SChand Composite Mathematics Class 7 Chapter 14 Perimeter and Area Exercise 14F

 Exercise 14 F

Question 1 

Find the area of each circle. In question 1 take $\pi=3 \frac{1}{7}$ or $\frac{22}{7}$ and in question 2 , take $\pi$ 3.14 

(i) Area $=\pi r^{2}$

Sol:: $\frac{22}{7} \times 14^{2}$ $\frac{22}{7} \times 14 \times 14=616 \mathrm{~cm}^{2}$

(ii) Diameter = 56cm

Sol:  Radius = 28cm 

$\because$ Area $=\pi r^{2}=\frac{22}{7} \times 28 \times 28=2464 \mathrm{~cm}^{2}$

 

(iii) Diameter $=2 \frac{4}{5} \mathrm{~cm}=\frac{14}{5} \mathrm{~cm}$

Radius $=\frac{D}{2}=\frac{\frac{14}{5}}{2}=\frac{14}{5 \times 2}=\frac{14}{10} \mathrm{cm}$

Area $=\pi r^{2}=\frac{22}{7} \times \frac{14}{10} \times \frac{14}{10}$

$\frac{154}{25}$ or $6 \frac{4}{25} \mathrm{~m}^{2}$

(iv) $R=1 \frac{3}{4} \mathrm{~cm}=\frac{7}{4} \mathrm{~cm}$

Sol: Area = $\pi r^{2}=\frac{22}{7} \times \frac{7}{4} \times \frac{7}{4}=\frac{77}{8}=9, \frac{5}{8} \mathrm{cm}^{2}$

Question 2

(i) radius $=10 \mathrm{~cm}$ Take $=\pi=3 \cdot 14$

Sol:  Area $=\pi r^{2}$

$=3.14 \times 10 \times 10=314\operatorname{cm}^{2}$

(ii) $D=12 \mathrm{~mm}$

$R=6 \mathrm{~mm}$

Sol:  Area $=\pi r^{2}=3.14 \times 6 \times 6=113.04 \mathrm{mm}^{2}$

(iii) diameter $=30 \mathrm{~km}$

$R=15 \mathrm{~km}$

Sol: Area $=\pi r^{2}=3.14 \times 15 \times 15=706.5 \mathrm{~km}^{2}$

(iv) $R=0.5 \mathrm{~m}$

Sol: Area = $\pi r^{2}=3.14 \times 0.5 \times 0.5=0.785 \mathrm{~m}^{2}$

Question 3

Taking $\pi=\frac{22}{7}$, find the radii of the circles whose areas are

(i) $\frac{22}{7} \mathrm{~cm}^{2}$

Sol: Area $=\frac{22}{7} \mathrm{~cm}^{2}$

$\pi r^{2}=\frac{22}{7} \Rightarrow r^{2}=\frac{22}{7} \times \frac{7}{22} \quad \Rightarrow r^{2}=1$

 r=1cm answer 

(ii) Area $=154 \mathrm{~cm}^{2}$

Sol: $\pi r^{2}=154 \Rightarrow r^{2}=\frac{154}{22} \times 7 \Rightarrow r^{2}=7 \times 7$

r= 7cm answer 

(iii) Area $=314 \frac{2}{7} \mathrm{~mm}^{2}$  $[\because 314 \times 7=2193+2=2200]$

Sol: $\pi \pi^{2}=\frac{2200}{7}$ 

$r^{2}=\frac{2200}{7} \times \frac{7}{22} \Rightarrow r^{2}=100$

$\mathrm{r}=10 \mathrm{~mm}$ Answer 

Question 4

Taking $\pi=3.14$, find the diameters of the circles whose areas are

(i) $2826 \mathrm{~m}^{2}$

(ii) $7.065 \mathrm{~mm}^{2}$

(iii) $254.34 \mathrm{~cm}^{2}$

Sol: (i) Area $=2826 \mathrm{~m}^{2}$

$\pi r^{2}=2826 \Rightarrow r^{2}=\frac{2326}{3.14} \quad \Rightarrow r^{2}=900$

Taking square root r= 30 m Answer D= 60m answer 

Sol:(ii) Area = 7.065 $\mathrm{mm}^{2}$

=$\pi r^{2}=7.065$

=$\pi^{2}=\frac{7.065}{3.14}$

$=r^{2}=2.25$

$r=1.5 \mathrm{~mm}$

$D=3.0 \mathrm{~mm}$

SChand Composite Mathematics Class 7 Chapter 14 Perimeter and Area Exercise 14E

 Exercise 14 E

Question 15

The circumference of the front wheel of a car  is 10 dm and that of the hind wheel in 16dm. How may revolutions more will the first wheel make than the second is covering a distance of 96 km?

