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Question Detail
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
1. 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}
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 perimeters of 5 squares are 24 cm, 32 cm, 40 cm, 76 cm and 80 cm respectively. The perimeter of another square equal in area to the sum of the area of these square is:
- 124 cm
- 120 cm
- 64 cm
- 56 cm
Answer: Option A
Explanation:
Clearly first we need to find the areas of the given squares, for that we need its side.
Side of sqaure = Perimeter/4
So sides are,
\begin{aligned}
\left(\frac{24}{4}\right),\left(\frac{32}{4}\right),\left(\frac{40}{4}\right),\left(\frac{76}{4}\right),\left(\frac{80}{4}\right) \\
= 6,8,10,19,20 \\
\text{Area of new square will be }\\
= [6^2+8^2+10^2+19^2+20^2] \\
= 36+64+100+361+400 \\
= 961 cm^2 \\
\text{Side of new Sqaure =}\sqrt{961} \\
= 31 cm \\
\text{Required perimeter =}(4\times31) \\
= 124 cm
\end{aligned}
5. 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}
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