Q: In trapezium ABCD with AB ∥ CD and AB < CD, suppose CD = 6 cm and the distance (height) between the parallel sides is 4 cm.\nIf area(ABCD) = 16 cm², find the length AB.

Introduction / Context: Trapezium (trapezoid) area is half the sum of parallel sides times the height. With area and one base given, solve for the other base. Given Data / Assumptions: CD = 6 cm (larger base), AB is smaller base. Height h = 4 cm; Area = 16 cm². Concept / Approach: Area formula: A = (1/2) * (AB + CD) * h. Step-by-Step Solution: 16 = (1/2) * (AB + 6) * 416 = 2 * (AB + 6) ⇒ AB + 6 = 8 ⇒ AB = 2 cm Verification / Alternative check: Plug back: A = (1/2)*(2+6)*4 = (1/2)*8*4 = 16 cm², correct. Final Answer: 2 cm

Question 1

The diagonals of two squares are in the ratio 2 : 5.
Find the ratio of their areas.

Accepted Answer

Introduction / Context: For squares, area is proportional to the square of any linear dimension (side, diagonal). Hence, the ratio of areas equals the square of the ratio of diagonals. Given Data / Assumptions: Diagonal ratio d1 : d2 = 2 : 5. Concept / Approach: Area ∝ (diagonal)^2; thus A1 : A2 = (2^2) : (5^2) = 4 : 25. Step-by-Step Solution: A1 : A2 = (2 : 5)^2 = 4 : 25 Why Other Options Are Wrong: They do not square both numbers or square only one term. Final Answer: 4:25

Question 2

A square of side a has the same area as a triangle with base a.
What is the altitude (height) of the triangle?

Accepted Answer

Introduction / Context: This is a direct comparison of area formulas between a square and a triangle built on the same base length a. Translate equality of areas into equality of formulas and solve for height. Given Data / Assumptions: Square area = a^2. Triangle area = (1/2) * base * height = (1/2) * a * h. Concept / Approach: Set a^2 = (1/2) * a * h and solve for h. Step-by-Step Solution: a^2 = (1/2) * a * h ⇒ h = 2a Final Answer: 2a

Question 3

A square has side length 14 feet.
Find its perimeter and its area (choose the correct numeric value from the options).

Accepted Answer

Introduction / Context: For a square with side s, perimeter is 4s and area is s^2. The options provide single values; the distinctive correct numeric result here is the area (since 14^2 is a perfect square in the list). Given Data / Assumptions: s = 14 ft. Concept / Approach: Compute both quantities: Perimeter = 4*14 = 56; Area = 14^2 = 196. From the options, 196 appears (area), while 56 does not. Step-by-Step Solution: Perimeter = 56 ftArea = 196 sq ft Why Other Options Are Wrong: 176, 186, 206 are not equal to either 56 or 196. Final Answer: 196 (area in sq ft; perimeter is 56 ft).

Question 4

On the same base and with the same altitude, consider a parallelogram (area P), a triangle (area R), and a rectangle (area A).
Which relation is always true?

Accepted Answer

Introduction / Context: For plane figures sharing the same base length b and altitude h, their areas follow standard formulas: parallelogram P = b*h, rectangle A = b*h, triangle R = (1/2)*b*h. Comparing these immediately yields the fixed relations among P, A, and R. Given Data / Assumptions: All figures share base b and altitude h. Concept / Approach: Write each area in terms of b and h and compare. Step-by-Step Solution: Parallelogram: P = b * hRectangle: A = b * hTriangle: R = (1/2) * b * hTherefore P = A and P = 2R, so in particular P = 2R. Why Other Options Are Wrong: P=R: False; triangle area is half. P=A/2: False; parallelogram equals rectangle, not half. Final Answer: P=2R

Question 5

A man walks diagonally across a square plot instead of along two edges.
Approximately what percentage distance is saved by taking the diagonal path?

