Q: Percent error propagation in area of a square:\nIf the measured side of a square is 2% too large, what is the percentage error in the calculated area?

Introduction / Context: Area of a square varies with the square of its side. A small percentage error in the side propagates roughly twice to area, but exactly it follows the square of the multiplicative factor. Given Data / Assumptions: Measured side = 1.02 * true side (2% excess). Area scales with side^2. Concept / Approach: New area factor = (1.02)^2 = 1.0404 → 4.04% excess in area. Hence the percent error in area is +4.04%. Step-by-Step Solution: Let true side = s, true area = s^2.Measured side = 1.02s → computed area = (1.02s)^2 = 1.0404 s^2.Percent error = (1.0404 − 1) * 100% = 4.04%. Verification / Alternative check: Linear approximation gives ~2 * 2% = 4%, close to exact 4.04%, validating the calculation. Why Other Options Are Wrong: 5.04%, 6.04%, 7.04%, 3.96% do not match the exact square of 1.02. Common Pitfalls: Doubling the percentage and ignoring the small quadratic term, or subtracting instead of adding when error is in excess. Final Answer: 4.04%

Q: Sector of a circle (pizza slice):\nA circular pizza of diameter 14 inches is cut into slices each subtending a central angle of 45°. Find the area of one slice (in square inches).

Introduction / Context: The area of a circular sector equals the corresponding fraction of the full circle’s area, with fraction = (central angle)/360°. Given Data / Assumptions: Diameter = 14 in → radius r = 7 in. Central angle = 45°. Concept / Approach: Full area = πr^2 = 49π. Sector area = (45/360) * 49π = (1/8) * 49π. Step-by-Step Solution: Area(slice) = (1/8) * 49π = 49π / 8.Using π ≈ 3.1416 → 49π/8 ≈ 153.938/8 ≈ 19.242 ≈ 19.25. Verification / Alternative check: Eight equal slices would total back to 49π (~153.94), confirming each at ~19.24 in^2. Why Other Options Are Wrong: 16.25, 18.25, 17.25, 20.25 are not equal to 49π/8 when evaluated. Common Pitfalls: Using diameter instead of radius in πr^2 or using 45/180 instead of 45/360 for the sector fraction. Final Answer: 19.25

Q: Circumradius of a right triangle with sides 9 cm, 12 cm, 15 cm:\nFind the circumradius (radius of the circumscribed circle) of the triangle.

Introduction / Context: For a right triangle, the circumradius equals half the hypotenuse. Recognizing the Pythagorean triple (9, 12, 15) streamlines the answer. Given Data / Assumptions: Sides: 9, 12, 15 cm (hypotenuse = 15 cm). Concept / Approach: In a right triangle, the circumcenter is the midpoint of the hypotenuse; hence R = hypotenuse / 2. Step-by-Step Solution: R = 15 / 2 = 7.5 cm. Verification / Alternative check: General formula R = (abc) / (4Δ). With a = 9, b = 12, c = 15 and Δ = (1/2)*9*12 = 54: R = (9*12*15)/(4*54) = 1620/216 = 7.5 cm. Why Other Options Are Wrong: 6 cm, 4.5 cm, 8 cm, 5 cm do not equal half the hypotenuse. Common Pitfalls: Using inradius formula or confusing circumradius with the radius of the incircle. Final Answer: 7.5 cm

Q: Open-top cuboid from a 7 cm square sheet with corner squares of area 0.25 cm^2 removed:\nFind the volume (in cubic centimetres) of the resulting open-top box after folding.

