Mensuration Questions and Answers

Answer: C

Convert the radian measure to degrees.

\[ 1.9 \text{ rad} = 1.9 \times 57.2958 = 108.86^{\circ} \]

\[ \sin 108.86^{\circ} = \sin(180^{\circ} - 108.86^{\circ}) = \sin 71.14^{\circ} = 0.946 \]

The angle is in the second quadrant, where the sine is positive and still close to its maximum - so a value near 1 is expected.

Answer: D

A chord divides a circle into two regions called segments. The larger, on the side of the centre, is the major segment; the smaller is the minor segment.

A quadrant is bounded by two radii, not a chord, so it is not the answer here.

Answer: C

Convert the radian measure to degrees first.

\[ \frac{\pi}{7} = \frac{180^{\circ}}{7} = 25.71^{\circ} \]

\[ \tan 25.71^{\circ} = 0.481 \]

Below \( 45^{\circ} \) the tangent is less than 1, which rules out any option above it.

Answer: B

Walking entirely around a semicircular garden means covering the curved arc and the straight diameter.

\[ \text{perimeter} = \pi r + 2r = r(\pi + 2) \]

With \( \pi = \tfrac{22}{7} \), so \( \pi + 2 = \tfrac{36}{7} \):

\[ r \times \frac{36}{7} = 324 \quad\Longrightarrow\quad r = \frac{324 \times 7}{36} = 63 \text{ m} \]

\[ A = \frac{1}{2}\pi r^{2} = \frac{1}{2} \times \frac{22}{7} \times 3969 = 6237 \text{ m}^{2} \]

Omitting the straight edge would make the perimeter \( \pi r \) alone and give a radius of 103 m - the diameter is a real part of the boundary he walks.

Answer: D

Volume of the prism = 1056 cm³
WKT, Volume of prism = Base x Height
Base area = Area of trapezium
Area of trapezium = (1/2)(a + b)*h = (1/2)[8 + 14]*8
= (1/2)*22* 8 = 88 cm²

Volume of prism = (Area of trapezium) x Height
1056 = 88 * height
Height of prism = 12 cm.

Answer: C

Let the radii of the smaller and the larger circles be s m and l m respectively. 

2∏s = 264 and 2∏l = 352 

s = 264/2∏ and l = 352/2∏ 

Difference between the areas = ∏l² - ∏s² 

= ∏{176²/∏² - 132²/∏²} 

= 176²/∏ - 132²/∏ 

= (176 - 132)(176 + 132)/∏ 

= (44)(308)/(22/7) = (2)(308)(7) = 4312 sq m

Answer: B

Convert the volume to cubic centimetres to match the base area.

\[ 2.6 \text{ m}^{3} = 2.6 \times 1{,}000{,}000 = 2{,}600{,}000 \text{ cm}^{3} \]

\[ \text{depth} = \frac{2{,}600{,}000}{6500} = 400 \text{ cm} = 4 \text{ m} \]

A cubic metre is a million cubic centimetres, not ten thousand - the factor that decides the answer.

Answer: D

Length of the first carpet = (1.44)(6) = 8.64 cm

Area of the second carpet = 8.64(1 + 40/100) 6 (1 + 25/100)

= 51.84(1.4)(5/4) sq m = (12.96)(7) sq m

Cost of the second carpet = (45)(12.96 * 7) = 315 (13 - 0.04) = 4095 - 12.6 = Rs. 4082.40

Answer: C

\[ \text{arc} = \frac{\theta}{360} \times 2\pi r \]

\[ = \frac{140}{360} \times 2 \times \frac{22}{7} \times 21 \]

\[ = \frac{7}{18} \times 132 = \frac{924}{18} = 51.33 \approx 51.31 \text{ cm} \]

The whole circumference is 132 cm, and \( 140^{\circ} \) is a little under four tenths of a turn - a quick check on the size of the answer.

Answer: D

\[ \pi \text{ radians} = 180^{\circ} \]

\[ \frac{\pi}{3} = \frac{180^{\circ}}{3} = 60^{\circ} \]

The useful landmarks are \( \tfrac{\pi}{6} = 30^{\circ} \), \( \tfrac{\pi}{4} = 45^{\circ} \), \( \tfrac{\pi}{3} = 60^{\circ} \) and \( \tfrac{\pi}{2} = 90^{\circ} \).