Question Detail

The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:

Answer: Option D

Explanation:

Let original length = x metres and original breadth = y metres.

\begin{aligned}
\text{Original area } = \text{xy } m^2 \\
\text{New Length }= \frac{120}{100}x = \frac{6}{5}x \\
\text{New Breadth }= \frac{120}{100}y = \frac{6}{5}y \\
=>\text{New Area }= \frac{6}{5}x * \frac{6}{5}y \\
=>\text{New Area }= \frac{36}{25}xy \\

\text{Area Difference} = \frac{36}{25}xy - xy \\
= \frac{11}{25}xy \\

Increase \% = \frac{Differnce}{Actual}*100 \\
= \frac{11xy}{25}*\frac{1}{xy}*100 = 44\%

\end{aligned}

1. The height of an equilateral triangle is 10 cm. find its area.

\begin{aligned} \frac{120}{\sqrt{3}} cm^2 \end{aligned}
\begin{aligned} \frac{110}{\sqrt{3}} cm^2 \end{aligned}
\begin{aligned} \frac{100}{\sqrt{3}} cm^2 \end{aligned}
\begin{aligned} \frac{90}{\sqrt{3}} cm^2 \end{aligned}

Answer: Option C

Explanation:

Let each side be a cm, then
\begin{aligned}
\left(\frac{a}{2}\right)^2+{10}^2 = a^2 \\
<=>\left(a^2-\frac{a^2}{4}\right) = 100 \\
<=> \frac{3a^2}{4} = 100 \\
a^2 = \frac{400}{3} \\
Area = \frac{\sqrt{3}}{4}*a^2 \\
= \left(\frac{\sqrt{3}}{4}*\frac{400}{3}\right)cm^2 \\
= \frac{100}{\sqrt{3}}cm^2
\end{aligned}

2. A farmer wishes to start a 100 sq. m. rectangular vegetable garden. Since he has only 30 meter barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. Then find the dimension of the garden.

10 m * 5 m
15 m * 5 m
20 m * 5 m
25 m * 5 m

Answer: Option C

Explanation:

From the question, 2b+l = 30
=> l = 30-2b
\begin{aligned}
Area = 100m^2 \\
=> l \times b = 100 \\
=> b(30-2b) = 100 \\
b^2 - 15b + 50 = 0 \\
=>(b-10)(b-5)=0 \\
\end{aligned}
b = 10 or b = 5
when b = 10 then l = 10
when b = 5 then l = 20
Since the garden is rectangular so we will take value of breadth 5.
So its dimensions are 20 m * 5 m

3. The area of incircle of an equilateral triangle of side 42 cm is :

\begin{aligned} 462 cm^2 \end{aligned}
\begin{aligned} 452 cm^2 \end{aligned}
\begin{aligned} 442 cm^2 \end{aligned}
\begin{aligned} 432 cm^2 \end{aligned}

Answer: Option A

Explanation:

\begin{aligned}
\text{Radius of incircle} = \frac{a}{2\sqrt3} \\
= \frac{42}{2\sqrt3} \\
= 7\sqrt{3} \\
\text{Area of incircle =} \\
\frac{22}{7}*49*3 = 462 cm^2
\end{aligned}

4. The difference of the areas of two squares drawn on two line segments in 32 sq. cm. Find the length of the greater line segment if one is longer than the other by 2 cm.

9 cm
8 cm
7 cm
6 cm

Answer: Option C

Explanation:

Let the lengths of the line segments be x and x+2 cm
then,
\begin{aligned}
{(x+2)}^2 - x^2 = 32 \\
x^2 + 4x + 4 - x^2 = 32 \\
4x = 28 \\
x = 7 cm
\end{aligned}

5. The area of a square is 69696 cm square. What will be its diagonal ?

373.196 cm
373.110 cm
373.290 cm
373.296 cm

Answer: Option D

Explanation:

If area is given then we can easily find side of a square as,
\begin{aligned}Side = \sqrt{69696} \\
= 264 cm \\
\text{we know diagonal =}\sqrt{2}\times side \\
= \sqrt{2}\times 264 \\
= 1.414 \times 264 \\
= 373.296 cm \end{aligned}