Mensuration Questions and Answers

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 \( \tfrac{\pi}{2} = 90^{\circ} \).

Answer: C

A cube has six identical square faces.

\[ \text{TSA} = 6a^{2} = 6 \times 1^{2} = 6 \text{ cm}^{2} \]

The 1 cm² option is the area of a single face and the 4 cm² the four side faces, leaving out the top and bottom.

Answer: D

Work out the water needed and the rate at which it arrives.

\[ V = 200 \times 150 \times 2 = 60{,}000 \text{ m}^{3} \]

\[ \text{pipe cross-section} = 1.5 \times 1.25 = 1.875 \text{ m}^{2} \]

\[ \text{flow} = 1.875 \times 20{,}000 = 37{,}500 \text{ m}^{3}\text{/h} \]

\[ t = \frac{60{,}000}{37{,}500} = 1.6 \text{ h} = 96 \text{ minutes} \]

The speed of 20 kmph becomes 20,000 metres per hour, which is the length of water column delivered each hour.

Answer: A

The angle is in radians, so convert it to degrees first.

\[ 1.1 \text{ rad} = 1.1 \times \frac{180}{\pi} = 63.03^{\circ} \]

\[ \tan 63.03^{\circ} = 1.964 \]

Since \( 63^{\circ} \) is above \( 45^{\circ} \), the tangent must exceed 1 - which eliminates every other option at once.

Answer: C

Melting conserves volume, and both cones share the same base radius, so their volumes add through the heights.

\[ V = \frac{1}{3}\pi (2.1)^{2}(4.1 + 4.3) = \frac{1}{3}\pi (4.41)(8.4) \]

\[ = \frac{1}{3}\pi (37.044) = 12.348\pi \text{ cm}^{3} \]

Setting this equal to the sphere's volume:

\[ \frac{4}{3}\pi R^{3} = 12.348\pi \quad\Longrightarrow\quad R^{3} = 9.261 \]

\[ R = 2.1 \text{ cm} \quad\Longrightarrow\quad d = 4.2 \text{ cm} \]

The question asks for the diameter, so the radius must be doubled.