Q: Trapezium from area and difference: The difference between the two parallel sides of a trapezium is 4 cm, the perpendicular distance is 19 cm, and the area is 475 sq cm. Find the lengths of the parallel sides.

Introduction / Context: Solving a trapezium from area and difference of bases involves forming and solving simultaneous linear equations. Given Data / Assumptions: Let bases be a and b with a > b a - b = 4 cm h = 19 cm Area A = 475 sq cm Concept / Approach: Area formula: A = (1/2)(a + b)h. Use A and h to get (a + b), then combine with (a - b) to solve a and b. Step-by-Step Solution: (1/2)(a + b) * 19 = 475 → a + b = (475 * 2) / 19 = 950 / 19 = 50a - b = 4Add: 2a = 54 → a = 27; then b = 23 Verification / Alternative check: Area check: (1/2)*(27+23)*19 = (1/2)*50*19 = 475. Difference 27 - 23 = 4. Consistent. Why Other Options Are Wrong: Other pairs do not satisfy both the sum 50 and difference 4 simultaneously. Common Pitfalls: Mixing which base is larger or miscomputing 950/19 (ensure exact division to 50). Final Answer: 27 and 23

Question 1

Rectangle transformation: The length of a rectangle is twice its breadth. If length is decreased by 5 cm and breadth is increased by 5 cm, the area increases by 75 sq cm. Find the original length of the rectangle.

Accepted Answer

Introduction / Context: This algebra problem involves relating original dimensions to changed dimensions and comparing areas to find unknowns. Given Data / Assumptions: L = 2B (original) (L - 5) * (B + 5) = L * B + 75 All lengths in cm; rectangle properties standard. Concept / Approach: Expand the changed-area expression, simplify, and use L = 2B to solve a linear system for B and L. Step-by-Step Solution: (L - 5)(B + 5) = LB + 75LB + 5L - 5B - 25 = LB + 755L - 5B - 25 = 75 → 5(L - B) = 100 → L - B = 20Given L = 2B → 2B - B = 20 → B = 20L = 2B = 40 Verification / Alternative check: Original area LB = 40*20 = 800. New area (35*25) = 875. Increase = 75 sq cm, matches the condition. Why Other Options Are Wrong: 10cm, 20cm, 30cm, 35cm do not satisfy both L = 2B and L - B = 20 simultaneously. Common Pitfalls: Dropping terms during expansion or mixing the sign in L - B; also, assuming both changes are equal but forgetting L = 2B. Final Answer: 40cm

Question 2

Area of a rectangle is 460 sq m. If the length is 15% more than the breadth, find the breadth of the rectangular field in metres.

Accepted Answer

Introduction / Context: The question examines percentage-based relation between sides and area calculation inversion to retrieve a side length. Given Data / Assumptions: Area A = 460 sq m L = 1.15 * B Concept / Approach: Area A = L * B = 1.15 * B^2. Solve for B via B^2 = A / 1.15. Step-by-Step Solution: 1.15 * B^2 = 460B^2 = 460 / 1.15 = 400B = 20 m (positive dimension) Verification / Alternative check: L = 1.15 * 20 = 23 m. Area check: 23 * 20 = 460 sq m. Correct. Why Other Options Are Wrong: 30 m, 40 m, 50 m, 25 m produce areas inconsistent with 460 given the 15% relation. Common Pitfalls: Using 1.15 incorrectly as 15 or 0.15; or taking square root before dividing by 1.15. Final Answer: 20 m

Question 3

Wire reshaped: A circular wire is bent to form a rhombus with side 11 cm (perimeter conserved). Find the diameter of the original circular wire.

Accepted Answer

Introduction / Context: When a wire is reshaped, length (perimeter) remains constant. We convert between a circle’s circumference and a rhombus’ perimeter. Given Data / Assumptions: Rhombus side = 11 cm → Perimeter = 4 * 11 = 44 cm Same length becomes circle circumference Use π = 22/7 unless stated Concept / Approach: Circumference C = π * d. With C known (44 cm), solve for diameter d. Step-by-Step Solution: C = 44 cmd = C / π = 44 / (22/7) = 44 * 7 / 22 = 14 cm Verification / Alternative check: Radius r = 7 cm. Circumference 2πr = 2 * 22/7 * 7 = 44 cm, matches the wire length. Why Other Options Are Wrong: 28 cm is twice the correct diameter; 7 cm is the radius; 3.5 cm is quarter of the diameter; 22 cm is unrelated here. Common Pitfalls: Using 2πr formula but substituting d for r or forgetting perimeter equality during shape change. Final Answer: 14 cm

Question 4

Circle comparison: For a circle of radius 5 units, the area is numerically what percent of its circumference?