Accepted Answer

Introduction / Context: Distance saved equals (edge path − diagonal) / edge path. In a square of side s, the edge path from one corner to the opposite via two sides is 2s, while the straight diagonal distance is s√2. Given Data / Assumptions: Square side = s (symbolic; cancels out). Concept / Approach: Compute percent saving = [(2s − s√2) / (2s)] * 100% = [1 − (√2)/2]*100%. Step-by-Step Solution: Saving fraction = 1 − √2/2 ≈ 1 − 0.7071 ≈ 0.2929Saving percent ≈ 29.29% ≈ 30% Verification / Alternative check: Set s = 1. Edge path = 2, diagonal ≈ 1.4142, saving ≈ 0.5858 ≈ 29.3% of 2. Why Other Options Are Wrong: 10%, 20%: Underestimates. 40%: Overestimates. Final Answer: 30%

Question 6

The area of a rectangle is 80 cm². If its length is 16 cm, find its width.

Accepted Answer

Introduction / Context: Use the rectangle area formula A = L * W to find the missing dimension when area and length are given. Given Data / Assumptions: A = 80 cm²; L = 16 cm. Concept / Approach: W = A / L. Step-by-Step Solution: W = 80 / 16 = 5 cm Final Answer: 5cm

Question 7

The area of a square is 49 mm².
What is the length of each side?

Accepted Answer

Introduction / Context: The side of a square equals the positive square root of its area. Given Data / Assumptions: A = 49 mm². Concept / Approach: Side s = √A. Step-by-Step Solution: s = √49 = 7 mm Final Answer: 7mm

Question 8

Find the area of the rectangle with length L = 5 inches and width W = 3 inches.

Accepted Answer

Introduction / Context: Rectangle area A = L * W. Multiply the given dimensions to obtain the area in square inches. Given Data / Assumptions: L = 5 in; W = 3 in. Step-by-Step Solution: A = 5 * 3 = 15 square inches Final Answer: 15

Question 9

A wheelchair slalom has a rectangular course of length 30 m and breadth 16 m.
Find the area of the course.

Accepted Answer

Introduction / Context: Area of a rectangle is the product of its length and breadth (width). Given Data / Assumptions: Length = 30 m; Breadth = 16 m. Step-by-Step Solution: Area = 30 * 16 = 480 square metres Final Answer: 480

Question 10

In trapezium ABCD with AB ∥ CD and AB < CD, suppose CD = 6 cm and the distance (height) between the parallel sides is 4 cm.
If area(ABCD) = 16 cm², find the length AB.

Accepted Answer

Introduction / Context: Trapezium (trapezoid) area is half the sum of parallel sides times the height. With area and one base given, solve for the other base. Given Data / Assumptions: CD = 6 cm (larger base), AB is smaller base. Height h = 4 cm; Area = 16 cm². Concept / Approach: Area formula: A = (1/2) * (AB + CD) * h. Step-by-Step Solution: 16 = (1/2) * (AB + 6) * 416 = 2 * (AB + 6) ⇒ AB + 6 = 8 ⇒ AB = 2 cm Verification / Alternative check: Plug back: A = (1/2)*(2+6)*4 = (1/2)*8*4 = 16 cm², correct. Final Answer: 2 cm

Question 11

A circle has area 346.5 cm².
Find its circumference (in centimetres).

Accepted Answer

Introduction / Context: Convert area to radius using A = pi * r^2, then compute circumference C = 2 * pi * r. With A given as a multiple compatible with pi = 22/7, exact arithmetic is straightforward. Given Data / Assumptions: A = 346.5 cm²; use pi = 22/7. Concept / Approach: r^2 = A / pi; then C = 2 * pi * r. Step-by-Step Solution: r^2 = 346.5 / (22/7) = 346.5 * 7 / 22 = 110.25r = √110.25 = 10.5 cmC = 2 * (22/7) * 10.5 = 66 cm Verification / Alternative check: Back-compute area: pi*r^2 = (22/7)*(10.5)^2 = (22/7)*110.25 = 346.5 cm². Final Answer: 66

Question 12

Total surface area of a cube (side = 7.5 cm)
Find the total surface area (in cm²) of a cube whose edge length is 7.5 cm.