Introduction / Context: Cutting equal squares from each corner of a sheet and folding up the flaps creates an open-top cuboid. Its height equals the side of the removed square; base dimensions shrink by twice that amount. Given Data / Assumptions: Square sheet: side 7 cm. Each cut-out square has area 0.25 cm^2 → side = √0.25 = 0.5 cm. Concept / Approach: Final box: length = 7 − 2*0.5 = 6 cm; breadth = 7 − 2*0.5 = 6 cm; height = 0.5 cm. Volume = l * b * h. Step-by-Step Solution: l = 6, b = 6, h = 0.5.Volume = 6 * 6 * 0.5 = 18 cm^3. Verification / Alternative check: Unit consistency: base area 36 cm^2 times height 0.5 cm gives 18 cm^3. Why Other Options Are Wrong: 21, 16, 20, 24 do not match 6*6*0.5. Common Pitfalls: Using the area 0.25 as the side directly, or forgetting to subtract twice the cut size from each dimension. Final Answer: 18

Q: Cyclic quadrilateral with equal opposite angles (rectangle case):\nA quadrilateral inscribed in a circle has equal opposite angles (hence it is a rectangle). If its adjacent sides are 6 cm and 8 cm, find the area of the circumcircle.

Introduction / Context: In a cyclic quadrilateral, equal opposite angles imply each is 90°, i.e., the figure is a rectangle. The circle circumscribing a rectangle has diameter equal to the rectangle’s diagonal. Given Data / Assumptions: Rectangle sides: 6 cm and 8 cm. Diagonal d = √(6^2 + 8^2) = 10 cm. Circumcircle diameter = diagonal; hence radius r = d/2 = 5 cm. Concept / Approach: Area of the circle = πr^2 = π * 25 = 25π sq cm. Step-by-Step Solution: Compute diagonal by Pythagoras: 10 cm.Radius r = 5 cm → Area = 25π sq cm. Verification / Alternative check: The rectangle’s vertices lie on a circle with diameter equal to the diagonal; this is a standard property of right angles in semicircles. Why Other Options Are Wrong: 64π, 36π, 49π, 50π do not equal π * 5^2. Common Pitfalls: Taking side as diameter or forgetting that the rectangle’s circumcircle radius depends on the diagonal, not any single side. Final Answer: 25π sq cm

Q: Parallel segment in a triangle and area of the lower trapezium:\nPoints P on AB and Q on AC satisfy PQ ∥ BC. If AP:PB = 2:3 and area(ΔAPQ) = 8 sq cm, find the area of trapezium PQCB.

Introduction / Context: A segment parallel to the base of a triangle creates a smaller similar triangle at the top. Areas scale with the square of the linear ratio. We use the given small-triangle area to recover the full triangle’s area, then subtract. Given Data / Assumptions: P on AB, Q on AC, PQ ∥ BC. AP:PB = 2:3 → AP/AB = 2/5. Area(ΔAPQ) = 8 sq cm. Concept / Approach: ΔAPQ ~ ΔABC with linear ratio 2/5. Therefore, area(APQ) : area(ABC) = (2/5)^2 = 4/25. Step-by-Step Solution: area(ABC) = area(APQ) * (25/4) = 8 * (25/4) = 50 sq cm.Area(trapezium PQCB) = area(ABC) − area(APQ) = 50 − 8 = 42 sq cm. Verification / Alternative check: Proportional reasoning aligns with standard results from the Basic Proportionality Theorem / similarity of triangles. Why Other Options Are Wrong: 50 is the whole triangle; 18 and 14 are too small; 32 does not match the derived difference. Common Pitfalls: Using linear ratio directly on area (forgetting the square), or misreading AP:PB as AP:AB. Final Answer: 42 sq cm

Question 1

Percentage decrease in area after shrinkage in both dimensions:
A towel loses 20% of its length and 10% of its breadth on bleaching. By what percentage does its area decrease?

Accepted Answer

Introduction / Context: Area of a rectangle scales multiplicatively with changes in both dimensions. Here, each side shrinks by a different percentage, so combine the factors to find the net change in area. Given Data / Assumptions: Length factor = 1 − 0.20 = 0.8. Breadth factor = 1 − 0.10 = 0.9. Concept / Approach: New area factor = 0.8 * 0.9 = 0.72 → a 28% decrease relative to the original area (since 1 − 0.72 = 0.28 = 28%). Step-by-Step Solution: Original area A.New area = 0.8A * 0.9 = 0.72A.Percentage decrease = (A − 0.72A) / A * 100 = 28%. Verification / Alternative check: Try a numeric example: let original be 100 unit^2 → new 72 unit^2 → fall is 28 units → 28%. Why Other Options Are Wrong: 18%, 38%, 48%, 30% do not equal the product-based reduction combining −20% and −10% simultaneously. Common Pitfalls: Adding percentage decreases (20 + 10 = 30) is incorrect because area is two-dimensional, requiring multiplicative combination of the factors. Final Answer: 28%

Question 2

Three-side fencing of a rectangular field (one side open 20 ft):
A rectangular field is fenced on three sides, leaving one 20-foot side uncovered. If the area is 680 sq ft, how many feet of fencing are required?

