Area Questions and Answers

Answer: A

Let original length = x metres and original breadth = y metres
Original Area=xy m²
New Length =120100x=65x
New Breadth=120100y=65y
⇒New Area=65x * 65y   
⇒New Area=3625xy
Area Difference=3625xy−xy                            
=1125xy
Increase%=DifferenceActual * 100   
=11xy25 * 1xy * 100
=44%

Answer: C

Express the length in terms of the breadth and substitute.

\[ l = 1.15b \quad\Longrightarrow\quad 1.15b \times b = 460 \]

\[ b^{2} = \frac{460}{1.15} = 400 \quad\Longrightarrow\quad b = 20 \text{ m} \]

The length is then \( 1.15 \times 20 = 23 \) m - offered as a distractor for anyone who solves for the wrong side. Checking: \( 23 \times 20 = 460 \) m².

Answer: B

Only three sides are fenced, and the 20 ft side is left open.

\[ \text{other side} = \frac{680}{20} = 34 \text{ ft} \]

The open side is one of the pair of 20 ft sides, so the fencing covers the other 20 ft side and both 34 ft sides:

\[ 20 + 34 + 34 = 88 \text{ ft} \]

Fencing all four sides would need the full perimeter of 108 ft; leaving out one 20 ft side saves exactly 20 ft.

Answer: A

We know perimeter of square = 4(side)
So Side of first square = 40/4 = 10 cm
Side of second square = 32/4 = 8 cm
Area of third Square = 10*10 - 8*8
= 36 cm
So side of third square = 6 (because area of square = side*side)
Perimeter=4*Side=4*6=24 cm

Answer: B

A square's diagonal and side are related by \( d = a\sqrt{2} \), so the area can be written directly in terms of the diagonal:

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

The factor \( \tfrac{1}{2} \) is common to both squares and cancels:

\[ \frac{A_{1}}{A_{2}} = \frac{d_{1}^{2}}{d_{2}^{2}} = \frac{2^{2}}{5^{2}} = \frac{4}{25} \]

So the ratio is \( 4 : 25 \). Any two similar figures have areas in the square of their linear ratio, whichever length is used to compare them.

Answer: D

The compound wall covers one side, so the wire spans the opposite side and both ends. Let the fenced-opposite side be \( L \) and each end \( W \).

\[ L + 2W = 30, \qquad L \times W = 100 \]

Substituting \( L = 30 - 2W \):

\[ (30 - 2W)W = 100 \quad\Longrightarrow\quad 2W^{2} - 30W + 100 = 0 \]

\[ W^{2} - 15W + 50 = 0 \quad\Longrightarrow\quad (W - 5)(W - 10) = 0 \]

\( W = 5 \) gives \( L = 20 \); \( W = 10 \) gives \( L = 10 \). Only 20 m × 5 m appears among the options.

Checking: \( 20 + 5 + 5 = 30 \) m of wire and \( 20 \times 5 = 100 \) m².

Answer: A

Convert to a single unit before dividing.

\[ \text{courtyard} = 2500 \times 1600 = 4{,}000{,}000 \text{ cm}^{2} \]

\[ \text{brick} = 20 \times 10 = 200 \text{ cm}^{2} \]

\[ n = \frac{4{,}000{,}000}{200} = 20{,}000 \]

A square metre holds 10,000 cm², so the 400 m² courtyard is four million square centimetres - the factor most often lost is that square metre conversion.

Answer: B

Both dimensions shrink, so multiply the two surviving fractions.

\[ \frac{A'}{A} = (1 - 0.20)(1 - 0.10) = 0.80 \times 0.90 = 0.72 \]

\[ \text{decrease} = 1 - 0.72 = 0.28 = 28\% \]

Adding the two losses gives 30%, which overstates the shrinkage by the product \( 0.20 \times 0.10 = 2\% \) - the part counted twice.

Answer: D

We know Perimeter = 2(l+b), right
So,
2(l+b) = 340
As we have to make 1 meter boundary around this, so
Area of boundary = ((l+2)+(b+2)-lb)
= 2(l+b)+4 = 340+4 = 344
So required cost will be = 344 * 10 = 3440

Answer: C

Turn each perimeter into a side, square it, add, then work back.

\[ \text{sides} = \frac{24}{4},\; \frac{32}{4},\; \frac{40}{4},\; \frac{76}{4},\; \frac{80}{4} = 6,\; 8,\; 10,\; 19,\; 20 \]

\[ \text{total area} = 36 + 64 + 100 + 361 + 400 = 961 \text{ cm}^{2} \]

\[ \text{side of new square} = \sqrt{961} = 31 \text{ cm} \]

\[ \text{perimeter} = 4 \times 31 = 124 \text{ cm} \]

Perimeters do not simply add: their sum is 252 cm, twice the answer, because area grows with the square of the side.