Sol: Circumference of front wheel =10dm 

Circumference of back wheel = 16 dm 

Distance to be cover= 96 km or 9,60,000 dm

More no . of revolution by front  wheel = Rev. by front  wheel - Rev by back wheel 

 

$\left(\frac{960,000}{10}\right)-\left(\frac{960,000}{16}\right)$

$=96000-60,000$

=36000 

Question 16

There is a circular pond and foot- path runs along its boundary. A man walks around it, exactly once, keeping close to the edge. If his step is 55cm long and he takes exactly 400 steps to go round the pond . what is the diameter of the pond ? 

Sol: Circumference of the pond = $55 \times 400$ $=22,000 \mathrm{~cm}$ or 220 m 

$\left[\begin{array}{c}C=2 \pi r \\ \text { or } \\ \pi d\end{array}\right]$

$\pi d=220$

$d=\frac{220 \times 7}{22} \quad \Rightarrow d=70 \mathrm{cm}$

Question 17

A piece of wire is bent in the shape of an equilateral triangle of each side 6.6cm. It is rebent to form a circular ring. what is the diameter of the ring? 

[Hint. Perimeter of triangle $=$ Circumference of circular ring, i.e., $\pi \times d=3 \times 6.6]$

(image to be added )

Sol: Circumference of circular ring= Perimeter of triangle 

$\pi d=3 \times 6.6$

$d=\frac{3 \times 6.6}{22} \times 7 \Rightarrow d=6.3$ cm

Question 18

A wire is wound to form a circle of radius 56 cm. If the same wire is converted into a square. Find the side of this square. 

Sol: Perimeter of square = Circumference of circle 

$4 \times$ side $=2 \times \frac{22}{7} \times 56$

Side of the square = $2 \times \frac{22}{7} \times \frac{56}{4}=88 \mathrm{~cm}$

SChand Composite Mathematics Class 7 Chapter 14 Perimeter and Area Exercise 14D

 Exercise 14 D 

Question 9 

A rectangular field is 24 m long and 15 m wide. How many triangular flower beds each of base 3m and altitude 4m can be laid in this field? 

Sol: No of triangular flower =  $\frac{Area of rectangular field}{Area of triangular flower}$

$=\frac{l \times B}{\frac{1}{2} \times \text { Base } \times \text { Height }}$

$=\frac{24 \times 15}{\frac{1}{2} \times 3 \times 4}=60$

Question 10

A right angled triangle has the largest side as 13 cm and one of the sides containing the right angle as 12cm. It's area in $\mathrm{Cm}^{2}$ is :

(IMAGE TO BE ADDED)

Sol:

 $\begin{aligned} H^{2} &=B^{2}+p^{2} \\ \Rightarrow P^{2} &=H^{2}-B^{2} \\ &=13^{2}-12^{2}=169-144=25 \end{aligned}$

$\Rightarrow p^{2}=25^{\circ}$

$\because \quad p=5 \mathrm{~cm}$

Area of Triangle = $\frac{1}{2} \times$ Base $\times$ height $=\frac{1}{2} \times 12 \times 5$

= 30 $\mathrm{Cm}^{2}$

Question 11

The area of a right angled triangle is 40 times its base. what is its height ? 

Sol: Area = $40 \times B$ 

Area of Triangle = $\frac{1}{2} \times$ Base $\times$ Height

Height $=\frac{2 \times \text { Area }}{\text { Base }}=\frac{2 \times 40 \times B}{B}$

= 80 cm 

Question 12

In two triangles, the ratio of their areas is 4:3 and that of their height is 3:4 .find the ratio of their bases.