Accepted Answer

Introduction / Context: We have a rectangle with one known side (20 ft) left unfenced. Using the area, deduce the other side, then total the lengths to be fenced on the remaining three sides. Given Data / Assumptions: One side (width) = 20 ft is open (unfenced). Area = 680 sq ft. Rectangle sides: let length = L, breadth = 20 ft. Concept / Approach: From area, L = Area / breadth. Fenced sides are the two breadths plus the other length: total = 2L + 20. Step-by-Step Solution: L = 680 / 20 = 34 ft.Fencing required = 2 * 34 + 20 = 68 + 20 = 88 ft. Verification / Alternative check: Full perimeter would be 2(L + 20) = 2(34 + 20) = 108 ft; removing the open side (20 ft) yields 108 − 20 = 88 ft, same result. Why Other Options Are Wrong: 22, 44, 66 are too small; 108 corresponds to full perimeter, not three sides only. Common Pitfalls: Accidentally fencing all four sides or subtracting the wrong side length from the perimeter. Final Answer: 88

Question 3

Altitude of an equilateral triangle with side 33 cm (repaired options):
Find the exact length of the altitude (height) of an equilateral triangle whose side is 33 cm.

Accepted Answer

Introduction / Context: (Recovery–First applied) The original options (4.5, 3.5, 2.5, 6.5) were inconsistent with an equilateral triangle of side 33 cm. The correct altitude formula is well-known, so we minimally repair the options to include the exact expression and a sensible approximation. Given Data / Assumptions: Equilateral triangle side s = 33 cm. Altitude h = (√3/2) * s. Concept / Approach: Altitude in an equilateral triangle splits it into two 30–60–90 right triangles; the height equals (√3/2) times the side length. Step-by-Step Solution: h = (√3/2) * 33 = 33√3 / 2 cm.Numerically: √3 ≈ 1.732 → h ≈ 33 * 0.866 = 28.578 cm ≈ 28.6 cm. Verification / Alternative check: Use Pythagoras in the half-triangle: (33/2)^2 + h^2 = 33^2 → h^2 = 33^2 − (16.5)^2 = (33 − 16.5)(33 + 16.5) = 16.5 * 49.5 = 816.75 → h ≈ 28.6 cm, consistent with 33√3/2. Why Other Options Are Wrong: 24.0 cm, 19.1 cm, 16.5 cm do not match the exact altitude formula; 28.6 cm is the rounded value, while 33√3 / 2 cm is exact. Common Pitfalls: Confusing altitude with medians in non-equilateral triangles; here all coincide and the altitude formula is precise. Final Answer: 33√3 / 2 cm

Question 4

Perimeter comparison for equal areas:
A square and a (non-square) rectangle have equal areas, with perimeters p and q respectively. Determine the correct relation between p and q.

Accepted Answer

Introduction / Context: (Recovery–First applied) The square minimizes perimeter for a given area among all rectangles. For equal areas, a non-square rectangle must have strictly larger perimeter than the square. We repair options to reflect the correct inequality, choosing the strict form when the rectangle is not a square. Given Data / Assumptions: Square area = Rectangle area. Rectangle is not a square (otherwise perimeters would be equal). Concept / Approach: For fixed area A, by AM–GM (or calculus), the rectangle with minimum perimeter is the square. Thus p (square) is minimal, and any non-square rectangle has q > p. Step-by-Step Solution (idea): Let the rectangle have sides x and A/x. Perimeter = 2(x + A/x) ≥ 4√A with equality at x = √A (i.e., square).Hence for a non-square rectangle, 2(x + A/x) > 4√A, so q > p. Verification / Alternative check: Pick A = 100. Square 10×10 has p = 40. Rectangle 5×20 has q = 50 > 40. Why Other Options Are Wrong: p > q contradicts the extremal property; p = q holds only if the rectangle is also a square; “none” is unnecessary since the strict inequality is known. Common Pitfalls: Assuming equal areas imply equal perimeters; perimeter depends on side distribution, not area alone. Final Answer: p < q

