NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area - Download PDF

Question: 8 A door of length $2m$ and breadth $1m$ is fitted in a wall. The length of the wall is $4.5m$ and the breadth is $3.6m$ (Fig11.6). Find the cost of white washing the wall, if the rate of white washing the wall is $Rs.20per{m}^2$

Answer: It is given that the length of door is $2m$ and breadth is $1m$ and the length of the wall is $4.5m$ and the breadth is $3.6m$

Now, we know that

Area of rectangle is $=length\times breath$

Thus, the area of the wall is

$\Rightarrow 4.5\times 3.6=16.2m^2$

And

Area of the door is

$\Rightarrow 2\times 1=2m^2$

Now, Area to be painted = Area of wall - Area of door = 16.2 - 2 = 14.2 $m^2$

Now, the cost of whitewashing the wall, at the rate of $Rs.20per{m}^2$ is

$\Rightarrow 14.2\times 20=284Rs$

Therefore, the cost of whitewashing the wall, at the rate of $Rs.20per{m}^2$ is Rs 284

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Topic 11.2.2

Question 1: Each of the following rectangles of length 6 cm and breadth 4 cm is composed of congruent polygons. Find the area of each polygon.

Answer:

The total area of rectangle $=6\times 4=24c{m}^2$

i) The first rectangle is divided into 6 equal parts. So the area of each part will be one-sixth of the total area

$area=\frac{6\times 4}{6}=4c{m}^2$

ii) The rectangle is divided into 4 equal parts, area of each part= one-forth area of rectangle= 6 cm ²

iii) and (iv) are divided into two equal parts. Area of each part will be one half the total area of rectangle= 12 cm ²

v) Area of rectangle is divided into 8 equal parts. Area of one part is one-eighth of the total area of rectangle = 3 cm ²

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Topic 11.3

Question:1(i) Find the area of following parallelograms:

Answer: Area of the parallelogram is the product of base and height

$area=8\times 3.5=28c{m}^2$

Question: 1(ii) Find the area of following parallelograms:

Answer: Area of the parallelogram is the product of base and height

$area=8\times 2.5=20c{m}^2$

Question:(iii) Find the area of the following parallelograms:

In a parallelogram $ABCD$ , $AB=7.2cm$ and the perpendicular from $C$ on $AB$ is $4.5cm$

Answer: Area of parallelogram = the product of base and height

$=7.2\times 4.5=32.4c{m}^2$

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.2

Question: 1(a) Find the area of the following parallelograms:

Answer: We know that

Area of parallelogram $=base\times height$

Here,

Base of parallelogram = 7cm

and

Height of parallelogram = 4 cm

$\Rightarrow 7\times 4=28c{m}^2$

Therefore, the area of the parallelogram is $28c{m}^2$

Question: 1(b) Find the area of the following parallelograms:

Answer:

We know that

Area of parallelogram $=base\times height$

Here,

Base of parallelogram = 5cm

and

Height of parallelogram = 3 cm

$\Rightarrow 3\times 5=15c{m}^2$

Therefore, the area of the parallelogram is $15c{m}^2$

Question: 1(c) Find the area of the following parallelograms:

Answer: We know that

Area of parallelogram $=base\times height$

Here,

Base of parallelogram = 2.5cm

and

Height of parallelogram = 3.5 cm

$\Rightarrow 3.5\times 2.5=8.75c{m}^2$

Therefore, area of parallelogram is $8.75c{m}^2$

Question: 1(d) Find the area of the following parallelograms:

Answer: We know that

Area of parallelogram $=base\times height$

Here,

Base of parallelogram = 5cm

and

Height of parallelogram = 4.8 cm

$\Rightarrow 5\times 4.8=24c{m}^2$

Therefore, the area of a parallelogram is $24c{m}^2$

Question: 1(e) Find the area of the following parallelograms:

Answer: We know that

Area of parallelogram $=base\times height$

Here,

Base of parallelogram = 2 cm

and

Height of parallelogram = 4.4 cm

$\Rightarrow 2\times 4.4=8.8c{m}^2$

Therefore, the area of a parallelogram is $8.8c{m}^2$

Question: 2(a) Find the area of each of the following triangles:

