Mensuration Questions and Answers

Answer: D

The group with all 3-dimensional shapes is cube, cuboid, sphere, cylinder.

Three-dimensional shapes have length, breadth and height and occupy space: cube, cuboid, sphere, cylinder and cone are all solids.

A semi-circle, a circle and a triangle are flat, two-dimensional shapes. Each of the other options includes at least one of them, so only the last option is made up entirely of solids.

Answer: B

The circumference is 352 cm.

\[ \begin{aligned} \pi r^{2} &= 176 \times 56 = 9856 \\ r^{2} &= 9856 \times \frac{7}{22} = 3136 \\ r &= 56\ \text{cm} \\ C &= 2 \times \frac{22}{7} \times 56 = 352\ \text{cm} \end{aligned} \]

396 cm would need a radius of 63 cm, 440 cm a radius of 70 cm and 308 cm a radius of 49 cm; none of these gives an area equal to the rectangle's 9856 square cm.

Answer: B

The true statement is 2F + 3V - 2E = 10.

A triangular prism has 5 faces (2 triangles and 3 rectangles), 6 vertices and 9 edges. Substituting:

\[ \begin{aligned} 4F + 2V - 3E &= 20 + 12 - 27 = 5 \\ 2F + 3V - 2E &= 10 + 18 - 18 = 10 \\ 2F + 4V - 3E &= 10 + 24 - 27 = 7 \\ 3F + 2V - 2E &= 15 + 12 - 18 = 9 \end{aligned} \]

Only the second equation holds. A quick check on the counts is Euler's formula, F + V - E = 5 + 6 - 9 = 2.

Answer: C

The wheel covers 550 metres.

In one round the wheel covers its circumference:

\[ \begin{aligned} 2\pi r &= 2 \times \tfrac{22}{7} \times 25 = \tfrac{1100}{7} \text{ cm} \\ 350 \times \tfrac{1100}{7} &= 55000 \text{ cm} = 550 \text{ m} \end{aligned} \]

1100 m comes from using the diameter 50 cm as the radius, and 275 m from using \(\pi r\) instead of \(2\pi r\).

Answer: A

The top view is figure (1), a row of four squares.

The solid is one cube deep: four cubes stand in a row on the bottom layer, with more cubes stacked above some of them. Looking down from above, each of the four columns shows a single square face, whatever its height, so the top view is a row of four squares.

Figure (4) is the front view, which shows the stacking. Figure (2) leaves out the short column on the right, and figure (3) shows only one cube.

Answer: B

The height of the water is 20 cm.

Volume of a cuboid = base area × height, so

\[ \begin{aligned} h &= \frac{140\ \text{m}^{3}}{700\ \text{m}^{2}} \\ &= 0.2\ \text{m} = 20\ \text{cm} \end{aligned} \]

The trap is the unit: the division gives metres, and 0.2 m must be converted to centimetres by multiplying by 100. The other values do not satisfy volume = area × height.

Answer: A

Each small square has an area of 4 \(cm^{2}\).

The big square has area \(4 \times 4 = 16\ cm^{2}\). Cutting it into 4 equal squares divides the area equally:

\[ \frac{16\ cm^{2}}{4} = 4\ cm^{2} \]

Each small square has side 2 cm. The option 16 \(cm^{2}\) is the whole square, 1 \(cm^{2}\) treats each small side as 1 cm, and 8 \(cm^{2}\) halves the area instead of quartering it.

Answer: B

The length is 60 cm.

The whole wire becomes the perimeter of the rectangle:

\[ \begin{aligned} 2(l + w) &= 180 \\ l + 30 &= 90 \\ l &= 60\ \text{cm} \end{aligned} \]

The option 90 cm is the half-perimeter (length plus width), 120 cm is twice the length, from subtracting both widths (180 - 60) but forgetting to halve, and 45 cm does not fit the perimeter.

Answer: B

The side is 4 units.

For a square of side \(s\), perimeter is \(4s\) and area is \(s^{2}\). Setting them equal:

\[ \begin{aligned} s^{2} &= 4s \\ s &= 4 \quad (s \neq 0) \end{aligned} \]

Check: perimeter \(= 16\) and area \(= 16\). With side 2 the perimeter is 8 but the area 4; with side 8 the perimeter is 32 but the area 64; with side 1 the perimeter is 4 but the area 1. So only 4 units works.

Answer: A

The breadth is 82 m.

For a rectangle, perimeter \(= 2(l + b)\). With perimeter 410 m and length 123 m:

\[ \begin{aligned} 2(123 + b) &= 410 \\ 123 + b &= 205 \\ b &= 82\ \text{m} \end{aligned} \]

Check: \(2(123 + 82) = 2 \times 205 = 410\). The other options give perimeters of 418, 422 and 430 m.