Question 1

Maintain area with a 25% longer length: If the length of a rectangle is increased by 25%, by what percentage must the breadth decrease so that the area remains unchanged?

Accepted Answer

Introduction / Context:Area of a rectangle is length * breadth. If one dimension is scaled, the other must scale reciprocally to preserve the product. A 25% increase corresponds to a scale factor of 1.25; the breadth must be multiplied by 1/1.25 to keep area constant. Given Data / Assumptions:Initial area A = L * B. New length L′ = 1.25L. Required B′ such that L′ * B′ = A. Concept / Approach:Set L′ * B′ = L * B ⇒ (1.25L)*B′ = L*B ⇒ B′ = B/1.25. Convert B′ = 0.8B into a percentage decrease from B: decrease = (1 − 0.8)*100% = 20% decrease. Step-by-Step Solution: Scale factor for length = 1.25.Reciprocal factor for breadth = 1/1.25 = 0.8.Percentage decrease = 20%. Verification / Alternative check:Example: L = 100, B = 100 ⇒ A = 10000. New L′ = 125; B′ = 80; area 125*80 = 10000 (unchanged). Why Other Options Are Wrong:25%, 30%, 40% do not correspond to the exact reciprocal adjustment of 1.25. Common Pitfalls:Subtracting 25% directly from breadth; area constraints demand multiplicative reciprocity, not additive subtraction. Final Answer:20%

Question 2

Lateral area to breadth of a cuboid: The area of four walls (lateral surface) of a cuboid is 400 cm². If its length is 15 cm and height is 8 cm, find the breadth (in cm).

Accepted Answer

Introduction / Context:The “area of 4 walls” of a cuboid is its lateral surface area (LSA), given by 2h(l + b), where l is length, b is breadth, and h is height. Knowing LSA, l, and h allows solving for b directly. Given Data / Assumptions: LSA = 400 cm². l = 15 cm; h = 8 cm. Concept / Approach:Use LSA = 2h(l + b). Rearrange to isolate (l + b) = LSA / (2h). Then b = (LSA / (2h)) − l. Substitute known values carefully to avoid arithmetic errors. Step-by-Step Solution: 2h = 2*8 = 16.l + b = 400 / 16 = 25.b = 25 − 15 = 10 cm. Verification / Alternative check:Compute LSA back: 2*8*(15 + 10) = 16*25 = 400 cm², confirming the value. Why Other Options Are Wrong:12, 20, 24 correspond to mis-divisions (e.g., dividing by 8 instead of 16) or forgetting to subtract l from the sum. Common Pitfalls:Confusing total surface area with lateral surface area; using 2(l + b) without multiplying by h and 2. Final Answer:10

Question 3

Similar triangles – perimeters to sides: ΔABC ~ ΔPQR and Perimeter(ABC) : Perimeter(PQR) = 5 : 9. If PQ = 45 cm, find AB (in cm).

Accepted Answer

Introduction / Context:In similar triangles, all corresponding linear measures (sides, heights, perimeters) are in the same ratio. If perimeter ratio is 5:9, then any corresponding side ratio is also 5:9 in the same order (ABC to PQR). Given Data / Assumptions: ΔABC ~ ΔPQR; Perimeter ratio ABC:PQR = 5:9. Correspondence: AB ↔ PQ (standard order). PQ = 45 cm. Concept / Approach:Side scale factor (ABC relative to PQR) is 5/9. Hence AB = (5/9)*PQ. Substitute PQ = 45 to get AB. Ensure the order of similarity is respected (ABC to PQR). Step-by-Step Solution: Scale = 5/9.AB = (5/9)*45 = 25 cm. Verification / Alternative check:Any other corresponding side would scale the same way; consistency across all sides and perimeters is guaranteed by similarity. Why Other Options Are Wrong:15 and 20 correspond to incorrect scale factors; 16 is unrelated; 25 uniquely follows 5/9 of 45. Common Pitfalls:Inverting the ratio (using 9/5) or mixing the triangle order. Final Answer:25

Question 4

Trapezium midpoints of diagonals: In a trapezium with parallel sides 8 cm and 4 cm, M and N are the midpoints of the two diagonals. Find the length MN (in cm).