Answer: We know that

Area of triangle $=\frac{1}{2}\times base\times height$

Here,

Base of triangle = 4 cm

and

Height of triangle =3 cm

$\Rightarrow \frac{1}{2}\times 4\times 3=6c{m}^2$

Therefore, area of triangle is $6c{m}^2$

Question: 2(b) Find the area of the following triangles:

Answer: We know that

Area of triangle $=\frac{1}{2}\times base\times height$

Here,

Base of triangle = 5 cm

and

Height of triangle =3.2 cm

$\Rightarrow \frac{1}{2}\times 5\times 3.2=8c{m}^2$

Therefore, the area of the triangle is $8c{m}^2$

Question: 2(c) Find the area of the following triangles:

Answer: We know that

Area of triangle $=\frac{1}{2}\times base\times height$

Here,

Base of triangle = 3 cm

and

Height of triangle =4 cm

$\Rightarrow \frac{1}{2}\times 3\times 4=6c{m}^2$

Therefore, area of triangle is $6c{m}^2$

Question: 2(d) Find the area of the following triangles:

Answer: We know that

Area of triangle $=\frac{1}{2}\times base\times height$

Here,

Base of triangle = 3 cm

and

Height of triangle =2 cm

$\Rightarrow \frac{1}{2}\times 2\times 3=3c{m}^2$

Therefore, the area of the triangle is $3c{m}^2$

Question: 3 Find the missing values:

Answer: We know that

Area of parallelogram $=base\times height$

a) Here, base and area of parallelogram is given

$\Rightarrow 246=20\times height$

$\Rightarrow height=\frac{246}{20}=12.3cm$

b) Here height and area of parallelogram is given

$\Rightarrow 154.5=15\times base$

$\Rightarrow base=\frac{154.5}{15}=10.3cm$

c) Here height and area of parallelogram is given

$\Rightarrow 48.72=8.4\times base$

$\Rightarrow base=\frac{48.72}{8.4}=5.8cm$

d) Here base and area of parallelogram is given

$\Rightarrow 16.38=15.6\times height$

$\Rightarrow height=\frac{16.38}{15.6}=1.05cm$

Question: 4 Find the missing values:

Answer: We know that

Area of triangle $=\frac{1}{2}\times base\times height$

a) Here, the base and area of the triangle is given

$\Rightarrow 87=\frac{1}{2}\times 15\times height$

$\Rightarrow height=\frac{87}{7.5}=11.6cm$

b) Here height and area of the triangle is given

$\Rightarrow 1256=\frac{1}{2}\times 31.4\times base$

$\Rightarrow base=\frac{1256}{15.7}=80mm$

c) Here base and area of the triangle is given

$\Rightarrow 170.5=\frac{1}{2}\times 22\times height$

$\Rightarrow height=\frac{170.5}{11}=15.5cm$

Question: 5(a) $PQRS$ is a parallelogram (Fig 11.23). $QM$ is the height from $Q$ to $SR$ and $QN$ is the height from $Q$ to $PS$ . If $SR=12cm$ and

$QM=7.6cm.$ Find:

the area of the parallelogram $PQRS$

Answer: We know that

Area of parallelogram $=base\times height$

Here,

Base of parallelogram = 12 cm

and

Height of parallelogram = 7.6 cm

$\Rightarrow 12\times 7.6=91.2c{m}^2$

Therefore, area of parallelogram is $91.2c{m}^2$

Question: 5(b) $PQRS$ is a parallelogram (Fig 11.23). $QM$ is the height from $Q$ to $SR$ and $QN$ is the height from $Q$ to $PS$ . If $SR=12cm$ and

$QM=7.6cm.$ Find:

$QN$ , if $PS=8cm$ .

