Question Detail

One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.

\begin{aligned} 120m^2 \end{aligned}
\begin{aligned} 130m^2 \end{aligned}
\begin{aligned} 140m^2 \end{aligned}
\begin{aligned} 150m^2 \end{aligned}

Answer: Option A

Explanation:

\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2

\end{aligned}

1. There is a plot which is rectangular in shape and it area is 460 square metres. What will be breadth of the plot if length of plot is 15% more than the breadth of plot ?

8 mtr
10 mtr
15 mtr
20 mtr

Answer: Option D

Explanation:

\begin{aligned} l*b = 460 m^2 \end{aligned} .. (i)
Lets suppose breadth of plot is = b rnLength will be,\begin{aligned}l=b*\frac{100+15}{100}\\=\frac{115b}{100}\end{aligned} .. (ii)

From (i) and (ii), we get :
\begin{aligned}
\frac{115b}{100}*b = 460 \\
b^2 = \frac{46000}{115} = 400 \\
=> b = \sqrt{400} = 20 meter
\end{aligned}

2. Find the circumference of a circle, whose area is 24.64 meter sqaure

17.90m
17.80m
17.60m
17.40m

Answer: Option C

Explanation:

\begin{aligned}
\text{Area of Square =} \pi*r^2\\
=> \pi*r^2 = 24.64 \\
=> r^2 = \frac{24.64}{22}*7 \\
=> r^2 = 7.84 \\
=> r = \sqrt{7.84} \\
=> r = 2.8 \\
\text{Circumference =}2\pi*r\\
= 2*\frac{22}{7}*2.8 \\
= 17.60m

\end{aligned}

3. The base of a triangle is 15 cm and height is 12 cm. The height of another triangle of double the area having the base 20 cm is :

22 cm
20 cm
18 cm
10 cm

Answer: Option C

Explanation:

\begin{aligned}
\text{Area of triangle, A1 = }\frac{1}{2}*base*height \\
= \frac{1}{2}*15*12 = 90 cm^2 \\
\text{Area of second triangle =} 2*A1 \\
= 180 cm^2 \\
\frac{1}{2}*20*height = 180 \\
=> height = 18 cm

\end{aligned}

4. 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

5. The sides of a triangle are in the ratio of \begin{aligned}\frac{1}{2}:\frac{1}{3}:\frac{1}{4}\end{aligned}. If the perimeter is 52 cm, then find the length of the smallest side.

12 cm
14 cm
16 cm
18 cm

Answer: Option A

Explanation:

\begin{aligned}
\text{Ratio of sides =}\frac{1}{2}:\frac{1}{3}:\frac{1}{4} \\
=6:4:3\\
Perimeter = 52 cm \\
\text{So sides are =} \\
\left( 52*\frac{6}{13}\right)cm,\left( 52*\frac{4}{13}\right)cm, \left( 52*\frac{3}{13}\right)cm
\end{aligned}
a = 24 cm, b = 16 cm and c = 12 cm
Length of the smallest side = 12 cm