Accepted Answer

Introduction / Context:There is a standard trapezium property: the segment joining the midpoints of the diagonals is parallel to the bases, and its length equals half the difference of the parallel sides’ lengths. This is frequently examined in geometry aptitude questions. Given Data / Assumptions: Parallel sides (bases) are 8 cm and 4 cm. M and N are midpoints of the diagonals. Concept / Approach:Apply the property directly: MN = |base1 − base2| / 2. Since the segment is always parallel to the bases, orientation does not affect length—only the base lengths matter. Step-by-Step Solution: Difference of bases = |8 − 4| = 4.MN = 4 / 2 = 2 cm. Verification / Alternative check:Construct a coordinate model (e.g., bases on horizontal lines) and verify via midpoint calculations; the result matches the property. Why Other Options Are Wrong:6 and 12 mistake sum for difference; 1 cm halves the wrong quantity; only 2 cm matches the theorem exactly. Common Pitfalls:Confusing mid-segment between bases (which is the average of bases) with the midpoint-of-diagonals segment (which is half the difference). Final Answer:2 cm

Question 5

Equilateral triangle inscribed in a circle: An equilateral triangle of side 6 cm is inscribed in a circle. Find the radius of the circle.

Accepted Answer

Introduction / Context:The circumradius R of an equilateral triangle with side a is given by R = a/√3. This follows from standard geometry or from R = a/(√3) derived via central angles and chord relations. Given Data / Assumptions:Side a = 6 cm; the triangle is equilateral and inscribed (i.e., the circle is circumcircle). Concept / Approach:Apply the circumradius formula R = a/√3 directly. Keep the radical form in simplest terms to match options. Step-by-Step Solution: R = a/√3 = 6/√3 = (6√3)/3 = 2√3 cm. Verification / Alternative check:Using R = a/(√3) is equivalent to R = (a√3)/3; substituting a = 6 yields the same 2√3 cm. Why Other Options Are Wrong:3√2 and 4√3 are too large; √3 is half the correct value. Common Pitfalls:Confusing inradius r = a√3/6 with circumradius R = a/√3. Final Answer:2√3 cm

Question 6

Right isosceles triangle relation: ABC is an isosceles right triangle with ∠C = 90°. For any point D on hypotenuse AB, evaluate AD² + BD² in terms of CD.

Accepted Answer

Introduction / Context:This identity leverages coordinate geometry in a right isosceles setup (legs equal, right angle at C). Placing the triangle conveniently on axes makes it easy to parametrize any point D on the hypotenuse AB and compute squared distances to A and B, then compare to CD². Given Data / Assumptions: Right angle at C; AC = BC = t. A = (t, 0), B = (0, t), C = (0, 0). D lies on AB, which joins (t, 0) to (0, t). Concept / Approach:Parametrize D as A + u(B − A) = (t(1 − u), tu). Then compute AD², BD², CD². Simplify each in terms of u and t. Compare AD² + BD² with CD² to find the constant factor independent of u. Step-by-Step Solution: AD² = u²|AB|² = u²(2t²).BD² = (1 − u)²|AB|² = (1 − u)²(2t²).Sum: AD² + BD² = 2t²(u² + (1 − u)²) = 4t²(u² − u + 1/2).CD² = (t(1 − u))² + (tu)² = 2t²(u² − u + 1/2).Hence AD² + BD² = 2 * CD². Verification / Alternative check:Pick u = 1/2 (midpoint) to sanity check: then AD = BD, and AD² + BD² clearly doubles CD² by computation. Why Other Options Are Wrong:CD², 3CD², 4CD² contradict the derived identity valid for any D on AB. Common Pitfalls:Placing the triangle incorrectly or assuming D is a special point; the relation holds for any D on AB. Final Answer:2CD2

Question 7

Regular hexagon ABCDEF (side 12 cm): Find the area (in cm²) of triangle ECD formed by vertices E, C, and D.

Accepted Answer

Introduction / Context:A regular hexagon of side s can be embedded in a coordinate plane so that exact lengths and heights are easily read. Choosing standard coordinates shows that triangle CDE has a simple base–height structure tied directly to s. Given Data / Assumptions: Regular hexagon ABCDEF with side s = 12 cm. Standard placement: A(s,0), B(s/2, √3 s/2), C(−s/2, √3 s/2), D(−s,0), E(−s/2, −√3 s/2), F(s/2, −√3 s/2). Concept / Approach:With these coordinates, CE is vertical with length √3 s, and D lies horizontally s/2 to the left of the CE line. Thus the altitude from D to CE equals s/2, and the base CE equals √3 s. Area = (1/2)*base*height = (1/2)*(√3 s)*(s/2) = (√3/4) s². Step-by-Step Solution: Base CE = √3 * s = 12√3.Perpendicular distance from D to CE = s/2 = 6.Area = (1/2) * (12√3) * 6 = 36√3 cm². Verification / Alternative check:Since each central triangle of the hexagon has area (√3/4)s², triangle CDE equals exactly one such central triangle (36√3 for s = 12). Why Other Options Are Wrong:18√3 and 24√3 are too small; 42√3 is too large. Common Pitfalls:Using side^2 directly without identifying the correct base–height pair; mislabelling vertices. Final Answer:36√3

Question 8

Two tangent circles and common tangents: Two circles touch at X. A common external tangent touches them at Y and Z. The tangent through X meets YZ at A, and XA = 16 cm. Find YZ (in cm).

