Area Questions and Answers

Answer: A

Find the barn's area, then its dimensions, then its perimeter.

\[ A_{\text{farm}} = 240^{2} = 57{,}600 \text{ m}^{2} \]

\[ A_{\text{barn}} = \frac{57{,}600}{8} = 7200 \text{ m}^{2} \]

With breadth \( b \) and length \( b + 10 \):

\[ b(b + 10) = 7200 \quad\Longrightarrow\quad b^{2} + 10b - 7200 = 0 \]

\[ b = \frac{-10 + \sqrt{100 + 28800}}{2} = \frac{-10 + 170}{2} = 80 \text{ m} \]

So the barn is 80 m by 90 m.

\[ \text{perimeter} = 2(80 + 90) = 340 \text{ m} \]

\[ \text{cost} = 340 \times 1.5 = \text{Rs. } 510 \]

The wall is priced by length, so the area of 7200 m² is only an intermediate step - pricing it by area would give Rs. 10,800.

Answer: A

Check the triangle inequality before computing anything: the sum of any two sides must exceed the third.

\[ 1 + 2 = 3 \]

This is equality, not inequality, so the three points are collinear and no triangle encloses any area.

Heron's formula confirms it. With \( s = \tfrac{1+2+3}{2} = 3 \):

\[ A = \sqrt{3(3-1)(3-2)(3-3)} = \sqrt{3 \times 2 \times 1 \times 0} = 0 \]

The zero factor \( (s - c) \) is the algebra registering the degenerate case.

Answer: D

Turn the two facts into a sum and a product.

\[ 2(l + b) = 50 \quad\Longrightarrow\quad l + b = 25 \]

\[ lb = 150 \]

So \( l \) and \( b \) are the roots of

\[ x^{2} - 25x + 150 = 0 \quad\Longrightarrow\quad (x - 15)(x - 10) = 0 \]

\[ l = 15 \text{ m}, \qquad b = 10 \text{ m} \]

Checking the options directly also works: only 15 and 10 give both a sum of 25 and a product of 150.

Answer: C

Two parallelograms on the same base and between the same parallels have the same height, and area is base times height:

\[ A = \text{base} \times \text{height} \]

Both the base and the height are shared, so the two areas are equal and the ratio is \( 1 : 1 \).

The condition "between the same parallels" is doing the real work here: a rhombus with the same side but a different angle has area \( a^{2}\sin\theta \), which is less than the square's \( a^{2} \) unless \( \theta = 90^{\circ} \).

Answer: C

Divide the total area between the slabs, then take the square root.

\[ \text{area of one slab} = \frac{72}{50} = 1.44 \text{ m}^{2} \]

\[ \text{side} = \sqrt{1.44} = 1.2 \text{ m} = 120 \text{ cm} \]

The slabs are square, so the side is the square root of the area - and the answer is asked for in centimetres, hence the factor of 100.

Answer: C

Let the side of smaller square at the extreme right be a 

Then side of the larger square is 3a 

Total area of the field = a² + a² +a² + (3a)² +(3a)² = 30a²

Therefore, 30 a² = 4320 => a² = 432 / 3 = 144 => a = 12 

Smaller side of the field = 36 

Larger side of the field = (3a) × 3 + a = 10 a = 10 × 12 = 120 

Original Dimensions of the field = 120m × 36m

Answer: B

Fencing follows the circumference, and the diameter is given directly.

\[ C = \pi d = \frac{22}{7} \times 28 = 88 \text{ m} \]

\[ \text{cost} = 88 \times 1.50 = \text{Rs. } 132 \]

Using \( 2\pi d \) - effectively treating 28 as the radius - would double the answer to Rs. 264.

Answer: C

\[ 2(l + b) = 600 \quad\Longrightarrow\quad l + b = 300 \]

\[ l = 300 - 100 = 200 \text{ m} \]

Halving the perimeter first is essential - subtracting the breadth from the full 600 m would give 500 m.

Answer: B

Walking diagonally means covering the rectangle's diagonal, which forms a right triangle with the two sides.

\[ b = \sqrt{20^{2} - 16^{2}} = \sqrt{400 - 256} = \sqrt{144} = 12 \text{ m} \]

This is the 3–4–5 triple scaled by four: 12, 16, 20.

Answer: C

The area is unchanged, so the new breadth follows from it directly.

\[ A = 18 \times 10 = 180 \text{ cm}^{2} \]

\[ b' = \frac{180}{25} = 7.2 \text{ cm} \]

Length and breadth are inversely proportional when the area is fixed: the length rose by a factor of \( \tfrac{25}{18} \), so the breadth falls by the same factor, \( 10 \times \tfrac{18}{25} = 7.2 \).