Q: A rectangle has diagonal 25 cm and breadth (width) 7 cm.\nFind its perimeter in centimetres.

Introduction / Context: For a rectangle with length L and breadth W, the diagonal d satisfies d^2 = L^2 + W^2. From L, W we obtain the perimeter P = 2(L + W). Here d and W are given, so L follows from Pythagoras. Given Data / Assumptions: d = 25 cm, W = 7 cm. Concept / Approach: Compute L = √(d^2 − W^2) and then P = 2(L + W). Step-by-Step Solution: L = √(25^2 − 7^2) = √(625 − 49) = √576 = 24 cmPerimeter P = 2(L + W) = 2(24 + 7) = 62 cm Verification / Alternative check: Check the triple: 7, 24, 25 is a classic Pythagorean triple. Hence computations are consistent. Why Other Options Are Wrong: 124 and 82 are double-counting or mixing values; 41 is half of 82 and not a valid perimeter given L and W. Final Answer: 62

Q: For a regular hexagon (6 sides), by how many degrees is the measure of each interior angle greater than the corresponding exterior angle?

Introduction / Context: In any regular polygon with n sides, each exterior angle measures 360°/n, and each interior angle measures 180° − 360°/n. Their difference is straightforward to compute. Here n = 6 (regular hexagon). Given Data / Assumptions: n = 6. Concept / Approach: Compute exterior and interior angles explicitly and subtract: interior − exterior. Step-by-Step Solution: Exterior angle = 360°/6 = 60°Interior angle = 180° − 60° = 120°Difference = 120° − 60° = 60° Verification / Alternative check: The sum of one interior and its exterior is always 180°, so difference = 180° − 2*exterior = 180° − 120° = 60°, same result. Why Other Options Are Wrong: 90°, 100°, 108° do not match the regular hexagon values; 108° is often confused with the interior angle of a regular pentagon. Common Pitfalls: Using 360°/n for the interior angle by mistake, or mixing n = 5 vs n = 6 values. Final Answer: 60°

Q: A right circular cone has curved surface area (lateral area) 99 cm² and slant height 9 cm.\nFind its diameter in centimetres.

Introduction / Context: The curved surface area (CSA) of a right circular cone is CSA = π * r * l, where r is the base radius and l is the slant height. Given CSA and l, we can recover r and then the diameter 2r. Given Data / Assumptions: CSA = 99 cm² Slant height l = 9 cm Concept / Approach: Compute r = CSA / (π * l). Using π = 22/7 yields an exact value here. Then diameter D = 2r. Step-by-Step Solution: r = 99 / (π * 9) = 99 / (9 * 22/7) = 99 * 7 / 198 = 3.5 cmDiameter D = 2 * 3.5 = 7 cm Verification / Alternative check: CSA back-check: π * r * l = (22/7) * 3.5 * 9 = (22/7) * 31.5 = 99 cm², consistent. Why Other Options Are Wrong: 3.5 is the radius (not diameter); 10.5 and 14 assume incorrect r; 7 is the correct diameter. Final Answer: 7

Question 1

Equal areas: circle vs square — compare diameter and diagonal:
If a circle and a square have equal areas, what is (diameter of circle / diagonal of square)^2?

Accepted Answer

Introduction / Context: Equate the areas to relate the circle radius to the square side. Then express the requested squared ratio in those terms. Using π = 22/7 gives a clean rational result matching options. Given Data / Assumptions: Area(circle) = Area(square). Let circle radius = r, square side = s. Use π = 22/7 for final numeric comparison to options. Concept / Approach: πr^2 = s^2 ⇒ (diameter/diagonal)^2 = (2r / (s√2))^2 = (4r^2)/(2s^2) = 2r^2/s^2 = 2/π. Step-by-Step Solution: From equality: s^2 = πr^2.(diameter/diagonal)^2 = 2r^2 / s^2 = 2 / π.With π = 22/7, 2/π = 2 / (22/7) = 14/22 = 7/11. Verification / Alternative check: Direct substitution with any r (e.g., r = 1) produces the same 2/π expression; rationalizing with 22/7 yields 7/11. Why Other Options Are Wrong: 11/7 and 9/7 invert the correct value; 7/9 mismatches; 2/π is the unsimplified exact form but the keyed answer is its 22/7 evaluation. Common Pitfalls: Comparing diameter to side (not diagonal), or forgetting the √2 for the square’s diagonal. Final Answer: 7/11

Question 2

Sector angle from area and radius:
The area of a circular sector is 462 sq cm and the circle’s radius is 21 cm. Find the sector’s central angle (in degrees).