Answer: We know that

Area of parallelogram $=base\times height$

Here,

Base of parallelogram = 12 cm

and

Height of parallelogram = 7.6 cm

$\Rightarrow 12\times 7.6=91.2c{m}^2$

Now,

The area is also given by $QN\times PS$

$\Rightarrow 91.2=QN\times 8$

$\Rightarrow QN=\frac{91.2}{8}=11.4cm$

Therefore, value of QN is $11.4cm$

Question: 6 $DL$ and $BM$ are the heights on sides $AB$ and $AD$ respectively, of parallelogram $ABCD$ (Fig 11.24). If the area of the parallelogram is $1470c{m}^2.$ $AB=35cm$ and $AD=49cm,$ find the length of $BM$ and $DL.$

Answer: We know that

Area of parallelogram $=base\times height$

Here,

Base of parallelogram(AB) = 35 cm

and

Height of parallelogram(DL) = h cm

$\Rightarrow 1470=35\times h$

$\Rightarrow h=\frac{1470}{35}=42cm$

Similarly,

Area is also given by $AD\times BM$

$\Rightarrow 1470=49\times BM$

$\Rightarrow BM=\frac{1470}{49}=30cm$

Therefore, the value of BM and DL are is 30cm and 42cm respectively

Question:7 $\mathrm{\Delta }ABC$ is right angled at $A$ (Fig 11.25). $AD$ is perpendicular to $BC$ . If $AB=5cm,$ $BC=13cm$ and $AC=12cm,$ Find the area of

$\mathrm{\Delta }ABC$ . Also find the length of $AD$ .

Answer: We know that

Area of triangle $=\frac{1}{2}\times base\times height$

Now,

When base = 5 cm and height = 12 cm

Then, the area is equal to

$\Rightarrow \frac{1}{2}\times 5\times 12=30c{m}^2$

Now,

When base = 13 cm and height = AD area remain same

Therefore,

$\Rightarrow 30=\frac{1}{2}\times 13\times AD$

$\Rightarrow AD=\frac{60}{13}cm$

Therefore, the value of AD is $\frac{60}{3}cm$ and the area is equal to $30c{m}^2$

Question: 8 $\mathrm{\Delta }ABC$ is isosceles with $AB=AC=7.5cm$ and $BC=9cm$ (Fig 11.26). The height $AD$ from $A$ to $BC,$ is $6cm,$ Find the area of

$\mathrm{\Delta }ABC.$ What will be the height from $C$ to $AB$ i.e., $CE$ ?

Answer: We know that

Area of triangle $=\frac{1}{2}\times base\times height$

Now,

When base(BC) = 9 cm and height(AD) = 6 cm

Then, the area is equal to

$\Rightarrow \frac{1}{2}\times 9\times 6=27c{m}^2$

Now,

When base(AB) = 7.5 cm and height(CE) = h , area remain same

Therefore,

$\Rightarrow 27=\frac{1}{2}\times 7.5\times CE$

$\Rightarrow CE=\frac{54}{7.5}=7.2cm$

Therefore, value of CE is 7.2cm and area is equal to $27c{m}^2$

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Topic 11.5.1

Question:(a) In Fig Which square has a larger perimeter?

Answer: The outer square has a larger perimeter. Since each side of the inner square forms a triangle. The length of the third side of a triangle is less than the sum of the other two lengths.

Question:(b) In Fig 11.31, Which is larger, perimeter of smaller square or the circumference of the circle?

Answer: The arc length of the circle is slightly greater than the side length of the inner square. Therefore circumference of the inner circle is greater than the inner square

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.3

Question: 1(a) Find the circumference of the circles with the following radius: (Take $\pi =\frac{22}{7}$ )

$14cm$

Answer: We know that

Circumference of a circle is $=2\pi r$

$\Rightarrow 2\pi r=2\times \frac{22}{7}\times 14=88cm$

Therefore, the circumference of the circle is 88 cm

Question: 1(b) Find the circumference of the circles with the following radius: (Take $\pi =\frac{22}{7}$ )

$28mm$

Answer: We know that

Circumference of circle is $=2\pi r$

$\Rightarrow 2\pi r=2\times \frac{22}{7}\times 28=176mm$

Therefore, the circumference of the circle is 176 mm

Question: 1(c) Find the circumference of the circles with the following radius: (Take $\pi =\frac{22}{7}$ )