Question 5

Percent error propagation in area of a square:
If the measured side of a square is 2% too large, what is the percentage error in the calculated area?

Accepted Answer

Introduction / Context: Area of a square varies with the square of its side. A small percentage error in the side propagates roughly twice to area, but exactly it follows the square of the multiplicative factor. Given Data / Assumptions: Measured side = 1.02 * true side (2% excess). Area scales with side^2. Concept / Approach: New area factor = (1.02)^2 = 1.0404 → 4.04% excess in area. Hence the percent error in area is +4.04%. Step-by-Step Solution: Let true side = s, true area = s^2.Measured side = 1.02s → computed area = (1.02s)^2 = 1.0404 s^2.Percent error = (1.0404 − 1) * 100% = 4.04%. Verification / Alternative check: Linear approximation gives ~2 * 2% = 4%, close to exact 4.04%, validating the calculation. Why Other Options Are Wrong: 5.04%, 6.04%, 7.04%, 3.96% do not match the exact square of 1.02. Common Pitfalls: Doubling the percentage and ignoring the small quadratic term, or subtracting instead of adding when error is in excess. Final Answer: 4.04%

Question 6

Sector of a circle (pizza slice):
A circular pizza of diameter 14 inches is cut into slices each subtending a central angle of 45°. Find the area of one slice (in square inches).

Accepted Answer

Introduction / Context: The area of a circular sector equals the corresponding fraction of the full circle’s area, with fraction = (central angle)/360°. Given Data / Assumptions: Diameter = 14 in → radius r = 7 in. Central angle = 45°. Concept / Approach: Full area = πr^2 = 49π. Sector area = (45/360) * 49π = (1/8) * 49π. Step-by-Step Solution: Area(slice) = (1/8) * 49π = 49π / 8.Using π ≈ 3.1416 → 49π/8 ≈ 153.938/8 ≈ 19.242 ≈ 19.25. Verification / Alternative check: Eight equal slices would total back to 49π (~153.94), confirming each at ~19.24 in^2. Why Other Options Are Wrong: 16.25, 18.25, 17.25, 20.25 are not equal to 49π/8 when evaluated. Common Pitfalls: Using diameter instead of radius in πr^2 or using 45/180 instead of 45/360 for the sector fraction. Final Answer: 19.25

Question 7

Circumradius of a right triangle with sides 9 cm, 12 cm, 15 cm:
Find the circumradius (radius of the circumscribed circle) of the triangle.

Accepted Answer

Introduction / Context: For a right triangle, the circumradius equals half the hypotenuse. Recognizing the Pythagorean triple (9, 12, 15) streamlines the answer. Given Data / Assumptions: Sides: 9, 12, 15 cm (hypotenuse = 15 cm). Concept / Approach: In a right triangle, the circumcenter is the midpoint of the hypotenuse; hence R = hypotenuse / 2. Step-by-Step Solution: R = 15 / 2 = 7.5 cm. Verification / Alternative check: General formula R = (abc) / (4Δ). With a = 9, b = 12, c = 15 and Δ = (1/2)*9*12 = 54: R = (9*12*15)/(4*54) = 1620/216 = 7.5 cm. Why Other Options Are Wrong: 6 cm, 4.5 cm, 8 cm, 5 cm do not equal half the hypotenuse. Common Pitfalls: Using inradius formula or confusing circumradius with the radius of the incircle. Final Answer: 7.5 cm

Question 8

Open-top cuboid from a 7 cm square sheet with corner squares of area 0.25 cm^2 removed:
Find the volume (in cubic centimetres) of the resulting open-top box after folding.

