Area Questions and Answers

Q: The perimeter of a semi circle is 144 cm then the radius is?

\[ P = r(\pi + 2) = r \times \frac{36}{7} \]\[ 144 = \frac{36r}{7} \quad\Longrightarrow\quad r = \frac{144 \times 7}{36} = 28 \text{ cm} \]Using \( \pi r \) alone would give \( r = 45.8 \) cm - the diameter is part of a semicircle's boundary and must be included.

Q: The radius of a circle is increased by 1%. Find how much % does its area increases?

Area varies as the square of the radius.\[ \frac{A'}{A} = (1.01)^{2} = 1.0201 \]\[ \text{increase} = 0.0201 = 2.01\% \]The exact rule is \( 2x + \tfrac{x^{2}}{100} = 2 + 0.01 \). For small changes the area rises by roughly twice the percentage change in the radius, with the tiny square term as the correction.

Q: The perimeter of a square is 48cm. The area of rectangle is 4cm2 less than the area of the square. If length of the rectangle is 14cm, then its perimeter is

Work from the square to the rectangle.\[ \text{side of square} = \frac{48}{4} = 12 \text{ cm} \quad\Longrightarrow\quad A = 144 \text{ cm}^{2} \]\[ A_{\text{rectangle}} = 144 - 4 = 140 \text{ cm}^{2} \]\[ \text{breadth} = \frac{140}{14} = 10 \text{ cm} \]\[ \text{perimeter} = 2(14 + 10) = 48 \text{ cm} \]The rectangle's perimeter coincidentally matches the square's, even though its area is smaller - equal perimeters do not imply equal areas, and the square is always the larger for a given perimeter.

Q: A rectangular farm has to be fenced one long side, one short side and the diagonal. If the cost of fencing is Rs.10 per meter. The area of farm is 1200 m² and the short side is 30 m long. How much would the job cost?

Find the long side from the area, then the diagonal.\[ \text{long side} = \frac{1200}{30} = 40 \text{ m} \]\[ \text{diagonal} = \sqrt{40^{2} + 30^{2}} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \text{ m} \]\[ \text{fencing} = 40 + 30 + 50 = 120 \text{ m} \]\[ \text{cost} = 120 \times 10 = \text{Rs. } 1200 \]The 30–40–50 triangle is the 3–4–5 triple scaled tenfold, so the diagonal comes out whole.

Important Instructions

41.

The area of a rectangular park is 2400 sq. metres. If the length of the park is decreased by 5 metres, the area decreases by 200 sq. metres. The perimeter of the park is

A.200 m
B.150cm
C.180cm
D.100cm

Answer: A

Cutting 5 m from the length removes a strip 5 m long and \( b \) wide.

\[ 5b = 200 \quad\Longrightarrow\quad b = 40 \text{ m} \]

\[ l = \frac{2400}{40} = 60 \text{ m} \]

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

The lost strip has the full breadth of the park, which is what makes the breadth recoverable from the area change alone.

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42.

The perimeter of a semi circle is 144 cm then the radius is?

A.25 cm
B.28 cm
C.30 cm
D.35 cm

Answer: B

\[ P = r(\pi + 2) = r \times \frac{36}{7} \]

\[ 144 = \frac{36r}{7} \quad\Longrightarrow\quad r = \frac{144 \times 7}{36} = 28 \text{ cm} \]

Using \( \pi r \) alone would give \( r = 45.8 \) cm - the diameter is part of a semicircle's boundary and must be included.

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43.

The radius of a circle is increased by 1%. Find how much % does its area increases?

A.1.01%
B.5.01%
C.3.01%
D.2.01%

Answer: D

Area varies as the square of the radius.

\[ \frac{A'}{A} = (1.01)^{2} = 1.0201 \]

\[ \text{increase} = 0.0201 = 2.01\% \]

The exact rule is \( 2x + \tfrac{x^{2}}{100} = 2 + 0.01 \). For small changes the area rises by roughly twice the percentage change in the radius, with the tiny square term as the correction.

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44.

How many minimum number of square tiles are needed to cover a rectangular floor of dimension 5.25m and 5.10m completely?