Accepted Answer

Introduction / Context: Cubes have six congruent square faces. A standard mensuration task is to compute total surface area (TSA) from the edge length, then ensure unit consistency (cm² here). Given Data / Assumptions: Edge length a = 7.5 cm Total surface area formula for a cube: TSA = 6 * a^2 Answer required in square centimetres (cm²). Concept / Approach: Each face is a square with area a^2. Since there are 6 faces in a cube, TSA = 6 * a^2. Compute a^2 first, then multiply by 6, taking care with decimal multiplication. Step-by-Step Solution: a^2 = 7.5 * 7.5 = 56.25 cm^2 TSA = 6 * 56.25 = 337.5 cm^2 Verification / Alternative check: Break 7.5 into 15/2 ⇒ a^2 = (15/2)^2 = 225/4 = 56.25; multiplying by 6 yields 337.5 exactly, confirming the decimal computation. Why Other Options Are Wrong: 364.5 and 446 arise from mis-multiplication; 684 mistakenly doubles 342; only 337.5 matches 6 * a^2. Common Pitfalls: Using lateral surface area (4 * a^2) instead of total, or squaring 7.5 incorrectly (e.g., 7.5 * 7 = 52.5) leads to wrong values. Final Answer: 337.5

Question 13

Circle vs rectangle with perimeter 50 cm
The area of a circle equals the area of a rectangle whose perimeter is 50 cm and whose length exceeds its breadth by 3 cm. Find the diameter of the circle (in cm).

Accepted Answer

Introduction / Context: This problem links a rectangle (constrained by perimeter and a length–breadth relation) to a circle with equal area. We first determine the rectangle’s dimensions and area, then equate it to πr^2 to obtain the circle’s diameter. Given Data / Assumptions: Perimeter of rectangle P = 50 cm l = b + 3 (length exceeds breadth by 3 cm) Area(circle) = Area(rectangle) Use π = 22/7 for exact integers here. Concept / Approach: From 2(l + b) = 50 ⇒ l + b = 25. Combine with l = b + 3 to solve for l and b. Then area of the rectangle equals πr^2, allowing solution for r and thus diameter 2r. Step-by-Step Solution: l + b = 25, l = b + 3 ⇒ 2b + 3 = 25 ⇒ b = 11 cm, l = 14 cm Area(rect) = 11 * 14 = 154 cm^2 πr^2 = 154 ⇒ with π = 22/7 ⇒ r^2 = 154 * 7 / 22 = 49 ⇒ r = 7 cm Diameter = 2r = 14 cm Verification / Alternative check: Using π ≈ 3.14 gives r ≈ 7.0, keeping diameter ≈ 14 cm; integer exactness with 22/7 confirms it. Why Other Options Are Wrong: 7 cm is radius, not diameter; 21 cm and 28 cm do not match the area equivalence; only 14 cm is correct. Common Pitfalls: Forgetting to halve perimeter to get l + b or mixing up radius/diameter leads to wrong choices. Final Answer: 14 cm

Question 14

Square from its diagonal (14 cm)
The diagonal of a square is 14 cm. Compute the area of the square (in cm²).

Accepted Answer

Introduction / Context: The diagonal of a square relates to its side by s√2. From the diagonal, we find the side and then the area s^2. This avoids unnecessary approximation because √2 cancels neatly when squaring. Given Data / Assumptions: Diagonal d = 14 cm Relation: d = s * √2 ⇒ s = d / √2 Area = s^2 Concept / Approach: Compute s from the diagonal, then square to get area: s^2 = (d^2) / 2, a convenient identity derived from d = s√2. Step-by-Step Solution: s^2 = d^2 / 2 = 14^2 / 2 = 196 / 2 = 98 cm^2 Verification / Alternative check: Explicitly: s = 14 / √2 = 7√2; s^2 = (7√2)^2 = 49 * 2 = 98, which matches the shortcut above. Why Other Options Are Wrong: 49 sq cm is side^2 if s were 7; 196 sq cm is d^2, not area; 77 sq cm has no basis here. Common Pitfalls: Using area = (1/2)*d^2 is correct for squares; forgetting this identity or mishandling √2 often causes errors. Final Answer: 98 sq cm

Question 15

Two externally touching circles
Two circles touch externally. The distance between their centres is 12 cm. If the radius of one circle is 7 cm, find the diameter (in cm) of the other circle.