Accepted Answer

Introduction / Context:This configuration uses tangent properties from a common external point. From any external point, the lengths of tangents drawn to a given circle are equal. Here A is a common external point to both circles, with one tangent line AX through X (touching both circles at X) and another common external tangent touching at Y and Z. Given Data / Assumptions: AX is tangent to both circles at X; XA = 16 cm. From point A, AY is tangent to the first circle and AZ to the second. Concept / Approach:For a fixed circle and point A, all tangent lengths from A are equal. Hence, for the first circle, AY = AX = 16, and for the second circle, AZ = AX = 16. The segment YZ along the common external tangent is AY + AZ when measured between tangency points, giving YZ = 16 + 16 = 32 cm. Step-by-Step Solution: AY = AX = 16.AZ = AX = 16.YZ = AY + AZ = 32 cm. Verification / Alternative check:Power-of-a-point consistency at A confirms equal tangent lengths to each circle. Why Other Options Are Wrong:18, 24, 16 undercount one or both tangent lengths. Common Pitfalls:Thinking YZ equals |AY − AZ|; here both are equal to AX. Final Answer:32

Question 9

All sides congruent – which quadrilateral? Choose the figure in which all four sides are congruent (equal in length).

Accepted Answer

Introduction / Context:Among common quadrilaterals, rectangles and general parallelograms do not require all sides to be equal; they require opposite sides equal. A rhombus requires all four sides equal (congruent), while angles need not be right angles. A kite has two pairs of adjacent equal sides, not all four. Given Data / Assumptions:The question asks for a quadrilateral with all sides congruent, independent of angles. Concept / Approach:Recall definitions: rectangle = opposite sides equal, all angles 90°; parallelogram = opposite sides equal and parallel; rhombus = all sides equal; kite = two pairs of adjacent equal sides. Step-by-Step Solution: Only the rhombus satisfies the “all sides equal” requirement by definition. Verification / Alternative check:Square is a special case of a rhombus (also a rectangle), but the generic name asked here is “rhombus”. Why Other Options Are Wrong:Rectangle/parallelogram lack the “all sides equal” constraint; kite has only adjacent pairs equal. Common Pitfalls:Answering “rectangle” because of right angles; equality of all sides is the key, not angles. Final Answer:Rhombus

Question 10

Rope boundary cost from area of circular ground: A circular cricket ground has area 24.64 hectares. Find the cost of laying a rope boundary at Rs 5.40 per metre.

Accepted Answer

Introduction / Context:From the area of a circle we can find its radius, then its circumference, which is the boundary length to be roped. Converting hectares to square metres ensures consistent units for the final cost calculation in rupees per metre. Given Data / Assumptions: Area = 24.64 hectares = 24.64 * 10,000 = 246,400 m². Use π = 22/7 (yields exact integers here). Rate = Rs 5.40 per metre. Concept / Approach:Area A = πr² ⇒ r = √(A/π). With π = 22/7, r becomes exact. Boundary length = circumference = 2πr. Cost = (boundary length)*(rate). Step-by-Step Solution: A/π = 246,400 / (22/7) = 246,400 * 7 / 22 = 11200 * 7 = 78,400.r = √78,400 = 280 m.Circumference = 2πr = 2*(22/7)*280 = 1760 m.Cost = 1760 * 5.40 = Rs 9,504. Verification / Alternative check:Using π ≈ 3.14 yields r ≈ 280 and cost ≈ Rs 9,495; the intended exact option with π = 22/7 is Rs 9,504. Why Other Options Are Wrong:9,600 and 9,876 overcount; 9,802 is a rounding with a different π assumption. Common Pitfalls:Forgetting to convert hectares to m²; mixing radius and diameter in circumference. Final Answer:Rs. 9,504

Question 11

If the radius of a circle is decreased by 50%, find the percentage decrease in its area.

Accepted Answer

Let original radius = R. New radius = 50 100 R = 50 100 R Original area = R 2 and new area = πR 2 3 πR 2 4 * 1 πR 2 * 100 Decrease in area = π R 2 2 = πR 2 4 = 75%

Question 12

The length of a rectangle is 12 cm. What is the width if the area is 72 cm ?

Accepted Answer

A=lb

Question 13

Perimeter of a square or rectangle – Given s = 3 cm (interpreted as square side): Find the perimeter when the given dimension is s = 3 cm, where s denotes the side of a square.