$21cm$

Answer: We know that

Circumference of circle is $=2\pi r$

$\Rightarrow 2\pi r=2\times \frac{22}{7}\times 21=132cm$

Therefore, circumference of circle is 132 cm

Question: 2(a) Find the area of the following circles, given that:

$radius=14mm$ (Take $\pi =\frac{22}{7}$ )

Answer: We know that

Area of circle is $=\pi r^2$

$\Rightarrow \pi r^2=\frac{22}{7}\times (14{)}^2=616m{m}^2$

Therefore, the area of the circle is $616m{m}^2$

Question: 2(b) Find the area of the following circles, given that:

$diameter=49m$

Answer: We know that

Area of circle is $=\pi r^2$

$\Rightarrow \pi r^2=\frac{22}{7}\times {\left(\frac{49}{2}\right)}^2=1886.5m^2$

Therefore, area of circle is $1886.5m^2$

Question: 2(c) Find the area of the following circles, given that:

$radius=5cm$

Answer: We know that

Area of circle is $=\pi r^2$

$\Rightarrow \pi r^2=\frac{22}{7}\times {\left(5\right)}^2=\frac{550}{7}c{m}^2$

Therefore, the area of the circle is $\frac{550}{7}c{m}^2$

Question: 3 If the circumference of a circular sheet is $154m,$ find its radius. Also, find the area of the sheet. (Take $\pi =\frac{22}{7}$ )

Answer: It is given that circumference of a circular sheet is 154 m

We know that

Circumference of circle is $=2\pi r$

$\Rightarrow 154=2\pi r$

$\Rightarrow 154=2\times \frac{22}{7}\times r$

$\Rightarrow r=\frac{49}{2}=24.5m$

Now,

Area of circle $=\pi r^2$

$\Rightarrow \frac{22}{7}\times {\left(24.5\right)}^2=1886.5m^2$

Therefore, the radius and area of the circle are 24.5 m and $1886.5m^2$ respectively

Question: 4 A gardener wants to fence a circular garden of diameter $21m$ . Find the length of the rope he needs to purchase if he makes $2$ rounds of fence. Also find the cost of the rope, if it costs $Rs.4permeter.$ (Take $\pi =\frac{22}{7}$ ).

Answer: It is given that diameter of a circular garden is $21m$ .

We know that

Circumference of circle is $=2\pi r$

$\Rightarrow 2\times \frac{22}{7}\times \frac{21}{2}=66m$

Now, length of the rope requires to makes $2$ rounds of fence is

$\Rightarrow 2\times$ circumference of circle

$\Rightarrow 2\times 66=132m$

Now, cost of rope at $Rs.4permeter$ is

$\Rightarrow 132\times 4=528Rs$

Therefore, length of the rope requires to makes $2$ rounds of fence is 132 m and cost of rope at $Rs.4permeter$ is Rs 528

Question: 5 From a circular sheet of radius $4cm$ , a circle of radius $3cm$ is removed. Find the area of the remaining sheet.(Take $\pi =3.14$ )

Answer: We know that

Area of circle $=\pi r^2$

Area of circular sheet with radius 4 cm $=\text{3}.14\times (4{)}^2=50.24c{m}^2$

Area of the circular sheet with radius 3 cm $=\text{3}.14\times (3{)}^2=28.26c{m}^2$

Now,

Area of remaining sheet = Area of circle with radius 4 cm - Area od circle with radius 3 cm

$=50.24-28.26=21.98c{m}^2$

Therefore, Area of remaining sheet is $21.98c{m}^2$

Question: 6 Saima wants to put a lace on the edge of a circular table cover of diameter $1.5m$ . Find the length of the lace required and also find its cost if one meter of the lace costs $Rs.15.$ (Take $\pi =3.14$ )

Answer: It is given that the diameter of a circular table is 1.5m.