Accepted Answer

Introduction / Context: Relates two basic measures of a circle: area and circumference, then converts their ratio to a percentage for a given radius. Given Data / Assumptions: r = 5 (units) Area A = πr^2 Circumference C = 2πr Concept / Approach: Compute A and C, then find (A/C)*100 percent. Note that π cancels out. Step-by-Step Solution: A = π * 5^2 = 25πC = 2π * 5 = 10πPercentage = (A / C) * 100 = (25π / 10π) * 100 = 2.5 * 100 = 250% Verification / Alternative check: Since A/C = r/2, for r = 5 we get 2.5 → 250%. Why Other Options Are Wrong: 150%, 350%, 450%, 200% do not match the exact ratio r/2 for r = 5. Common Pitfalls: Forgetting to cancel π or squaring the wrong quantity when computing area vs circumference. Final Answer: 250%

Question 5

Grazing area: A cow is tethered with a rope and can graze 9856 sq m. Assuming a circular grazing path, find the length of the rope.

Accepted Answer

Introduction / Context: This classic geometry problem links area of a circle to its radius, modeling the rope length as the radius of the grazing circle. Given Data / Assumptions: Area A = 9856 sq m Grazing path is circular Use π = 22/7 unless specified Concept / Approach: Area A = πr^2 → r = sqrt(A/π). The rope length equals r. Step-by-Step Solution: r^2 = A / π = 9856 / (22/7) = 9856 * 7 / 229856 / 22 = 448 → r^2 = 448 * 7 = 3136r = sqrt(3136) = 56 m Verification / Alternative check: Area with r = 56: πr^2 = (22/7)*56*56 = (22/7)*3136 = 22*448 = 9856. Perfect match. Why Other Options Are Wrong: 16m, 14m, 76m, 64m do not square to 3136 when used with π = 22/7 to produce 9856 m^2. Common Pitfalls: Using diameter instead of radius, or incorrect π handling during division. Final Answer: 56m

Question 6

Office area in “square units”: An office floor measures 13 units by 9 units (rectangular). Compute the total area in square units.

Accepted Answer

Introduction / Context: A straightforward rectangle area calculation using unit measures (no physical units given). Given Data / Assumptions: Length = 13 (units) Breadth = 9 (units) Area unit is “square units”. Concept / Approach: Area of a rectangle A = L * B. Step-by-Step Solution: A = 13 * 9 = 117 square units Verification / Alternative check: Break 13 as 10 + 3: 10*9 = 90; 3*9 = 27; sum 117. Why Other Options Are Wrong: 97, 107, 127, 81 are not equal to 13*9. Common Pitfalls: Adding instead of multiplying, or mixing up perimeter with area. Final Answer: 117

Question 7

Trapezium area: The parallel sides are 12 cm and 10 cm, and the perpendicular distance between them is 14 cm. Find the area of the trapezium.

Accepted Answer

Introduction / Context: Area of a trapezium uses the average of the parallel sides multiplied by the distance between them. Given Data / Assumptions: a = 12 cm, b = 10 cm (parallel sides) Distance (height) h = 14 cm Concept / Approach: Area A = (1/2) * (a + b) * h. Step-by-Step Solution: A = 0.5 * (12 + 10) * 14A = 0.5 * 22 * 14 = 11 * 14 = 154 sq.cm Verification / Alternative check: Compute average of bases: (12+10)/2 = 11; multiply by height 14 → 154. Why Other Options Are Wrong: 121, 164, 186, 140 do not equal 0.5*(22)*14. Common Pitfalls: Using difference instead of sum of bases, or forgetting the 1/2 factor. Final Answer: 154 sq.cm

Question 8

Square inscribed in a circle: If the area of the circle is 220 sq m, find the area of the square inscribed in it.