A.1190
B.1020
C.1080
D.1230

Answer: A

Square tiles must fit a whole number of times along both edges, so the tile's side must divide both dimensions. Fewest tiles means largest tiles, so take the highest common factor.

\[ 5.25 \text{ m} = 525 \text{ cm}, \qquad 5.10 \text{ m} = 510 \text{ cm} \]

\[ 525 = 3 \times 5^{2} \times 7, \qquad 510 = 2 \times 3 \times 5 \times 17 \]

\[ \text{HCF} = 3 \times 5 = 15 \text{ cm} \]

\[ n = \frac{525}{15} \times \frac{510}{15} = 35 \times 34 = 1190 \]

Converting to centimetres first is essential - the highest common factor of 5.25 and 5.10 is not a whole number of metres.

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45.

The perimeter of a square is 48cm. The area of rectangle is 4cm2 less than the area of the square. If length of the rectangle is 14cm, then its perimeter is

A.24cm
B.48cm
C.50cm
D.54cm

Answer: B

Work from the square to the rectangle.

\[ \text{side of square} = \frac{48}{4} = 12 \text{ cm} \quad\Longrightarrow\quad A = 144 \text{ cm}^{2} \]

\[ A_{\text{rectangle}} = 144 - 4 = 140 \text{ cm}^{2} \]

\[ \text{breadth} = \frac{140}{14} = 10 \text{ cm} \]

\[ \text{perimeter} = 2(14 + 10) = 48 \text{ cm} \]

The rectangle's perimeter coincidentally matches the square's, even though its area is smaller - equal perimeters do not imply equal areas, and the square is always the larger for a given perimeter.

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46.

The ratio of the radius of two circles is 1: 3, and then the ratio of their areas is?

A.1:3
B.1:6
C.1:9
D.1:5

Answer: C

Area varies as the square of the radius, so the ratio is squared.

\[ \frac{A_{1}}{A_{2}} = \frac{\pi r_{1}^{2}}{\pi r_{2}^{2}} = \left(\frac{1}{3}\right)^{2} = \frac{1}{9} \]

So the areas are in the ratio \( 1 : 9 \). Carrying the linear ratio straight through gives \( 1 : 3 \) - the circumferences, not the areas.

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47.

What is the area of a square field whose diagonal of length 20 m?

A.300 sq m
B.250 sq m
C.200 sq m
D.400 sq m

Answer: C

A square's diagonal and side satisfy \( d = a\sqrt{2} \), so

\[ a^{2} = \frac{d^{2}}{2} \]

\[ A = \frac{20^{2}}{2} = \frac{400}{2} = 200 \text{ m}^{2} \]

Squaring the diagonal without halving gives 400, offered as the last option; the diagonal cuts the square into two right triangles, and the halving reflects that.

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48.

A rectangular farm has to be fenced one long side, one short side and the diagonal. If the cost of fencing is Rs.10 per meter. The area of farm is 1200 m² and the short side is 30 m long. How much would the job cost?

A.Rs.700
B.Rs.1200
C.Rs.1400
D.Rs.1500

Answer: B

Find the long side from the area, then the diagonal.

\[ \text{long side} = \frac{1200}{30} = 40 \text{ m} \]

\[ \text{diagonal} = \sqrt{40^{2} + 30^{2}} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \text{ m} \]

\[ \text{fencing} = 40 + 30 + 50 = 120 \text{ m} \]

\[ \text{cost} = 120 \times 10 = \text{Rs. } 1200 \]

The 30–40–50 triangle is the 3–4–5 triple scaled tenfold, so the diagonal comes out whole.

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49.

The radius of a semi circle is 6.3 cm then its perimeter is?

A.35.4 cm
B.32.4 cm
C.32 cm
D.30 cm

Answer: B

The perimeter of a semicircle includes the straight diameter as well as the curved arc.

\[ P = \pi r + 2r = r(\pi + 2) \]

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

\[ P = 6.3 \times \frac{36}{7} = 0.9 \times 36 = 32.4 \text{ cm} \]

The arc alone measures \( \pi r = 19.8 \) cm; leaving out the 12.6 cm diameter is the standard omission.

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50.

Find the area of the quadrilateral of one of its diagonals is 20 cm and its off sets 9 cm and 6 cm?

A.120 sq m
B.150 sq m
C.110 sq m
D.300 sq m

Answer: B

A diagonal splits the quadrilateral into two triangles that share it as their base, with the offsets as their heights.

\[ A = \frac{1}{2} d (h_{1} + h_{2}) \]

\[ = \frac{1}{2} \times 20 \times (9 + 6) = 10 \times 15 = 150 \text{ cm}^{2} \]

The offsets are the perpendiculars dropped onto the diagonal from the two opposite corners, which is why they add before the halving.

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