We know that

Circumference of circle is $=2\pi r$

$\Rightarrow 2\times \frac{22}{7}\times \frac{1.5}{2}=4.71m$

Now, length of the lace required is

$\Rightarrow$ circumference of circle $=4.71m$

Now, cost of lace at $Rs.15permeter$ is

$\Rightarrow 4.71\times 15=70.65Rs$

Therefore, length of the lace required is 4.71 m and cost of lace at $Rs.15permeter$ is Rs 70.65

Question: 7 Find the perimeter of the adjoining figure, which is a semicircle including its diameter.

Answer: It is given that the diameter of semi-circle is 10 cm.

We know that

Circumference of semi circle is $=\pi r$

Circumference of semi-circle with diameter 10 cm including diameter is

$\Rightarrow (\frac{22}{7}\times \frac{10}{2})+10=15.7+10=25.7cm$

Therefore, Circumference of semi-circle with diameter 10 cm including diameter is 25.7 cm

Question: 8 Find the cost of polishing a circular table-top of diameter $1.6m$ , if the rate of polishing is $Rs.15/m^2$ . (Take $\pi =3.14$ )

Answer: It is given that the diameter of a circular table is 1.6m.

We know that

Area of circle is $=\pi r^2$

$\Rightarrow 3.14\times {\left(\frac{1.6}{2}\right)}^2=2.0096m^2$

Now, the cost of polishing at $Rs.15per{m}^2$ is

$\Rightarrow 2.0096\times 15=30.144Rs$

Therefore, the cost of polishing at $Rs.15per{m}^2$ is Rs 30.144

Question: 9 Shazli took a wire of length $44cm$ and bent it into the shape of a circle. Find the radius of that circle. Also, find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square? (Take $\pi =\frac{22}{7}$ )

Answer: It is given that the length of wire is 44 cm

Now, we know that

Circumference of the circle (C) = $2\pi r$

$\Rightarrow 44=2\pi r$

$\Rightarrow r=\frac{44\times 7}{2\times 22}=7cm$

Now,

Area of circle (A) = $\pi r^2$

$\Rightarrow \pi r^2=\frac{22}{7}\times (7{)}^2=154c{m}^2$ - (i)

Now,

Perimeter of square(P) = $4a$

$\Rightarrow 44=4a$

$\Rightarrow a=\frac{44}{4}=11cm$

Area of sqaure = $a^2$

$\Rightarrow a^2=(11{)}^2=121c{m}^2$ -(ii)

From equation (i) and (ii) we can clearly see that area of the circular-shaped wire is more than square-shaped wire

Question: 10. From a circular card sheet of radius $14cm$ , two circles of radius $3.5cm$ and a rectangle of length $3cm$ and breadth $1cm$ are removed.(as shown in the adjoining figure). Find the area of the remaining sheet. (Take $\pi =\frac{22}{7}$ )

Answer: It is given that radius of circular card sheet is $14cm$

Now, we know that

Area of circle (A) = $\pi r^2$

$\Rightarrow \pi r^2=\frac{22}{7}\times (14{)}^2=616c{m}^2$ - (i)

Now,

Area of circle with radius 3.5 cm is

$\Rightarrow \pi r^2=\frac{22}{7}\times (3.5{)}^2=38.5c{m}^2$

Area of two such circle is = $38.5\times 2=77c{m}^2$ -(ii)

Now, Area of rectangle = $l\times b$

$\Rightarrow l\times b=3\times 1=3c{m}^2$ -(iii)

Now, the remaining area is (i) - [(ii) + (iii)]

$\Rightarrow 616-[77+3]\Rightarrow 616-80=536c{m}^2$

Therefore, area of the remaining sheet is $536c{m}^2$

Question:11 A circle of radius $2cm$ is cut out from a square piece of an aluminium sheet of side $6cm$ . What is the area of the left over aluminium sheet?

