3 marks

Question 13 Marks

How many different rectangles can be made with a 48 cm long string? Find the possible pairs of length and breadth of the rectangles.

Answer

Length of the string to be made into rectangle $=48 cm$
$\therefore$ Perimeter of the rectangle $=48 cm$
$2 \times(1+b)=48 cm$
$1+b=\frac{48}{2} \quad$ Possible pairs of length and breadth are
$1+b=24 cm$
$(1,23),(2,22)(3,21),(4,20),(5,19),(6,18),(7,17),(8,16),(9,15) .(10,14),(11,13),(12,12)$
Number of different rectangles $=12$

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Question 23 Marks

Your garden is in the shape of a square of side 5 m. Each side is to be fenced with 2 rows of wire. Find how much amount is needed to fence the garden at ₹ 10 per metre

Answer

Perimeter of a square = (4 × side) units
Side = 5 m
∴ Perimeter = (4 × 5) m
= 20 m
Each side is fenced with 2 rows of wire
∴ Distance to be fenced = 2 × 20 m
= 40 m
Cost of fencing per metre = ₹ 10
∴ Cost of fencing 40 m = 40 × 10
= ₹ 400
Cost of fencing the garden = ₹ 400

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Question 33 Marks

What will be the area of a new square formed if the side of a square is made one-fourth?

Answer

Let the side of square is $s$ units then area $=(s \times s)$ units $^2$
If the side of the new square is made one fourth then side $=\left(\frac{1 \times 5}{4}\right)$ units
Then area $=\left(\frac{1 \times s}{4} \times \frac{1 \times s}{4}\right)$ units $^2$
$
=\frac{s \times s}{16}
$
$=\frac{1}{16}(s \times s)$ units ${ }_2$ Area of the new square is reduced to $\frac{1}{16}$ times to that of original area.

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Question 43 Marks

Draw a square B whose side is twice of the square A. Calculate the perimeter of the squares A and B

Answer

Perimeter of square B
= 4 × 4 m
= 16 m
Perimeter of square A
= 4 × 2 m
= 8 m
Perimeter of square B is twice that of square A.

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Question 53 Marks

Find the perimeter and the area of a right angled triangle whose sides are $6$ feet, $8$ feet and $10$ feet

Answer

Perimeter of a right angled triangle $=$ sum of three sides
$=(6+8+10)$ feets
$=24$ feetsArea of a right angled triangle $=\frac{1}{2} \times(b \times h)$ unit ${ }^2$ Here the longest side is $10$ feet which is the hypoteneous
$\therefore$ Sides containing right angle are 6 feet and 8 feetLet base $=6$ feet and height $=8$ feet
$\therefore \text { Area }=\frac{1}{2} \times 6 \times 8 \text { feet }^2 $
$=24 \text { sq.feet }$
$ \quad \text { Area }=24 \text { sq.feet }$

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Question 63 Marks

Find the perimeter and the area of the square whose side is 8 cm

Answer

Perimeter of a square $=(4 \times$ side $)$ units
Side $=8 cm$
$\therefore$ Perimeter $=4 \times 8 cm$
$=32 cm$
Perimeter $=32 cm$
Area of a square $=($ side $\times$ side $)$ unit $^2$
$=(8 \times 8) cm^2$
$=64 cm^2$
Area $=64 cm^2$

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Question 73 Marks

Find the perimeter and the area of the rectangle whose length is 6 m and breadth is 4 m

Answer

Perimeter of a rectangle $P=2(1+b)$ units
Length $I =6 m$
breadth $b =4 m$
$p=2 \times(6+4) m$
$=2 \times 10 m$
Perimeter $=20 m$
Area $=1 \times$ b unit $^2$
$=6 \times 4 m^2$
$=24 m^2$
Area $=24 m^2$

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Question 83 Marks

Find the perimeter and area of the following shape

Answer

Perimeter $=(4+4+4+4+4+4+4+4+4+4+4+4) cm$
$=48 cm$
$\text { Perimeter }=48 cm$
Area of 5 squares of side 4 cm
Area of a square $=\left(\right.$ side $\times$ side unit $^2$
$\therefore A=5 \times(4 \times 4) cm^2$
$=5 \times 16 cm^2$
$=80 cm^2$
$80 cm^2$

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Question 93 Marks

Find the perimeter and area of the following shape

Answer

$\text { Perimeter }=(4+5+4+5+4+5+4+5) cm$
$ =36 cm \quad \text { Area of a square of side } 3 cm +\text { Area of } 4 \text { right }$
$ \text { Perimeter }=36 cm$
$ \text { triangles } $
$=(3 \times 3)+\left[4 \times \frac{1}{2} \times 4 \times 3\right] cm ^2$
$=(9+24) cm ^2$
$=33\ cm ^2$
$\text { Area }=33 cm ^2$

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Question 103 Marks

Find the perimeter and area of the following shape

Answer

Perimeter $=(50+12+13+40+10+10+10+5) cm$
$=150 cm$
Perimeter $=150 cm$
Area $=$ Area of a rectangle + Area of a square + Area of a right
triangle.
$=(1 \times b)+(s \times s)+\left(\frac{1}{2} \times b \times h\right) c m^2$
$=(50 \times 5)+(10 \times 10)+\left(\frac{1}{2} \times 12 \times 5\right) cm ^2$
$=(250+100+30) cm ^2$
$=380 cm ^2$
Area $=380 cm ^2$

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Question 113 Marks

A field is in the shape of a right angled triangle whose base is 25 m and height 20 m. Find the cost of levelling the field at the rate of ₹ 45 per sq.m$^2$

Answer

Area of a right angled triangle $=\frac{1}{2} \times($ base $\times$ height $)$ unit ${ }^2$
base $=25 m$
height $=20 m$
$\therefore$ Area $=\frac{1}{2} \times(25 \times 20)^{\text {Cost of levelling per } m^2=\text{₹} 45}$
Area $=250 m ^2$
$\therefore$ Cost of levelling $250 m ^2$
$=250 \times 45$
$=\text{₹} 11,250$ Cost of levelling $=\text{₹} 11,250$

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Question 123 Marks

The scalene triangle has 40 cm as its perimeter and whose two sides are 13 cm and 15 cm, find the third side

Answer

Given two sides of a scalene triangle are 13 cm and 15 cm
Perimeter of the triangle = sum of three sides
40 = 13 + 15 + Third side
40 = 28 + Third side
∴ Third side = 40 – 28
= 12 cm
∴ The third side of the triangle = 12 cm

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Question 133 Marks

A square park has 40 m as its perimeter. What is the length of its side? Also find its area

Answer

Given perimeter $=40 mPerimeter$ of a square $=4 \times$ Length of a side
$40=4 \times$ Length of a side
$\therefore$ Length of its side $=\frac{40}{4} m$
$=10 m$
$\therefore$ Side of the park $=10 m$
$=(10 \times 10) m^2$
$=100 m ^2$
$\therefore$ Area of the Park $=100 m ^2$

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Question 143 Marks

The area of a rectangular shaped photo is 820 sq.cm. and its width is 20 cm. What is its length? Also find its perimeter

Answer

Area of a rectangle $=\left(\right.$ length $\times$ breadth) unit ${ }^2$ Here width $=20 cm$ (breadth)Area $=820 sq \cdot cm$
$\therefore 820 sq . cm =($ length $\times 20) cm ^2$
Length $=\frac{820}{20}$
$=41 cm$ Length of the photo $=41 cm$ Perimeter of a rectangle $=2(l+b)$ units
$=2(41+20) cm$
$=2 \times 61$
$=122 cm$ Perimeter of the photo $=122 cm$

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