Accepted Answer

Introduction / Context: When circles touch externally, the distance between centres equals the sum of their radii. From that, we can find the unknown radius and then the diameter (twice the radius). Given Data / Assumptions: Distance between centres = 12 cm Known radius r1 = 7 cm Let unknown radius r2 be required; diameter sought is 2 * r2. Concept / Approach: Externally touching ⇒ r1 + r2 = distance. Solve for r2, then compute 2r2 for the diameter. Step-by-Step Solution: r1 + r2 = 12 ⇒ 7 + r2 = 12 ⇒ r2 = 5 cm Diameter = 2 * 5 = 10 cm Verification / Alternative check: Substitute back: 7 + 5 = 12 (equals given centre distance), so the configuration is consistent. Why Other Options Are Wrong: 12 is the centre distance, not the diameter; 8 and 5 do not reflect the computed diameter; only 10 cm fits. Common Pitfalls: Confusing external with internal touching (where distance is the absolute difference of radii) changes the equation entirely. Final Answer: 10

Question 16

Rhombus from area and one diagonal
The area of a rhombus is 84 cm² and one diagonal measures 7 cm. Find the length of the other diagonal (in cm).

Accepted Answer

Introduction / Context: For a rhombus, area equals half the product of its diagonals. Thus, knowing the area and one diagonal allows us to directly compute the other diagonal using a simple algebraic rearrangement. Given Data / Assumptions: Area A = 84 cm² One diagonal d1 = 7 cm Unknown diagonal d2 Concept / Approach: Use A = (d1 * d2) / 2 ⇒ d2 = 2A / d1. Substitute the values to find d2 exactly. Step-by-Step Solution: d2 = (2 * 84) / 7 = 168 / 7 = 24 cm Verification / Alternative check: Check: (7 * 24)/2 = 84 cm², which matches the given area perfectly. Why Other Options Are Wrong: 12 and 36 halve or 1.5x the correct value; 48 doubles the right answer; only 24 satisfies A = (d1 * d2)/2. Common Pitfalls: Confusing the rhombus area with the parallelogram area (b * h) is unnecessary here; the diagonal formula is the direct route. Final Answer: 24

Question 17

Perimeter of a semicircle (radius = 28 cm)
Find the perimeter of a semicircle of radius 28 cm (use perimeter = πr + 2r).

Accepted Answer

Introduction / Context: The boundary length of a semicircle includes the curved arc plus the diameter. The standard formula is Perimeter = (1/2 of circumference) + diameter = πr + 2r. Given Data / Assumptions: Radius r = 28 cm Use π = 22/7 for exact integer outcomes. Concept / Approach: Compute the arc component πr and add the straight diameter 2r. Avoid using the full circumference 2πr, since we only have half the circle plus the diameter. Step-by-Step Solution: Arc (semicircle) = πr = (22/7) * 28 = 88 cm Diameter = 2r = 56 cm Perimeter = 88 + 56 = 144 cm Verification / Alternative check: Using π ≈ 3.14 gives 3.14*28 + 56 ≈ 87.92 + 56 ≈ 143.92, rounding to 144, consistent with exact form. Why Other Options Are Wrong: 288 doubles the arc; 121 and 242 are unrelated to πr + 2r with r = 28 cm; only 144 matches. Common Pitfalls: Forgetting to include the diameter or mistakenly halving 2πr to πr and then halving again causes errors. Final Answer: 144

Question 18

Cyclic quadrilateral angle bisectors meet the circle
ABCD is a quadrilateral inscribed in a circle of radius r. The internal bisectors of ∠DAB and ∠BCD meet the circumcircle again at X and Y, respectively. Find the length of XY.

Accepted Answer

Introduction / Context: This is a well-known geometry fact for a cyclic quadrilateral: the internal angle bisectors at opposite vertices (A and C) meet the circumcircle again at diametrically opposite points. Consequently, the chord joining those second intersection points is a diameter of the circle, whose length equals 2r, where r is the circumradius. Given Data / Assumptions: ABCD cyclic with circumradius r. The internal angle bisectors of ∠DAB and ∠BCD intersect the circumcircle again at X and Y. X and Y are antipodal (diametrically opposite) points. Concept / Approach: In a circle, the locus of points subtending equal angles at chord endpoints ties to angle-bisector properties. In a cyclic quadrilateral, the bisectors of ∠A and ∠C meet the circumcircle at endpoints of a diameter (this can be reasoned using directed arcs or properties of supplementary arcs in opposite angles). Hence XY is a diameter. Step-by-Step Solution: Opposite angles A and C are supplementary. Their bisectors intercept equal arcs implying X and Y are opposite ends of a diameter. Therefore, XY = diameter = 2r. Verification / Alternative check: Constructive proofs via inscribed angle theorem or using equal subtended angles at the circumference also arrive at XY as a diameter directly. Why Other Options Are Wrong: π(r r) and π(r r)/2 resemble area-like forms, not a length; (r + 2) is dimensionally inconsistent. Only 2r is a length equal to the diameter. Common Pitfalls: Treating the bisectors as intersecting at arbitrary points on the circle leads to overlooking the supplementary relationship of opposite angles that forces a diameter. Final Answer: 2r