(Take $\pi =3.14$ )

Answer: It is given that the radius of the circle is $2cm$

Now, we know that

Area of the circle (A) = $\pi r^2$

$\Rightarrow \pi r^2=3.14\times (2{)}^2=12.56c{m}^2$ - (i)

Now,

Now, Area of square = $a^2$

$\Rightarrow a^2=(6{)}^2=36c{m}^2$ -(ii)

Now, the remaining area is (ii) - (i)

$\Rightarrow 36-12.56=23.44c{m}^2$

Therefore, the area of the remaining aluminium sheet is $23.44c{m}^2$

Question: 12 The circumference of a circle is $31.4cm$ . Find the radius and the area of the circle? (Take $\pi =3.14$ )

Answer: It is given that circumference of circle is $31.4cm$

Now, we know that

Circumference of circle is = $2\pi r$

$\Rightarrow 31.4=2\pi r$

$\Rightarrow 31.4=2\times 3.14\times r$

$\Rightarrow r=5cm$

Now, Area of circle (A) = $\pi r^2$

$\Rightarrow \pi r^2=3.14\times (5{)}^2=78.5c{m}^2$

Therefore, radius and area of the circle are $5cmand78.5c{m}^2$ respectively

Question: 13 A circular flower bed is surrounded by a path $4m$ wide. The diameter of the flower bed is $66m$ . What is the area of this path? $(\pi =3.14)$

Answer: It is given that the diameter of the flower bed is $66m$

Therefore, $r=\frac{66}{2}=33m$

Now, we know that

Area of the circle (A) = $\pi r^2$

$\Rightarrow \pi r^2=3.14\times (33{)}^2=3419.46m^2$ -(i)

Now, Area of outer circle with radius(r ') = 33 + 4 = 37 cm is

$\Rightarrow \pi r^{\prime 2}=3.14\times (37{)}^2=4298.66m^2$ -(ii)

Area of the path is equation (ii) - (i)

$\Rightarrow 4298.66-3419.46=879.2m^2$

Therefore, the area of the path is $879.2m^2$

Question:14 A circular flower garden has an area of $314m^2$ . A sprinkler at the centre of the garden can cover an area that has a radius of $12m$ . Will the sprinkler water the entire garden? (Take $\pi =3.14$ )

Answer: It is given that the radius of the sprinkler is $12m$

Now, we know that

Area of the circle (A) = $\pi r^2$

Area cover by sprinkle is

$\Rightarrow \pi r^2=3.14\times (12{)}^2=452.16m^2$

And the area of the flower garden is $314m^2$

Therefore, YES sprinkler water the entire garden

Question:15 Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take $\pi =3.14$ )

Answer: We know that

Circumference of circle = $2\pi r$

Now, the circumference of the inner circle with radius (r) = $19-10=9m$ is

$\Rightarrow 2\pi r=2\times 3.14\times 9=56.52m$

And the circumference of the outer circle with radius (r ') = 19 m is

$\Rightarrow 2\pi r=2\times 3.14\times 19=119.32m$

Therefore, the circumference of inner and outer circles are $56.52mand119.32m$ respectively

Question:16 How many times a wheel of radius $28cm$ must rotate to go $352m$ ? (Take $\pi =\frac{22}{7}$ )

Answer: It is given that radius of wheel is $28cm$

Now, we know that

Circumference of circle = $2\pi r$

$\Rightarrow 2\pi r=2\times 3.14\times 28=175.84cm$

Now, number of rotation done by wheel to go 352 m is

$\Rightarrow \frac{352m}{175.84cm}=\frac{35200}{175.84}\cong 200$

Therefore, number of rotation done by wheel to go 352 m is 200

Question: 17 The minute hand of a circular clock is $15cm$ long. How far does the tip of the minute hand move in $1$ hour. (Take $\pi =3.14$ )

Answer: It is given that minute hand of a circular clock is $15cm$ long i.e. ( r = 15 cm)

Now, we know that one hour means a complete circle of minute hand

Now,

Circumference of circle = $2\pi r$

$\Rightarrow 2\pi r=2\times 3.14\times 15=94.2cm$

Therefore, distance cover by minute hand in one hour is 94.2 cm

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Topic 11.6

Question:(i) Convert the following:

$50c{m}^{2}inm{m}^2$

Answer: 1 cm = 10 mm

$1c{m}^2=1cm\times 1cm=10mm\times 10mm=100m{m}^2$

therefor

$50c{m}^2=50\times 100m{m}^2=5000m{m}^2$