Question Detail

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}

1. 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}

2. The length of a rectangle is three times of its width. If the length of the diagonal is \begin{aligned}8\sqrt{10}\end{aligned}, then find the perimeter of the rectangle.

60 cm
62 cm
64 cm
66 cm

Answer: Option C

Explanation:

Let Breadth = x cm,
then, Length = 3x cm
\begin{aligned}
x^2+{(3x)}^2 = {(8\sqrt{10})}^2 \\
=> 10x^2 = 640 \\
=> x = 8 \\
\end{aligned}
So, length = 24 cm and breadth = 8 cm
Perimeter = 2(l+b)
= 2(24+8) = 64 cm

3. A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m , then the altitude of the triangle is.

200 m
150 m
148 m
140 m

Answer: Option A

Explanation:

Let the triangle and parallelogram have common base b,
let the Altitude of triangle is h1 and of parallelogram is h2(which is equal to 100 m), then
\begin{aligned}
\text{Area of triangle =}\frac{1}{2}*b*h1\\
\text{Area of rectangle =}b*h2\\
\text{As per question }\\
\frac{1}{2}*b*h1 = b*h2 \\

\frac{1}{2}*b*h1 = b*100 \\

h1 = 100*2 = 200 m \\

\end{aligned}

4. 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}

5. The perimeters of two squares are 40 cm and 32 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares .

22 cm
24 cm
26 cm
28 cm

Answer: Option B

Explanation:

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