Accepted Answer

Introduction / Context:For a square, perimeter P = 4 * side. The prompt uses s to denote square side. Given Data / Assumptions: s = 3 cm is the side of a square. Concept / Approach:Apply P = 4s directly. Step-by-Step Solution: P = 4 * 3 = 12 cm. Verification / Alternative check:Add all sides: 3 + 3 + 3 + 3 = 12 cm. Why Other Options Are Wrong:13, 14, 15 are not multiples of 3 for a square perimeter. Common Pitfalls:Interpreting s as rectangle side without the other dimension. Final Answer:12

Question 14

Perimeter of a square – Given s = 3 cm: Assuming s is the side of a square, find its perimeter.

Accepted Answer

Introduction / Context:Duplicate of the previous pattern: perimeter of a square is 4 times its side. Given Data / Assumptions:s = 3 cm denotes square side. Concept / Approach:Use P = 4s. Step-by-Step Solution: P = 4 * 3 = 12 cm. Final Answer:12

Question 15

Area–perimeter ratio with side ratio – Find rectangle length: The ratio of area to perimeter of a rectangle is 60:11. Its length : breadth = 6 : 5. Find the rectangle’s length (in units).

Accepted Answer

Introduction / Context:Let L = 6x, B = 5x. Then area = 30x^2, perimeter = 22x. The given ratio equates 30x^2 : 22x to 60 : 11 to solve for x. Given Data / Assumptions: L : B = 6 : 5. Area : Perimeter = 60 : 11. Concept / Approach:30x^2 : 22x = 60 : 11 ⇒ (15x) : 11 = 60 : 11 ⇒ 15x = 60 ⇒ x = 4 ⇒ L = 24. Step-by-Step Solution: L = 6*4 = 24; B = 5*4 = 20.Check: Area = 480; Perimeter = 88 ⇒ 480:88 = 60:11 ✔️ Final Answer:24 units

Question 16

Parallelogram – Base twice height (repeat form): The base of a parallelogram equals twice its height. If its area is 338 sq cm, find the height (in cm).

Accepted Answer

Introduction / Context:Same as earlier: Area = base * height with base = 2h ⇒ 2h^2 = 338. Step-by-Step Solution: 2h^2 = 338 ⇒ h^2 = 169 ⇒ h = 13 cm. Final Answer:13 cm

Question 17

Trapezium with AB ∥ CD and AB = 2CD – Diagonals intersect at O: Find the ratio of areas [∆AOB] : [∆COD].

Accepted Answer

Introduction / Context:In a trapezium with parallel sides AB and CD, the triangles formed by the diagonals at the ends of the parallel sides are similar; their areas are in the square of the ratio of the parallel sides. Concept / Approach:If AB ∥ CD, then [∆AOB] : [∆COD] = (AB/CD)^2. Step-by-Step Solution: Given AB = 2CD ⇒ (AB/CD) = 2.Therefore area ratio = 2^2 = 4 : 1. Final Answer:4:1

Question 18

Orthocentre angle relation – ∠BOC = 100° (O is orthocentre): In triangle ABC, O is the orthocentre. If ∠BOC = 100°, find ∠BAC.

Accepted Answer

Introduction / Context:A key property: in any triangle, the angle ∠BHC at the orthocentre H equals 180° − ∠A. Here, O plays the role of H. Concept / Approach:Thus ∠BOC = 180° − ∠A ⇒ ∠A = 180° − ∠BOC. Step-by-Step Solution: ∠A = 180° − 100° = 80°. Final Answer:80°

Question 19

Circumcircle area of right triangle 6-8-10: Find the area of the circumcircle of a triangle with sides 6 cm, 8 cm, and 10 cm.

Accepted Answer

Introduction / Context:A 6-8-10 triangle is right-angled. For a right triangle, circumradius R = hypotenuse / 2. Area of circumcircle = πR^2. Step-by-Step Solution: Hypotenuse = 10 ⇒ R = 10/2 = 5 cm.Area = π * 5^2 = 25π.Using π = 22/7 ⇒ area = 25 * (22/7) = 550/7. Final Answer:550/7

Question 20

Triangle in a circle radius 14 cm with diameter as one side (Thales): Triangle PQR is inscribed in a circle of radius 14 cm. PQ is a diameter and PR = 10 cm. Find the area of ∆PQR.

Accepted Answer

Introduction / Context:By Thales’ theorem, an angle subtended by a diameter is a right angle. Thus ∆PQR is right-angled at R with hypotenuse PQ = diameter = 28 cm. Step-by-Step Solution: PR = 10; PQ = 28; let QR = x.By Pythagoras: 10^2 + x^2 = 28^2 ⇒ x^2 = 784 − 100 = 684 ⇒ x = √684 = 2√171 = 6√19.Area = (1/2) * PR * QR = 0.5 * 10 * 6√19 = 30√19. Final Answer:30√19

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