Accepted Answer

Introduction / Context: Cutting equal squares from each corner of a sheet and folding up the flaps creates an open-top cuboid. Its height equals the side of the removed square; base dimensions shrink by twice that amount. Given Data / Assumptions: Square sheet: side 7 cm. Each cut-out square has area 0.25 cm^2 → side = √0.25 = 0.5 cm. Concept / Approach: Final box: length = 7 − 2*0.5 = 6 cm; breadth = 7 − 2*0.5 = 6 cm; height = 0.5 cm. Volume = l * b * h. Step-by-Step Solution: l = 6, b = 6, h = 0.5.Volume = 6 * 6 * 0.5 = 18 cm^3. Verification / Alternative check: Unit consistency: base area 36 cm^2 times height 0.5 cm gives 18 cm^3. Why Other Options Are Wrong: 21, 16, 20, 24 do not match 6*6*0.5. Common Pitfalls: Using the area 0.25 as the side directly, or forgetting to subtract twice the cut size from each dimension. Final Answer: 18

Question 9

Cyclic quadrilateral with equal opposite angles (rectangle case):
A quadrilateral inscribed in a circle has equal opposite angles (hence it is a rectangle). If its adjacent sides are 6 cm and 8 cm, find the area of the circumcircle.

Accepted Answer

Introduction / Context: In a cyclic quadrilateral, equal opposite angles imply each is 90°, i.e., the figure is a rectangle. The circle circumscribing a rectangle has diameter equal to the rectangle’s diagonal. Given Data / Assumptions: Rectangle sides: 6 cm and 8 cm. Diagonal d = √(6^2 + 8^2) = 10 cm. Circumcircle diameter = diagonal; hence radius r = d/2 = 5 cm. Concept / Approach: Area of the circle = πr^2 = π * 25 = 25π sq cm. Step-by-Step Solution: Compute diagonal by Pythagoras: 10 cm.Radius r = 5 cm → Area = 25π sq cm. Verification / Alternative check: The rectangle’s vertices lie on a circle with diameter equal to the diagonal; this is a standard property of right angles in semicircles. Why Other Options Are Wrong: 64π, 36π, 49π, 50π do not equal π * 5^2. Common Pitfalls: Taking side as diameter or forgetting that the rectangle’s circumcircle radius depends on the diagonal, not any single side. Final Answer: 25π sq cm

Question 10

Parallel segment in a triangle and area of the lower trapezium:
Points P on AB and Q on AC satisfy PQ ∥ BC. If AP:PB = 2:3 and area(ΔAPQ) = 8 sq cm, find the area of trapezium PQCB.

Accepted Answer

Introduction / Context: A segment parallel to the base of a triangle creates a smaller similar triangle at the top. Areas scale with the square of the linear ratio. We use the given small-triangle area to recover the full triangle’s area, then subtract. Given Data / Assumptions: P on AB, Q on AC, PQ ∥ BC. AP:PB = 2:3 → AP/AB = 2/5. Area(ΔAPQ) = 8 sq cm. Concept / Approach: ΔAPQ ~ ΔABC with linear ratio 2/5. Therefore, area(APQ) : area(ABC) = (2/5)^2 = 4/25. Step-by-Step Solution: area(ABC) = area(APQ) * (25/4) = 8 * (25/4) = 50 sq cm.Area(trapezium PQCB) = area(ABC) − area(APQ) = 50 − 8 = 42 sq cm. Verification / Alternative check: Proportional reasoning aligns with standard results from the Basic Proportionality Theorem / similarity of triangles. Why Other Options Are Wrong: 50 is the whole triangle; 18 and 14 are too small; 32 does not match the derived difference. Common Pitfalls: Using linear ratio directly on area (forgetting the square), or misreading AP:PB as AP:AB. Final Answer: 42 sq cm

Question 11

Melting a solid sphere (radius 9 cm) into a thin wire (radius 1 mm):
Find the length of the wire (in metres). Take π = 22/7.

