Area Questions and Answers

Answer: D

All lengths in a square scale together: if the diagonal is three times larger, so is the side, and so is the perimeter.

\[ d = a\sqrt{2} \quad\Longrightarrow\quad \frac{d_{1}}{d_{2}} = 3 \;\Longrightarrow\; \frac{a_{1}}{a_{2}} = 3 \]

\[ \frac{P_{1}}{P_{2}} = \frac{4a_{1}}{4a_{2}} = 3 \quad\Longrightarrow\quad 3 : 1 \]

Only the areas would be squared, in the ratio 9 : 1. Perimeter, side and diagonal are all first powers of length and share one ratio.

Answer: C

Write the length in terms of the breadth and use the given difference.

\[ l = 1.6b \quad\Longrightarrow\quad l - b = 0.6b = 24 \]

\[ b = 40 \text{ cm}, \qquad l = 64 \text{ cm} \]

\[ A = 64 \times 40 = 2560 \text{ cm}^{2} \]

"60% more than the breadth" means \( 1.6b \), so the difference is 60% of the breadth, not of the length - taking 24 as 60% of the length would give \( l = 40 \) and a smaller rectangle.

Answer: D

\[ A = 5.5 \times 3.75 = 20.625 \text{ m}^{2} \]

\[ \text{cost} = 20.625 \times 800 \]

\[ = 20 \times 800 + 0.625 \times 800 = 16{,}000 + 500 = \text{Rs. } 16{,}500 \]

Splitting the multiplication into a whole part and a decimal part keeps the arithmetic manageable without a calculator.

Answer: C

Square tiles must divide both dimensions exactly, and the fewest tiles means the largest tile - the highest common factor.

\[ 15 \text{ m } 17 \text{ cm} = 1517 \text{ cm}, \qquad 9 \text{ m } 2 \text{ cm} = 902 \text{ cm} \]

\[ 1517 = 37 \times 41, \qquad 902 = 2 \times 11 \times 41 \]

\[ \text{HCF} = 41 \text{ cm} \]

\[ n = \frac{1517}{41} \times \frac{902}{41} = 37 \times 22 = 814 \]

The awkward 17 cm and 2 cm are deliberate: they force a common factor of 41 rather than a round number.

Answer: D

Find the missing perpendicular side by Pythagoras.

\[ \text{height} = \sqrt{10^{2} - 8^{2}} = \sqrt{100 - 64} = \sqrt{36} = 6 \]

\[ A = \frac{1}{2} \times 8 \times 6 = 24 \]

The hypotenuse is never a side of the right angle, so it cannot be used as a height - using 8 and 10 as the two legs would give 40.

Answer: A

\[ A = a^{2} = 15^{2} = 225 \text{ m}^{2} \]

The perimeter would be \( 4 \times 15 = 60 \) m - a different quantity with different units, and worth keeping distinct from the area.

Answer: B

Poles stand along the boundary, so first find its length.

\[ \text{perimeter} = 2(90 + 50) = 280 \text{ m} \]

\[ n = \frac{280}{5} = 56 \]

On a closed loop the number of gaps equals the number of poles - the last pole is 5 m from the first - so no "+1" is needed here. That extra pole is required only for a fence along an open line.

Answer: B

a = 100         a² = 10000

a = 125         a² = 15625

                         ----------------

        10000 --------- 5625

        100 -------? => 56.25 %

Answer: B

The 0.25 m border runs along all four sides, so each dimension loses twice that.

\[ \text{tiled area} = (4 - 0.5)(9 - 0.5) = 3.5 \times 8.5 = 29.75 \text{ m}^{2} \]

\[ \text{tile} = 0.25 \times 0.25 = 0.0625 \text{ m}^{2} \]

\[ n = \frac{29.75}{0.0625} = 476 \]

Subtracting only 0.25 m from each dimension instead of 0.5 is the usual slip - the space is left on both opposite sides.

Answer: B

Find the radius from the area, then use it for the circumference.

\[ \pi r^{2} = 616 \quad\Longrightarrow\quad r^{2} = \frac{616 \times 7}{22} = 196 \]

\[ r = 14 \text{ cm} \]

\[ C = 2\pi r = 2 \times \frac{22}{7} \times 14 = 88 \text{ cm} \]

The two formulas share \( \pi \) but not the power of \( r \), so the radius must be extracted before the circumference can be found.