Accepted Answer

Introduction / Context: This geometry problem links a circle and its inscribed square via the circle’s diameter as the square’s diagonal. Given Data / Assumptions: Area of circle A_c = 220 sq m Inscribed square has diagonal equal to circle’s diameter (2r) Use π = 22/7 unless stated. Concept / Approach: If the circle has radius r, then A_c = πr^2. For the inscribed square, diagonal d = 2r; area of square A_s = d^2 / 2 = (4r^2)/2 = 2r^2. Step-by-Step Solution: πr^2 = 220 → r^2 = 220 / (22/7) = 220 * 7 / 22 = 70A_s = 2r^2 = 2 * 70 = 140 sq m Verification / Alternative check: Diagonal d = 2r → area = (d^2)/2 = 2r^2, consistent with calculation above. Why Other Options Are Wrong: 49, 70, 150, 110 do not correspond to 2r^2 with r^2 = 70. Common Pitfalls: Mistaking side for diagonal or using r instead of 2r when converting circle-to-square dimensions. Final Answer: 140

Question 9

Infinite midpoint-triangle process: Starting with a right triangle with sides 12 cm, 16 cm, and 20 cm, a new right triangle is formed by joining midpoints of the sides; the process continues infinitely. Find the sum of the areas of all triangles for…

Accepted Answer

Introduction / Context: Joining midpoints of a triangle’s sides forms the medial triangle, whose area is one-quarter of the original. Repeating this generates a geometric series of areas. Given Data / Assumptions: Original right triangle sides: 12 cm, 16 cm, 20 cm Area of right triangle = (1/2) * product of perpendicular sides Each medial triangle has area = 1/4 of its parent We include the original triangle unless otherwise specified (standard for “all triangles so made”). Concept / Approach: The sum S of an infinite geometric series with first term a and ratio r (|r| < 1) is S = a / (1 - r). Here a = initial area, r = 1/4. Step-by-Step Solution: Original area a = (1/2) * 12 * 16 = 96 sq cmRatios: 96, 24, 6, 1.5, … (each next is 1/4 of previous)S = a / (1 - r) = 96 / (1 - 1/4) = 96 / (3/4) = 128 sq cm Verification / Alternative check: Partial sums: 96 + 24 = 120; + 6 = 126; + 1.5 = 127.5; approaching 128, confirming the limit. Why Other Options Are Wrong: 312, 412, 246 are unrelated totals; 96 is just the first area without the infinite continuation. Common Pitfalls: Confusing side-halving (which halves length) with area-halving; area reduces by a factor of 1/4, not 1/2. Final Answer: 128 sq.cm

Question 10

Trapezium from area and difference: The difference between the two parallel sides of a trapezium is 4 cm, the perpendicular distance is 19 cm, and the area is 475 sq cm. Find the lengths of the parallel sides.

Accepted Answer

Introduction / Context: Solving a trapezium from area and difference of bases involves forming and solving simultaneous linear equations. Given Data / Assumptions: Let bases be a and b with a > b a - b = 4 cm h = 19 cm Area A = 475 sq cm Concept / Approach: Area formula: A = (1/2)(a + b)h. Use A and h to get (a + b), then combine with (a - b) to solve a and b. Step-by-Step Solution: (1/2)(a + b) * 19 = 475 → a + b = (475 * 2) / 19 = 950 / 19 = 50a - b = 4Add: 2a = 54 → a = 27; then b = 23 Verification / Alternative check: Area check: (1/2)*(27+23)*19 = (1/2)*50*19 = 475. Difference 27 - 23 = 4. Consistent. Why Other Options Are Wrong: Other pairs do not satisfy both the sum 50 and difference 4 simultaneously. Common Pitfalls: Mixing which base is larger or miscomputing 950/19 (ensure exact division to 50). Final Answer: 27 and 23

Question 11

Fencing a rectangle: The length of a rectangular plot is 20 m more than its breadth. If total fencing cost at Rs 26.50 per metre is Rs 5300, find the length of the plot (in metres).