Accepted Answer

Introduction / Context: Volume is conserved during melting and reshaping. Equate the sphere’s volume to the cylinder (wire) volume and solve for the wire length. Unit consistency is essential (mm vs cm). Given Data / Assumptions: Sphere radius R = 9 cm. Wire radius r = 1 mm = 0.1 cm. π = 22/7. Concept / Approach: Sphere volume V_s = (4/3)πR^3. Wire (cylinder) volume V_c = πr^2L. Set V_s = V_c and solve L. Step-by-Step Solution: V_s = (4/3)π * 9^3 = (4/3)π * 729 = 972π cm^3.V_c = π * (0.1)^2 * L = 0.01πL.Equate: 972π = 0.01πL → L = 972 / 0.01 = 97200 cm = 972 m. Verification / Alternative check: Cancel π on both sides; convert cm to m at the end (divide by 100). Why Other Options Are Wrong: 1166.4 m, 1458 m, 777.6 m, 648 m arise from arithmetic or unit-conversion errors; 972 m matches exact conservation. Common Pitfalls: Forgetting to convert 1 mm to cm (0.1 cm), or neglecting to square the wire radius. Final Answer: 972 metres

Question 12

Cost of leveling a triangular plot at 20 paise per m^2:
Find the total cost to level a triangle with sides 72 m, 30 m, and 78 m.

Accepted Answer

Introduction / Context: Use Heron’s formula to compute the area from three sides, then multiply by the given rate (20 paise = Rs 0.20) to obtain cost. Given Data / Assumptions: Sides: 72 m, 30 m, 78 m. Rate: 20 paise per m^2 = Rs 0.20 per m^2. Concept / Approach: Heron’s formula: with semiperimeter s = (a + b + c)/2, area Δ = √[s(s − a)(s − b)(s − c)]. Cost = area * 0.20. Step-by-Step Solution: s = (72 + 30 + 78)/2 = 180/2 = 90.Δ = √[90 * (90 − 72) * (90 − 30) * (90 − 78)] = √(90 * 18 * 60 * 12).Compute: 90*18 = 1620; 60*12 = 720; product = 1,166,400; √ = 1080 m^2.Cost = 1080 * 0.20 = Rs 216. Verification / Alternative check: Right/acute check: sides satisfy triangle inequality; numerical square-root confirms an integer area 1080, making the rate multiplication exact. Why Other Options Are Wrong: Rs. 220, Rs. 200, Rs. 210, Rs. 224 differ from the exact multiplication of 1080 by Rs 0.20. Common Pitfalls: Treating 20 paise as Rs 20 or mixing units; or arithmetic mistakes in the product under the radical. Final Answer: Rs. 216

Question 13

The three interior angles of a triangle are in the ratio 1 : 1 : 4.
If the perimeter of the triangle is k times its largest side, what is the exact value of k?

Accepted Answer

Introduction / Context: This problem mixes triangle angle ratios with side-perimeter relations. By converting the ratio of angles into concrete angle measures and then using the Law of Sines (or well-known values for special angles), we can express each side in terms of the largest side and hence obtain the perimeter-to-largest-side factor k. Given Data / Assumptions: Angle ratio 1 : 1 : 4 implies angles 30°, 30°, and 120° (sum 180°). Largest angle is 120°, so the side opposite it is the largest side L. Perimeter P = k * L; we seek k. Concept / Approach: By the Law of Sines, side lengths are proportional to the sines of their opposite angles. Thus the two smaller sides are proportional to sin 30° = 1/2, while the largest side is proportional to sin 120° = sin 60° = √3/2. Therefore the ratio (smaller side):(largest side) = (1/2):(√3/2) = 1/√3. Step-by-Step Solution: Let the largest side be L.Each smaller side = L * (1/√3).Perimeter P = L + 2 * (L/√3) = L * (1 + 2/√3).Hence k = P/L = 1 + 2/√3. Verification / Alternative check: Assign a scale: let L correspond to √3/2. Then smaller sides are each 1/2. P = √3/2 + 1/2 + 1/2 = √3/2 + 1. Divide by L = √3/2 to get (√3/2 + 1)/(√3/2) = 1 + 2/√3, confirming the expression. Why Other Options Are Wrong: 1 - 2/√3: Incorrect sign; perimeter must exceed L. 2 + 2/√3: Overcounts; would imply each smaller side equals L. 2: Would require sum of smaller sides = L, which is false here. Common Pitfalls: Using sin 120° = −sin 60° (sign confusion); lengths use absolute values of sine. Assuming the triangle is acute; it is obtuse (120°). Final Answer: 1 + 2/√3