Accepted Answer

Introduction / Context: Sector area relates to the full circle area by the angle fraction θ/360. With tidy numbers, the computation becomes clean using π = 22/7. Given Data / Assumptions: Sector area A_sector = 462 cm^2. Radius r = 21 cm ⇒ full circle area A_circle = πr^2. Concept / Approach: A_sector = (θ/360) * πr^2 ⇒ θ = 360 * A_sector / (πr^2). Step-by-Step Solution: πr^2 = (22/7)*441 = 22*63 = 1386.θ = 360 * 462 / 1386 = 360 / 3 = 120°. Verification / Alternative check: Plug back: (120/360)*1386 = 462, consistent. Why Other Options Are Wrong: 90°, 60°, 30°, 150° do not satisfy the exact proportion with area 462. Common Pitfalls: Forgetting to divide by 360 or using diameter instead of radius in the area. Final Answer: 120°

Question 3

Capacity of a canal with isosceles trapezium cross-section:
Cross-section is an isosceles trapezium, 3 m wide at bottom and 5 m at top, depth 2 m, length 110 m. Find the maximum capacity (in cubic metres).

Accepted Answer

Introduction / Context: The canal’s capacity equals cross-sectional area times length. The cross-section is a trapezium with known parallel sides and height (depth). Given Data / Assumptions: Bottom width b1 = 3 m; top width b2 = 5 m. Depth h = 2 m. Length L = 110 m. Concept / Approach: Area(trapezium) = (1/2) * (b1 + b2) * h. Capacity = Area * L. Step-by-Step Solution: Area = (1/2) * (3 + 5) * 2 = 8 m^2.Capacity = 8 * 110 = 880 m^3. Verification / Alternative check: Units: (m^2)*m → m^3, correct. Why Other Options Are Wrong: 1760, 1650, 1056, 990 do not equal 8*110. Common Pitfalls: Using the average width incorrectly or forgetting to multiply by the length. Final Answer: 880 cubic metres

Question 4

In triangle PQR, the perpendicular bisectors OX, OY, and OZ meet at O (the circumcenter).
Given ∠QPR = 65° and ∠PQR = 60°, find the value (in degrees) of ∠QOR + ∠POR.

Accepted Answer

Introduction / Context: This problem uses the circumcenter O of triangle PQR. The perpendicular bisectors of the three sides meet at O. For any arc in a circle, the central angle at O is twice the inscribed angle that subtends the same chord. We will convert given vertex angles into corresponding central angles and then add them appropriately. Given Data / Assumptions: ∠QPR = 65° (angle at P) ∠PQR = 60° (angle at Q) O is the circumcenter; ∠QOR subtends arc QR; ∠POR subtends arc PR. Concept / Approach: In a circumcircle, a central angle equals twice the inscribed angle that subtends the same arc. Therefore, ∠QOR = 2 * ∠QPR and ∠POR = 2 * ∠PQR. Sum them to obtain the target value. Step-by-Step Solution: ∠QOR = 2 * ∠QPR = 2 * 65° = 130°∠POR = 2 * ∠PQR = 2 * 60° = 120°∠QOR + ∠POR = 130° + 120° = 250° Verification / Alternative check: Using triangle angle sum, ∠PRQ = 180° − (65° + 60°) = 55°. The three corresponding central angles would be 130°, 120°, and 110°, which sum to 360°, consistent with a full circle. Our selected pair (130° and 120°) adds to 250° as computed. Why Other Options Are Wrong: 180: Would require ∠QPR + ∠PQR = 90°, which is not the case. 210 or 125: Do not match 2*(65°) + 2*(60°). Common Pitfalls: Confusing the inscribed angle with the central angle (forgetting the factor 2). Using the wrong vertex angle for a given arc. Final Answer: 250

Question 5

In right triangle PQR with ∠PQR = 90°, the leg PQ = 10 cm and the hypotenuse PR = 26 cm.
Find the inradius r (radius of the incircle) in centimetres.