Accepted Answer

Introduction / Context: This is perimeter-from-cost, then dimensions-from-perimeter algebra, concluding with identification of the correct length among options. Given Data / Assumptions: Cost = Rs 5300 Rate = Rs 26.50 per m L = B + 20 Concept / Approach: Perimeter P = Cost / Rate. For a rectangle, P = 2(L + B). With L = B + 20, solve for B and then L. Step-by-Step Solution: P = 5300 / 26.5 = 200 mL + B = P/2 = 100(B + 20) + B = 100 → 2B = 80 → B = 40 → L = 60 Verification / Alternative check: Perimeter check: 2*(60+40) = 2*100 = 200; Cost check: 200*26.5 = 5300. All consistent. Why Other Options Are Wrong: 40, 20, 30, 50 are not the computed length; correct length is 60, which is not listed; therefore “none of these”. Common Pitfalls: Forgetting to divide perimeter by 2 or misdividing 5300 by 26.5 (it equals 200 exactly). Final Answer: none of these

Question 12

Mat for a room: A room is 15 ft by 12 ft. A mat is placed leaving a uniform 1.5 ft margin from each wall. Find the cost of the mat at Rs 3.50 per square foot.

Accepted Answer

Introduction / Context: This problem combines inner rectangle dimensions after margins with unit rate multiplication for cost. Given Data / Assumptions: Outer room: 15 ft by 12 ft Uniform margin: 1.5 ft on all sides Rate = Rs 3.50 per sq ft Concept / Approach: Inner mat dimensions subtract twice the margin from each side. Area = length * breadth. Cost = area * rate. Step-by-Step Solution: Inner length = 15 - 2*1.5 = 12 ftInner breadth = 12 - 2*1.5 = 9 ftArea = 12 * 9 = 108 sq ftCost = 108 * 3.5 = Rs 378 Verification / Alternative check: 3.5 * 100 = 350; 3.5 * 8 = 28; total 378. Correct. Why Other Options Are Wrong: 472.5, 496, 630, 315 do not match the exact computation 108 * 3.5. Common Pitfalls: Subtracting margin only once, or applying it to area directly rather than dimensions first. Final Answer: 378

Question 13

Squares and circles: Find the ratio of the areas of the incircle and the circumcircle of a square.

Accepted Answer

Introduction / Context: For a square, the incircle touches sides while the circumcircle passes through vertices. Their radii relate to the square side a, enabling an area ratio. Given Data / Assumptions: Square side = a Incircle radius r_in = a/2 Circumcircle radius r_out = a/√2 Concept / Approach: Area of a circle is πr^2. Ratio = (πr_in^2) : (πr_out^2) = r_in^2 : r_out^2. Step-by-Step Solution: r_in^2 = (a/2)^2 = a^2/4r_out^2 = (a/√2)^2 = a^2/2Ratio = (a^2/4) : (a^2/2) = 1/4 : 1/2 = 1 : 2 Verification / Alternative check: Set a = 2 to visualize: r_in = 1, r_out = √2; areas π and 2π give 1:2 directly. Why Other Options Are Wrong: 1:1, 1:3, 1:4, 2:3 do not arise from the algebraic ratio of squared radii for a square. Common Pitfalls: Mixing diameter and radius or forgetting area depends on r^2, not r. Final Answer: 1:2

Question 14

Rectangle from perimeter: The perimeter of a rectangle is 84 m and its length is 6 m more than its breadth. Find the area of the rectangle.

Accepted Answer

Introduction / Context: Standard perimeter-to-dimensions problem followed by area computation for a rectangle. Given Data / Assumptions: P = 84 m L = B + 6 m Concept / Approach: Use P = 2(L + B) to find L + B, then substitute L = B + 6 to solve. Area = L * B. Step-by-Step Solution: L + B = 84 / 2 = 42(B + 6) + B = 42 → 2B = 36 → B = 18 → L = 24Area = 24 * 18 = 432 sq m Verification / Alternative check: Perimeter check: 2*(24+18) = 2*42 = 84 m. Correct. Why Other Options Are Wrong: 330, 333, 362, 384 do not equal 24*18 given the perimeter condition. Common Pitfalls: Forgetting to halve the perimeter or miscomputing 24*18 (it is 432). Final Answer: 432 sq.mts

Question 15

Least number of square tiles: How many least number of equal square tiles are required to pave a floor 15.17 m by 9.02 m?

