Question Detail

If the radius of a circle is diminished by 10%, then the area is diminished by:

200%
210%
300%
310%

Answer: Option C

Explanation:

Let the original radius be R cm. New radius = 2R

\begin{aligned}
Area = \pi R^2 \\
\text{New Area =} \pi {2R}^2 \\
= 4\pi R^2 \\
\text{Increase in area =}(4\pi R^2 - \pi R^2) \\
= 3\pi R^2 \\
\text{Increase percent =} \frac{3\pi R^2}{\pi R^2}*100 \\
= 300 \%
\end{aligned}

1. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:

\begin{aligned} 152600 m^2\end{aligned}
\begin{aligned} 153500 m^2\end{aligned}
\begin{aligned} 153600 m^2\end{aligned}
\begin{aligned} 153800 m^2\end{aligned}

Answer: Option C

Explanation:

Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:

Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]

As per question length is 3x and width is 2x
We know perimeter of rectangle is 2(L+B)
So, 2(3x+2x) = 1600
=> x = 160
So Length = 160*3 = 480 meter
and Width = 160*2 = 320 meter

Finally, Area = length * breadth
= 480 * 320 = 153600

2. If the area of a square with the side a is equal to the area of a triangle with base a, then the altitude of the triangle is.

a
a/2
2a
None of above

Answer: Option C

Explanation:

\begin{aligned}
\text{We know area of square =}a^2 \\
\text{Area of triangle =}\frac{1}{2}*a*h \\
=> \frac{1}{2}*a*h = a^2 \\
=> h = 2a

\end{aligned}

3. 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}

4. The Diagonals of two squares are in the ratio of 2:5. find the ratio of their areas.

4:25
4:15
3:25
3:15

Answer: Option A

Explanation:

Let the diagonals of the squares be 2x and 5x.
Then ratio of their areas will be
\begin{aligned}
\text{Area of square} = \frac{1}{2}*{Diagonal}^2 \\
\frac{1}{2}*{2x}^2:\frac{1}{2}*{5x}^2 \\
4x^2:25x^2 = 4:25
\end{aligned}

5. 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}