Accepted Answer

Introduction / Context: The inradius r of a right triangle has a convenient formula in terms of side lengths: if the legs are a and b and the hypotenuse is c, then r = (a + b − c) / 2. We will first find the missing leg using the Pythagorean theorem, then apply the inradius formula. Given Data / Assumptions: Right angle at Q, so PR is the hypotenuse. PQ = 10 cm, PR = 26 cm. Concept / Approach: Use Pythagoras to get the other leg QR. For a right triangle, r = (a + b − c)/2, where c is the hypotenuse. Step-by-Step Solution: QR = √(PR^2 − PQ^2) = √(26^2 − 10^2) = √(676 − 100) = √576 = 24 cmLet a = 10, b = 24, c = 26.r = (a + b − c) / 2 = (10 + 24 − 26) / 2 = 8 / 2 = 4 cm Verification / Alternative check: Area K = (1/2) * a * b = (1/2) * 10 * 24 = 120. Also, K = r * s with semiperimeter s = (a + b + c)/2 = (10 + 24 + 26)/2 = 30. Then r = K/s = 120/30 = 4, consistent. Why Other Options Are Wrong: 9, 8, 6: Do not satisfy r = (a + b − c)/2 for the given triple 10–24–26. Common Pitfalls: Using (a + b + c)/2 instead of (a + b − c)/2; mixing up which side is the hypotenuse. Final Answer: 4

Question 6

A single straight line cuts two concentric circles (same center). The chord lengths formed on the inner and outer circles are 4 cm and 16 cm, respectively.
Find the difference between the squares of their radii (in cm²).

Accepted Answer

Introduction / Context: When a line at distance d from the center cuts a circle of radius r, the chord length L satisfies L = 2 * √(r^2 − d^2). With two concentric circles cut by the same line, d is common to both, enabling an easy elimination approach to find r2^2 − r1^2. Given Data / Assumptions: Inner circle chord length L1 = 4 cm. Outer circle chord length L2 = 16 cm. Both chords arise from the same line, so the perpendicular distance d from center to the line is the same. Concept / Approach: Use the relation (L/2)^2 = r^2 − d^2 for each circle, then subtract to eliminate d^2 and directly obtain the difference of squared radii. Step-by-Step Solution: For inner circle: (4/2)^2 = r1^2 − d^2 ⇒ 2^2 = r1^2 − d^2 ⇒ r1^2 − d^2 = 4For outer circle: (16/2)^2 = r2^2 − d^2 ⇒ 8^2 = r2^2 − d^2 ⇒ r2^2 − d^2 = 64Subtract: (r2^2 − d^2) − (r1^2 − d^2) = 64 − 4 ⇒ r2^2 − r1^2 = 60 Verification / Alternative check: Pick any d with 0 < d < r1; for example d = 1 gives r1^2 = 5, r2^2 = 65, difference = 60, confirming independence of d. Why Other Options Are Wrong: 240, 120, 90: Each assumes incorrect squaring or arithmetic; only 60 satisfies the chord–radius relation. Common Pitfalls: Forgetting to halve the chord length before squaring; mixing up r^2 − (L/2)^2 with (r − L/2)^2. Final Answer: 60

Question 7

A rectangle has diagonal 25 cm and breadth (width) 7 cm.
Find its perimeter in centimetres.