Accepted Answer

Introduction / Context: To minimize the number of equal square tiles, use the largest possible square tile whose side length equals the greatest common divisor (GCD) of the floor dimensions (expressed in the same unit). Given Data / Assumptions: Dimensions: 15.17 m and 9.02 m Convert to cm to use integers: 1517 cm and 902 cm Tile side = GCD(1517, 902) cm Concept / Approach: Number of tiles = (Area of floor) / (Area of one tile) = (L * B) / (g^2), where g = GCD(L, B) in the same unit. Step-by-Step Solution: Compute GCD: 1517 − 902 = 615; 902 − 615 = 287; 615 − 2*287 = 41; 287 = 7*41 → GCD = 41 cmTiles along length = 1517 / 41 = 37Tiles along breadth = 902 / 41 = 22Total tiles = 37 * 22 = 814 Verification / Alternative check: 41 cm = 0.41 m. Floor in tiles: (15.17/0.41) * (9.02/0.41) = 37 * 22 = 814, consistent. Why Other Options Are Wrong: 714, 614, 713, 902 arise from incorrect GCD or rounding; 814 is minimal and exact. Common Pitfalls: Using LCM, or choosing a smaller non-maximal tile side leading to more tiles, or mixing metres and centimetres. Final Answer: 814

Question 16

Fencing poles: A rectangular plot 90 m by 50 m is to be enclosed with poles placed every 5 m along the boundary. How many poles are needed in total?

Accepted Answer

Introduction / Context: Regular spacing of poles along a closed perimeter uses division of the total perimeter by spacing (when perimeter is an exact multiple). Given Data / Assumptions: Dimensions: 90 m by 50 m Spacing: 5 m Perimeter is a multiple of spacing (no partial gap at the end). Concept / Approach: Total number of poles = Perimeter / Spacing, when the start and end coincide on a closed loop and perimeter is exactly divisible by spacing. Step-by-Step Solution: Perimeter P = 2 * (90 + 50) = 280 mNumber of poles = 280 / 5 = 56 Verification / Alternative check: Each 5 m segment corresponds to one spacing; with 280 m, this yields exactly 56 segments/poles, no remainder. Why Other Options Are Wrong: 45, 55, 60, 65 reflect miscounting corners or not using total perimeter correctly. Common Pitfalls: Adding an extra pole for the starting point even when the spacing closes exactly, or miscomputing perimeter. Final Answer: 56

Question 17

Painting a rectangular boundary: The length and breadth of a rectangular field are in the ratio 8:7. If painting the boundary (perimeter) costs Rs 3000 at Rs 10 per metre, find the area of the field in square metres.

Accepted Answer

Introduction / Context: Cost-per-metre on a boundary gives perimeter. With side ratio known, we can solve dimensions and compute area. Given Data / Assumptions: Length : Breadth = 8 : 7 Total cost = Rs 3000 at Rs 10 per m → Perimeter P = 3000 / 10 = 300 m Concept / Approach: Let L = 8k, B = 7k. Then perimeter P = 2(L + B) = 2 * (15k) = 30k. Solve k from P, then area L*B. Step-by-Step Solution: 30k = 300 → k = 10L = 80 m, B = 70 mArea = 80 * 70 = 5600 sq m Verification / Alternative check: Perimeter check: 2*(80+70) = 300 m equals the derived perimeter from cost; consistent. Why Other Options Are Wrong: 1400, 4400, 3600, 4900 do not match the product 80*70 under the given perimeter constraint. Common Pitfalls: Using 8:7 as difference instead of ratio or forgetting to double the sum when computing perimeter. Final Answer: 5600 sq.m

Question 18

Rectangle to square transformation and perimeter:
If the length (L) of a rectangle is decreased by 4 cm and the width (W) is increased by 3 cm, the new figure becomes a square that has the same area as the original rectangle. Determine the perimete…

