Mensuration Questions and Answers

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

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: A

Correct answer: 1540 cu. cm

\[V = \pi r^2 h = \frac{22}{7} \times 7^2 \times 10 = \frac{22}{7} \times 490 = 1540 \text{ cu. cm}\]

The radius is squared but the height is not, so doubling the radius would quadruple the volume while doubling the height merely doubles it.

Answer: B

Correct answer: 100 sq. cm

For a square of side \(a\), the diagonal is \(a\sqrt{2}\).

\[a\sqrt{2} = 10\sqrt{2} \quad\Longrightarrow\quad a = 10 \text{ cm}\]

\[\text{Area} = a^2 = 100 \text{ sq. cm}\]

Equivalently, the area of a square is half the square of its diagonal: \(\tfrac{1}{2}(10\sqrt{2})^2 = 100\).

Answer: A

Correct answer: 240 sq. cm

\[\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}\]

\[= \frac{1}{2} \times (15 + 25) \times 12 = \frac{1}{2} \times 40 \times 12 = 240 \text{ sq. cm}\]

The height must be the perpendicular distance between the parallel sides, not the length of a slanting side.

Answer: A

Correct answer: 220 sq. cm

\[\text{Curved surface area} = \pi r l = \frac{22}{7} \times 7 \times 10 = 220 \text{ sq. cm}\]

Adding the base, \(\pi r^2 = 154\) sq. cm, would give a total surface area of 374 sq. cm, so read carefully which is asked.

Answer: B

Correct answer: Rs. 5000

Fencing follows the perimeter, not the area.

\[\text{Perimeter} = 2(60 + 40) = 200 \text{ m}\]

\[\text{Cost} = 200 \times 25 = \text{Rs. } 5000\]

Answer: C

Correct answer: 616 sq. cm

\[\text{Surface area} = 4\pi r^2 = 4 \times \frac{22}{7} \times 49 = 4 \times 154 = 616 \text{ sq. cm}\]

Notice that this is exactly four times the area of a great circle of the sphere, which is a useful check.

Answer: A

Same material means the same density, so weight is proportional to volume.

\[ V_{\text{solid}} = \frac{4}{3}\pi (3)^{3} = 36\pi \text{ cm}^{3} \;\longrightarrow\; 6 \text{ kg} \]

The hollow sphere has outer radius 9 cm and inner radius 6 cm:

\[ V_{\text{hollow}} = \frac{4}{3}\pi\left(9^{3} - 6^{3}\right) = \frac{4}{3}\pi (729 - 216) = \frac{4}{3}\pi (513) = 684\pi \text{ cm}^{3} \]

\[ \text{weight} = 6 \times \frac{684\pi}{36\pi} = 6 \times 19 = 114 \text{ kg} \]

The cavity must be subtracted; treating the object as a solid 18 cm sphere would give 162 kg.

Answer: D

Correct answer: 216 sq. cm

A cube has six identical square faces.

\[\text{Total surface area} = 6a^2 = 6 \times 6^2 = 6 \times 36 = 216 \text{ sq. cm}\]

The volume happens to be \(6^3 = 216\) cu. cm as well, which is a coincidence of the edge being 6.