Question 19

Canvas needed for a tent (cylinder + cone, r = 10 m)
A tent is formed by a cylinder of radius 10 m and height 5 m, surmounted by a cone of the same radius with slant height 15 m. How much canvas is required if 20% extra is allowed for folding/stitc…

Accepted Answer

Introduction / Context: Canvas required equals the lateral areas (outer surface only) since tents typically exclude the base/floor unless specified. Here a cylinder is topped by a cone. We compute each lateral area and then add 20% wastage for folds and stitching. Given Data / Assumptions: Cylinder: radius r = 10 m, height h = 5 m Cone: same radius r = 10 m, slant height l = 15 m π = 22/7; base is not included (canvas excludes floor). Extra allowance = 20% of computed lateral area. Concept / Approach: Lateral area (cylinder) = 2 * π * r * h. Lateral area (cone) = π * r * l. Sum, then multiply by 1.20 to account for extra canvas. Step-by-Step Solution: A_cyl = 2 * π * r * h = 2 * (22/7) * 10 * 5 = 100 * (22/7) = 2200/7 ≈ 314.286 m^2 A_cone = π * r * l = (22/7) * 10 * 15 = 150 * (22/7) = 3300/7 ≈ 471.429 m^2 A_total = 2200/7 + 3300/7 = 5500/7 ≈ 785.714 m^2 Canvas (with 20% extra) = 1.2 * 5500/7 = 6600/7 ≈ 942.857 m^2 Verification / Alternative check: If a floor were included, add πr^2 = (22/7) * 100 = 2200/7 ≈ 314.286, yielding ≈ 1320 m^2 even then. None of the given options match ≈ 943 m^2 or ≈ 1320 m^2. Why Other Options Are Wrong: All numeric choices are far larger than either interpretation (with or without base). Hence the correct selection within this set is “None of these.” Common Pitfalls: Including the base by default or using circumference * slant height for the cone incorrectly. Maintain unit consistency and apply the 20% only after summing areas. Final Answer: None of these

Question 20

Earthwork spread from a cylindrical ditch
A square field has area X m². A cylindrical ditch of radius 7 m and depth 2 m is dug; the removed earth is spread uniformly over the remaining part of the field (excluding the circular ditch), raising its l…

Accepted Answer

Introduction / Context: Conservation of volume applies: soil removed from the cylindrical ditch equals the volume added to the remaining area as a uniform layer. Care is needed to subtract the hole’s plan area when determining where the soil is spread. Given Data / Assumptions: Radius r = 7 m, depth h = 2 m Rise over remaining area = 0.77 m Use π = 22/7 for neat integers. Concept / Approach: Volume removed V = π r^2 h. Remaining area = X − π r^2. By conservation: V = (X − π r^2) * 0.77. Solve for X in m². Step-by-Step Solution: V = (22/7) * 7^2 * 2 = (22/7) * 49 * 2 = 22 * 14 = 308 m^3 π r^2 = (22/7) * 49 = 154 m^2 308 = (X − 154) * 0.77 ⇒ X − 154 = 308 / 0.77 = 400 ⇒ X = 554 m^2 Verification / Alternative check: Plug back: remaining area = 554 − 154 = 400; 400 * 0.77 = 308, which matches removed volume exactly. Why Other Options Are Wrong: Values other than 554 fail the exact conservation equation with π = 22/7. 554 is the unique match. Common Pitfalls: Spreading soil over the whole square (including the ditch) or using the wrong π approximation can skew the result. Final Answer: 554 sq. m

Area Questions