5 Marks Questions

Question 15 Marks

The perimeters of two squares are 40 m and 96 m, respectively. Find the perimeter of another square, equal in area to the sum of the first two squares.

Answer

Let side of two squares be $\text{a}_1$ and $\text{a}_2$.
Then, perimeter of first square $=4 \text{a}_1$
According to the question,
Perimeter = 40
$\therefore \quad 4 \text{a}_2=96$
$\Rightarrow \quad \text{a}_2=24$
Similarly, perimeter of second square $=4 \text{a}_2$
But according to the question,
Perimeter = 96
$\therefore \quad 4 \text{a}_2=96$
$\Rightarrow \quad \text{a}_2=24$
Let $\text{a}$ be the side of another square.
Now, sum of the areas of first and second square
= Area of first square + Area of second square
$=\text{a}_1^2+\text{a}_2^2$
Area $=(10)^2+(24)^2=100+576$
$\therefore$ Area $=676$
$\Rightarrow \quad \text{a}^2=676 \Rightarrow \text{a}=\sqrt{676}$
Then, $\text{a}=26 \text{ m}$
Perimeter of another square $=4 \text{a}=4 \times 26=104$
So, perimeter of another square is 104 m.

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Question 25 Marks

During a mass drill exercise, 6250 students of different schools are arranged in rows such that the number of students in each row is equal to the number of rows. In doing so, the instructor finds out that 9 children are left out. Find the number of children in each row of the square.

Answer

Total number of students $=6250$
Number of students forming a square $= 6250 -9= 6241$
Thus, 6241 students form a big square which has number of rows equal to the number of students in each row.
Let the number of students in each row be x, then the number of rows will be x.
Therefore, $x \times x=6241$
$\Rightarrow \quad x=\sqrt{6241}=79$
Hence, there are 79 students in each row of the square formed.
The value depicted here is that exercise is essential for good health. It makes our body fit and strong.

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Question 35 Marks

Find the square root of 324 by the method of repeated subtraction.

Answer

$\begin{array}{l} \text {Here, } 324-1=323, \qquad 323-3=320 \\ 320-5=315, \qquad\qquad~~ 315-7=308 \\ 308-9=299, \qquad\qquad~~ 299-11=288 \\ 288-13=275, \qquad\qquad 275-15=260 \\ 260-17=243, \qquad\qquad 243-19=224 \\ 224-21=203, \qquad\qquad 203-23=180 \\ 180-25=155, \qquad\quad~~~~ 155-27=128 \\ 128-29=99, \qquad\qquad~~ 99-31=68 \\ 68-33=35, \qquad\qquad~~~~ 35-35=0 \end{array}$
So, to get 0, we use 18 steps.
Hence, the square root of 324 is 18.

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Question 45 Marks

Find the least number that must be added to 1500, so as to get a perfect square. Also, find the square root of the perfect square.

Answer

We have, 1500

We see that $38^2<1500<39^2$
So, number to be added $=39^2-1500=$ $1521-1500=21$
Therefore, the perfect square is
$1500+21=1521$
and $\quad \sqrt{1521}=39$
So, the required number is 21 and the square root is 39.

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Question 55 Marks

A ladder 10 m long rests against a vertical wall. If the foot of the ladder is 6 m away from the wall and the ladder just reaches the top of the wall, how high is the wall?

Answer

Let AC be the ladder.
Therefore, AC =10 m
Let BC be the distance between the foot of the ladder and the wall.

Therefore, BC = 6m
$\triangle \text{ABC}$ is a right angled triangle, right angled at $\text B$.
By Pythagoras theorem,
$\text{AC}^2=\text{AB}^2+\text{BC}^2 \Rightarrow 10^2=\text{AB}^2+6^2$
$\begin{array}{ll}\Rightarrow & \text{AB}^2=10^2-6^2 \Rightarrow \text{AB}^2=100-36=64 \\ \therefore & \text{AB}=\sqrt{64}=8 \text{ m}\end{array}$
Hence, the wall is 8 m high.

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