Accepted Answer

Introduction / Context: For a rectangle with length L and breadth W, the diagonal d satisfies d^2 = L^2 + W^2. From L, W we obtain the perimeter P = 2(L + W). Here d and W are given, so L follows from Pythagoras. Given Data / Assumptions: d = 25 cm, W = 7 cm. Concept / Approach: Compute L = √(d^2 − W^2) and then P = 2(L + W). Step-by-Step Solution: L = √(25^2 − 7^2) = √(625 − 49) = √576 = 24 cmPerimeter P = 2(L + W) = 2(24 + 7) = 62 cm Verification / Alternative check: Check the triple: 7, 24, 25 is a classic Pythagorean triple. Hence computations are consistent. Why Other Options Are Wrong: 124 and 82 are double-counting or mixing values; 41 is half of 82 and not a valid perimeter given L and W. Final Answer: 62

Question 8

For a regular hexagon (6 sides), by how many degrees is the measure of each interior angle greater than the corresponding exterior angle?

Accepted Answer

Introduction / Context: In any regular polygon with n sides, each exterior angle measures 360°/n, and each interior angle measures 180° − 360°/n. Their difference is straightforward to compute. Here n = 6 (regular hexagon). Given Data / Assumptions: n = 6. Concept / Approach: Compute exterior and interior angles explicitly and subtract: interior − exterior. Step-by-Step Solution: Exterior angle = 360°/6 = 60°Interior angle = 180° − 60° = 120°Difference = 120° − 60° = 60° Verification / Alternative check: The sum of one interior and its exterior is always 180°, so difference = 180° − 2*exterior = 180° − 120° = 60°, same result. Why Other Options Are Wrong: 90°, 100°, 108° do not match the regular hexagon values; 108° is often confused with the interior angle of a regular pentagon. Common Pitfalls: Using 360°/n for the interior angle by mistake, or mixing n = 5 vs n = 6 values. Final Answer: 60°

Question 9

A right circular cone has curved surface area (lateral area) 99 cm² and slant height 9 cm.
Find its diameter in centimetres.

Accepted Answer

Introduction / Context: The curved surface area (CSA) of a right circular cone is CSA = π * r * l, where r is the base radius and l is the slant height. Given CSA and l, we can recover r and then the diameter 2r. Given Data / Assumptions: CSA = 99 cm² Slant height l = 9 cm Concept / Approach: Compute r = CSA / (π * l). Using π = 22/7 yields an exact value here. Then diameter D = 2r. Step-by-Step Solution: r = 99 / (π * 9) = 99 / (9 * 22/7) = 99 * 7 / 198 = 3.5 cmDiameter D = 2 * 3.5 = 7 cm Verification / Alternative check: CSA back-check: π * r * l = (22/7) * 3.5 * 9 = (22/7) * 31.5 = 99 cm², consistent. Why Other Options Are Wrong: 3.5 is the radius (not diameter); 10.5 and 14 assume incorrect r; 7 is the correct diameter. Final Answer: 7

Question 10

What is the measure of an interior angle of a regular decagon?

Accepted Answer

144°

Question 11

The circumference of a circle is 88 cm, find its area?

Accepted Answer

616 sq cm

Question 12

If the radius of a circumscribed circle in an equilateral triangle is 14 cm. Then what will be the radius (in cm) of the inscribed circle?

Accepted Answer

7

Question 13

What is the measure (in degree) of an exterior angle of a regular polygon of 9 sides?

Accepted Answer

40

Question 14

Δ DEF is right angled at E. If m∠D = 30°, what is the length of DE (in cm), if EF = 6√3 cm?

Accepted Answer

18

Question 15

The length and breadth of a rectangle are 24 cm and 7 cm respectively. Calculate its perimeter (in cm).

Accepted Answer

62

Question 16

What is the reflection of the point (-0.5, 6) in the x-axis?

Accepted Answer

(-0.5, -6)

Question 17

The areas of two similar triangles ΔABC and ΔPQR are 36 sq cms and 9 sq cms respectively. If PQ = 4 cm then what is the length of AB (in cm)?

Accepted Answer

8

Question 18

If the perimeter of a semi-circle is 108 cm, then find its radius (in cm).

Accepted Answer

21

Question 19

If the length of the side of an equilateral triangle is 8 cm, what is its area?

Accepted Answer

16√3 sq cm

Question 20

The lengths of the diagonals of a rhombus are 8 cm and 6 cm. The area of rhombus is:

Accepted Answer

24 cm2

Area Questions