 m.

Question 18

Two concentric circles have radii 68 cm and 22 cm. What is the area of the ring (annular region) bounded between the two circular boundaries? (Give your answer in terms of π.)

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Introduction: This problem involves the area of an annulus, which is the region between two concentric circles (circles with the same centre but different radii). The key idea is that the area of this ring is simply the difference between the areas of the larger and the smaller circles. Given Data / Assumptions: Radius of the larger circle R = 68 cm. Radius of the smaller circle r = 22 cm. Both circles are concentric. We need the area of the region between them, expressed in terms of π. Concept / Approach: The area of a circle is given by Area = πr². For two concentric circles, the annular area (ring area) is simply the difference of their areas. So, Annular area = πR² − πr² = π(R² − r²). We will compute R² and r², subtract, and then multiply by π. Step-by-Step Solution: Compute the squares:R² = 68² = 4624.r² = 22² = 484.Difference R² − r² = 4624 − 484 = 4140.Therefore, area of the annulus = π × 4140 = 4140π sq. cm. Verification / Alternative check: You can separately compute the two circle areas: larger circle = 4624π, smaller circle = 484π, and then subtract: 4624π − 484π = 4140π, which matches the earlier calculation. Both procedures lead to the same result, confirming that the computation is correct. Why Other Options Are Wrong: 4110π, 4080π, 4050π, and 4000π sq. cm correspond to incorrect differences of the squared radii. They would result from wrong squaring, incorrect subtraction, or mixing up one of the radii. Only 4140π sq. cm is consistent with the formula π(R² − r²) using R = 68 cm and r = 22 cm. Common Pitfalls: Common mistakes include subtracting the radii first and then squaring (which is incorrect), or forgetting to square one of the radii. Always remember that areas depend on the square of the radius, so you must compute R² and r² before subtracting. Another mistake is to try to plug in π approximately even though the question clearly asks for an answer in terms of π. Final Answer: The area of the region between the two concentric circles is 4140π sq. cm.

Question 19

A solid metal sphere of radius 6 cm is melted and recast into a long cylindrical wire of radius 0.2 cm. Assuming no loss of material, what is the length of the wire (in metres)?

Accepted Answer

Introduction: This question combines solid geometry and volume conservation. A sphere is melted and reshaped into a cylindrical wire. Since no material is lost, the volume of the sphere must equal the volume of the cylinder. From this equality we can compute the unknown length of the wire. Given Data / Assumptions: Radius of the sphere, R = 6 cm. Radius of the cylindrical wire, r = 0.2 cm. No loss of material during melting and recasting. We need to find the length L of the wire, and express it in metres. Concept / Approach: Use standard volume formulas: Volume of sphere = (4/3) * π * R³.Volume of cylinder = π * r² * L.Since the sphere is completely converted into the cylinder, we set these two volumes equal and solve for L. Finally we convert the length from centimetres to metres. Step-by-Step Solution: Compute volume of the sphere:Volume_sphere = (4/3) * π * 6³ = (4/3) * π * 216 = 288π cm³.Volume of the cylinder:Volume_cylinder = π * r² * L = π * (0.2)² * L = π * 0.04 * L.Set volumes equal:288π = π * 0.04 * L.Cancel π on both sides:288 = 0.04 * L.Solve for L:L = 288 / 0.04.0.04 = 4/100, so L = 288 * (100/4) = 288 * 25 = 7200 cm.Convert to metres: 7200 cm = 72 m. Verification / Alternative check: We can quickly check: radius of cylinder is very small (0.2 cm), so we expect a long wire. A length of 72 m = 7200 cm is reasonable when compared with the volume 288π. If we plug L = 72 m back in (as 7200 cm), the cylinder volume becomes π * 0.04 * 7200 = π * 288 = 288π cm³, which matches the original sphere volume exactly. Why Other Options Are Wrong: Lengths such as 81 m, 80 m, 75 m or 70 m would give cylinder volumes that either exceed or fall short of 288π cm³. Only 72 m keeps the volume exactly equal, which is necessary because no metal is lost in the process. Common Pitfalls: Typical mistakes include forgetting to cube the radius for the sphere, squaring the radius incorrectly for the cylinder, or not converting centimetres to metres at the end. Some learners also forget to cancel π from both sides, which complicates the algebra unnecessarily. Final Answer: The length of the cylindrical wire is 72 m.

Question 20

In a trapezium ABCD, sides AB and CD are parallel, and ∠ADC = 90°. If AB = 15 cm, CD = 40 cm and the diagonal AC = 41 cm, what is the area of trapezium ABCD in cm²?

Accepted Answer

Introduction: This geometry question deals with a right-angled trapezium, where one non-parallel side is perpendicular to the parallel sides. You are given the lengths of the two parallel sides and a diagonal, and asked to find the area of the trapezium. The key is to interpret the trapezium as composed of right triangles and a rectangle or to use coordinates to find the height. Given Data / Assumptions: AB ∥ CD in trapezium ABCD. ∠ADC = 90°. AB = 15 cm (shorter parallel side). CD = 40 cm (longer parallel side). Diagonal AC = 41 cm. We must find area(ABCD) in cm². Concept / Approach: Place the trapezium on a coordinate plane to find its height. Since ∠ADC = 90° and CD is parallel to AB, we can take CD as horizontal and AD as vertical. The diagonal AC length will then allow us to compute the vertical height of the trapezium. Finally, use the area formula for a trapezium: Area = (1/2) * (sum of parallel sides) * height. Step-by-Step Solution: Let D = (0, 0), C = (40, 0). Then CD = 40 cm on the x-axis.Since ∠ADC = 90°, AD is perpendicular to CD, so AD is vertical. Let A = (0, h).AB is parallel to CD, so AB is also horizontal at y = h. Since AB = 15 cm, B = (15, h).Diagonal AC connects A(0, h) to C(40, 0) and has length 41 cm.Use distance formula for AC:AC² = (40 − 0)² + (0 − h)² = 40² + h² = 1600 + h².Given AC = 41, so AC² = 41² = 1681.Thus 1600 + h² = 1681 ⇒ h² = 81 ⇒ h = 9 cm (height of trapezium).Area of trapezium = (1/2) * (AB + CD) * h = (1/2) * (15 + 40) * 9.Sum of parallel sides = 55, so area = (1/2) * 55 * 9 = 27.5 * 9 = 247.5 cm². Verification / Alternative check: You can view the trapezium as a large right triangle plus or minus a smaller right triangle, but all consistent constructions give height 9 cm and the same area. The use of Pythagoras with diagonal 41 and base 40 naturally produces a 9-40-41 triangle, which is a known Pythagorean triple, giving confidence in the height value. Why Other Options Are Wrong: Values 245, 240, 250 and 260 cm² correspond to incorrect assumed heights or arithmetic mistakes (for example, using the wrong base sum or miscomputing the Pythagorean relationship). Only 247.5 cm² is consistent with the height of 9 cm and the given sides. Common Pitfalls: Errors often occur when interpreting the right angle position, incorrectly assigning coordinates, or using AB or CD as the height instead of a perpendicular distance. Another common mistake is to forget that area of a trapezium uses the average of the parallel sides times the height, not simply base times height. Final Answer: The area of trapezium ABCD is 247.5 cm².

Area Questions