Question 14

What is the exact area (in square centimetres) of an equilateral triangle having side length 6 cm?

Accepted Answer

Introduction / Context: The area of an equilateral triangle with side a is a direct formula question, frequently tested for speed and accuracy. Remembering or deriving the formula quickly is key. Given Data / Assumptions: Side length a = 6 cm. Use the standard equilateral triangle area formula. Concept / Approach: The exact area is A = (√3/4) * a^2. This follows from splitting the triangle into two 30-60-90 right triangles and using the altitude a√3/2. Step-by-Step Solution: A = (√3/4) * a^2= (√3/4) * 36= 9√3 square centimetres Verification / Alternative check: Altitude h = a√3/2 = 6√3/2 = 3√3. Then A = (1/2) * base * height = (1/2) * 6 * 3√3 = 9√3, consistent. Why Other Options Are Wrong: 36√3: Multiplies by √3 without the 1/4 factor. 9 or 36: Misses the √3 factor entirely. Common Pitfalls: Dropping the √3 factor or mis-squaring the side (6^2 = 36). Final Answer: 9√3

Question 15

In a right triangle, the side opposite the right angle (hypotenuse) measures 9√3 cm.
What is the length of the side opposite the 30° angle?

Accepted Answer

Introduction / Context: In a 30°–60°–90° right triangle, side lengths have a fixed ratio relative to the hypotenuse. Recognizing this pattern avoids trigonometry and speeds computation. Given Data / Assumptions: Hypotenuse c = 9√3 cm. We need the side opposite the 30° angle. Concept / Approach: In a 30°–60°–90° triangle: (short leg):(long leg):(hypotenuse) = 1 : √3 : 2. The side opposite 30° is the short leg, equal to hypotenuse/2. Step-by-Step Solution: Short leg = c / 2 = (9√3) / 2 cm Verification / Alternative check: If short leg = (9√3)/2, then long leg = short leg * √3 = (9√3/2)*√3 = (9*3)/2 = 27/2, and hypotenuse = 2 * short leg = 9√3, consistent with the given value. Why Other Options Are Wrong: 9 cm, 6 cm: Treat √3 incorrectly or ignore the factor 1/2. 3√3 cm: This would be hypotenuse/3, not /2. Common Pitfalls: Confusing which leg is opposite 30°; it is always the shorter leg. Final Answer: (9√3)/2 cm

Question 16

An equilateral triangle has area 9√3 square centimetres.
Find the exact height (altitude) of the triangle.

Accepted Answer

Introduction / Context: Converting between area and altitude in an equilateral triangle relies on two linked formulas: A = (√3/4)a^2 and h = (√3/2)a. Use one to find side length, then compute the altitude. Given Data / Assumptions: Area A = 9√3 cm^2. Concept / Approach: First find the side a from A = (√3/4)a^2. Then use h = (√3/2)a to get the altitude. Step-by-Step Solution: 9√3 = (√3/4) * a^2 ⇒ a^2 = 36 ⇒ a = 6 cmh = (√3/2) * a = (√3/2) * 6 = 3√3 cm Verification / Alternative check: Compute A from a=6: (√3/4)*36 = 9√3 cm^2; consistent. Why Other Options Are Wrong: 6 cm, 9 cm: These are side-length-like numbers, not the altitude for A = 9√3. 6√3 cm: Overshoots by a factor of 2. Common Pitfalls: Mixing up the area and altitude formulas or skipping the square on a. Final Answer: 3√3 cm

Question 17

A circle has circumference 26.4 cm.
Find its exact area (in square centimetres).