Accepted Answer

Introduction / Context: Problems that connect rectangles and squares typically ask you to use both shape properties and area relations. Here, a rectangle, after a specific change to length and width, becomes a square whose area equals the original rectangle’s area. We are asked to find the original perimeter, which requires recovering the original dimensions first. Given Data / Assumptions: Original rectangle dimensions: length L, width W (in cm). New figure after changes: length decreased by 4 cm and width increased by 3 cm; this new figure is a square. The square’s area equals the original rectangle’s area. Concept / Approach: Let the square’s side be s. Because the changed figure is a square: s = L - 4 = W + 3. Also, areas are equal: L * W = s^2. Use these equations to solve for s, L, and W, then compute the perimeter 2(L + W). Step-by-Step Solution: Let s = L - 4 and s = W + 3 ⇒ L = s + 4 and W = s - 3.Area equality: L * W = s^2 ⇒ (s + 4)(s - 3) = s^2.Expand: s^2 + s - 12 = s^2 ⇒ s - 12 = 0 ⇒ s = 12.Then L = 12 + 4 = 16, W = 12 - 3 = 9.Perimeter = 2(L + W) = 2(16 + 9) = 2 * 25 = 50 cm. Verification / Alternative check: Original area: 16 * 9 = 144. Square of side 12 has area 144. Areas match; perimeter computed from valid L, W. Why Other Options Are Wrong: 20, 30, 40 are inconsistent with the derived dimensions (16, 9), which uniquely satisfy both “square after change” and “equal area”. Common Pitfalls: Confusing “becomes a square” with “has equal sides after change but different area”. Both conditions must hold.Setting (L - 4) = (W + 3) but forgetting to also enforce area equality. Final Answer: 50

Question 19

Wheel revolutions over a distance:
A wheel has diameter 40 cm. How many complete revolutions does it make while traveling 176 m in a straight line?

Accepted Answer

Introduction / Context: Counting revolutions converts a linear travel distance into “how many circumferences fit into that distance”. Each full revolution covers one circumference of the wheel. Given Data / Assumptions: Diameter d = 40 cm = 0.40 m. Total distance D = 176 m (straight line, no slipping). Revolutions n = total distance / circumference. Concept / Approach: The circumference C of a circle is C = π * d. Once C is known in meters, compute n = D / C and round only if the division is exact for the intended answer. In this case, it yields a near-integer due to π approximation in options. Step-by-Step Solution: C = π * d = π * 0.40 = 0.4π m.n = D / C = 176 / (0.4π) = 440 / π ≈ 140.096…Using π ≈ 22/7 ⇒ n = 440 / (22/7) = 440 * 7 / 22 = 140 exactly. Verification / Alternative check: Circumference with π = 22/7 is 0.4 * 22/7 = 8.8/7 ≈ 1.257 m; 176 / 1.257 ≈ 140. Why Other Options Are Wrong: 240, 40, 340 produce incorrect distances when multiplied by the circumference; only 140 aligns with 176 m using standard π approximation. Common Pitfalls: Forgetting to convert centimeters to meters.Using diameter directly as the distance per revolution instead of circumference. Final Answer: 140

Question 20

Annulus (pool plus walkway):
A circular swimming pool is surrounded by a concrete ring 4 ft wide. The area of the concrete ring is 11/25 of the area of the (inner) pool. Find the pool’s radius in feet.

Accepted Answer

Introduction / Context: This is an annulus (ring) problem: the region between two concentric circles (outer edge of walkway and pool edge). We are given a ratio between the ring’s area and the pool’s area and must find the inner radius (pool radius). Given Data / Assumptions: Pool radius = r (ft). Walkway width = 4 ft ⇒ outer radius = r + 4. Area of ring = π((r + 4)^2 − r^2) = 11/25 of πr^2 (pool area). Concept / Approach: Set up the area equation using ring area and simplify. Cancel π to reduce algebra. Solve the resulting quadratic for r and select the positive root. Step-by-Step Solution: π((r + 4)^2 − r^2) = (11/25) * πr^2(r + 4)^2 − r^2 = 11r^2 / 25r^2 + 8r + 16 − r^2 = 11r^2 / 25 ⇒ 8r + 16 = 11r^2 / 25Multiply by 25: 200r + 400 = 11r^2 ⇒ 11r^2 − 200r − 400 = 0Discriminant Δ = 200^2 + 4*11*400 = 57600 ⇒ √Δ = 240r = (200 + 240) / (2*11) = 440/22 = 20 (ft) (negative root discarded). Verification / Alternative check: Plug r = 20: Ring area = π((24)^2 − 20^2) = π(576 − 400) = 176π; Pool area = π * 400 = 400π; Ratio = 176/400 = 11/25. Why Other Options Are Wrong: 10ft, 30ft, 40ft do not satisfy the precise ratio 11/25 when substituted. Common Pitfalls: Using outer diameter instead of outer radius in the ring formula.Forgetting to cancel π and mishandling algebra. Final Answer: 20ft

Area Questions