Accepted Answer

Introduction / Context: Relating circumference to area requires computing the radius from C = 2 * pi * r, then substituting into A = pi * r^2. Using the common fraction pi = 22/7 keeps arithmetic exact here. Given Data / Assumptions: C = 26.4 cm. Use pi ≈ 22/7 for clean ratios. Concept / Approach: Find r from C = 2 * pi * r, then compute A = pi * r^2. Step-by-Step Solution: r = C / (2 * pi) = 26.4 / (2 * 22/7) = 26.4 * 7 / 44 = 4.2 cmA = pi * r^2 = (22/7) * (4.2)^2 = (22/7) * 17.64 = 22 * 2.52 = 55.44 cm^2 Verification / Alternative check: Back-compute: With r = 4.2, C = 2 * pi * r = 2 * (22/7) * 4.2 = 26.4 cm; consistent. Why Other Options Are Wrong: They correspond to incorrect r values due to arithmetic or pi substitution errors. Common Pitfalls: Using pi ≈ 3 or 3.14 without keeping consistency across both computations, leading to rounding drift. Final Answer: 55.44

Question 18

A circle has circumference 22 cm.
Compute its area in square centimetres.

Accepted Answer

Introduction / Context: Again convert circumference to radius using C = 2 * pi * r, then area A = pi * r^2. This case is tailored to the exact fraction pi = 22/7. Given Data / Assumptions: C = 22 cm; take pi = 22/7. Concept / Approach: Isolate r, then square to compute area. Step-by-Step Solution: r = 22 / (2 * 22/7) = 22 * 7 / 44 = 3.5 cmA = pi * r^2 = (22/7) * (3.5)^2 = (22/7) * 12.25 = 22 * 1.75 = 38.5 cm^2 Verification / Alternative check: C back = 2*pi*r = 2*(22/7)*3.5 = 22 cm. Why Other Options Are Wrong: They come from mis-squaring 3.5 or misplacing the 1/7 factor. Final Answer: 38.5

Question 19

Find the exact area (in square units) of triangle AOB with vertices A(0, 8), O(0, 0), and B(5, 0).

Accepted Answer

Introduction / Context: With right-angled triangles aligned to the axes, area is quickly found by half the product of legs. Here AO is vertical and OB is horizontal, making a right angle at O. Given Data / Assumptions: AO length = 8 units (along y-axis). OB length = 5 units (along x-axis). Concept / Approach: Area formula: A = (1/2) * base * height for right triangles. Step-by-Step Solution: A = (1/2) * 5 * 8 = 20 square units Verification / Alternative check: Shoelace formula with points (0,0), (5,0), (0,8) also gives 20. Why Other Options Are Wrong: They are arithmetic distractors unrelated to (1/2)*5*8. Final Answer: 20 sq. units

Question 20

In triangle ABC, medians AD, BE, and CF intersect at centroid G.
If area(ΔABC) = 36 cm^2, find area(ΔCGE).

Accepted Answer

Introduction / Context: Medians of a triangle intersect at the centroid, partitioning the triangle into six smaller triangles of equal area. Recognizing this partition lets us read off areas of sub-triangles like ΔCGE directly. Given Data / Assumptions: AD, BE, CF are medians; E is midpoint of AC. G is the centroid (intersection of the three medians). Area(ΔABC) = 36 cm^2. Concept / Approach: Centroid divides the triangle into six equal-area small triangles formed by the three medians. Thus each small triangle has area (total area)/6. Step-by-Step Solution: Each small sub-triangle area = 36 / 6 = 6 cm^2ΔCGE is one of these six equal small triangles.Therefore area(ΔCGE) = 6 cm^2 Verification / Alternative check: Draw medians; around each vertex, two small triangles meet; all six are congruent in area by symmetry and midpoint properties. Why Other Options Are Wrong: 9, 12, 18: These assume thirds or halves, not sixths. Common Pitfalls: Forgetting that three medians create 6, not 4, parts; confusing centroid 2:1 segment property with area partition. Final Answer: 6 sq cm

Area Questions