Sol: Ratio of areas of triangles $A_{1}: A_{2}$ is $4 x: 3 u$

Ratio of their height $H_{1}: H_{2}$ is $3 y: 4 y$

now area of triangle = $\frac{1}{2} \times$ Base $\times$ Height

Base $B_{1}=\frac{2 \times \text { Area of } \Delta\left(A_{1}\right)}{\text { Height }\left(H_{1}\right)}$

Basc B $2=\frac{2 \times \text { Area of } \triangle\left(A_{2}\right)}{\text { Height }\left(\mathrm{H}_{2}\right)}$

$\frac{B_{1}}{B_{2}}=\frac{2 \times \frac{A_{1}}{H_{1}}}{2 \times \frac{A_{2}}{H_{2}}}$

$\frac{4 x}{3 y} \times \frac{4 y}{3x}$

$\frac{B_{1}}{B_{2}}=\frac{16}{9}$

$B_{1}: B_{2}=16: 9$

SChand Composite Mathematics Class 7 Chapter 14 Perimeter and Area Exercise 14C

  Exercise 14 C

Question 1 

Find the area of each parallelogram.

$\begin{array}{|l|l|l|l|l|l|}\hline & \text { (i) } & \text { (ii) } & \text { (iii) } & \text { (iv) } & \text { (v) } \\\hline \text { Base } & 8 \mathrm{~cm} & 12 \mathrm{~mm} & 6.5 \mathrm{~m} & 1 \mathrm{~m}5 \cdot \mathrm{cm} & 4.2 \mathrm{dm} \\\hline \text { Height } & 3 \mathrm{~cm} & 8.7 \mathrm{~cm} & 4.8 \mathrm{~m} & 45\mathrm{~cm} & 25 \mathrm{~cm} \\\hline\end{array}$

 (i) Area = $=B \times H$

$=8 \times 3$

$=24 \mathrm{~cm}^{2}$

(ii) Base $212 \mathrm{~mm}$

$1 c m=10 \mathrm{~mm}$

B= 1.2cm 

Area = $=B \times H$

$=1.2 \times 3.7=10.44\mathrm{~cm}2$

(iii) 

$\begin{aligned} A &=B \times H \\ &=6.5 \times 4.8 \\ &=31.20 \\ & m^{2} \end{aligned}$

(iv) $1 m=100 \mathrm{~cm}$

$B=105 \mathrm{~cm}$

Area $=105 \times 45^{-}$

$=4725 \mathrm{~cm}^{2}$ $=0.4725 \mathrm{~m}^{2}$

Question 2

(i) Area of ||gm ABCD = $48 \mathrm{~cm}^{2}$ DE = 6cm AB =? 

(DIAGRAM TO BE ADDED)

Sol: Area $=B \times 4$

$\begin{aligned}&48=D E \times A B \\&A B=\frac{43}{6} \\&A B=8 \mathrm{~cm}\end{aligned}$

(ii) 

Area of $\| \mathrm{gm}$ PQRS $=252 \mathrm{~cm}^{2}$ PQ = 9CM RT= ? 

Sol: 

$\begin{aligned} \text { Area } &=P Q \times RT \\ \Rightarrow 252 &=9 \times R T \\ \Rightarrow R T &=\frac{252}{9} \Rightarrow R T=28 \mathrm{~cm} \end{aligned}$

Question 3

The side of a rhombs is 7.2cm and its altitude is 5cm. Find its area. 

[Hint. Since a rhombus is a parallelogram with all its sides equal, the formula for area of a ||gm is applicable to it also]

Sol: Area = Base $ \times $ Height (altitude)

$=7.2 \times 5$

$=36 \mathrm{~cm}^{2}$

Question 4

The adjacent sides of a parallelogram are 36 cm and 27cm in length . if the perpendicular distance between the shorter sides is 12 cm, find the distacne between the longer sides. 

(IMAGE TO BE ADDED)

Sol: Given AE = 12CM 

Area of parallelogram ABCD = BASE $\times$ height 

$=D C \times A E$

$=27 \times 12 \mathrm{~cm}^{2}$

As we know Area of ||gm ABCD = $ BC \times AF $ = $DC \times AE $

$\Rightarrow A F=\frac{27 \times 12}{36} \Rightarrow A F=9 \mathrm{~cm}$,

Question 5

The area of a parallelogram and a square are the same. If the perimeter of the square is $160 \mathrm{~m}$ and the height of the parallelogram is $20 \mathrm{~m}$, find the length of the corresponding base of the parallelogram.

Sol: Perimeter of square = 160 m 

$4 \times$ side $=160 \Rightarrow$ side $=\frac{160}{4}$ =  Side = 40 m

Area of square = side $^{2}$ $=40^{2}=1600 \mathrm{~m}^{2}$

Area of parallelogram = Base $ \times $ Height $=1600$

$\Rightarrow$ Base $=\frac{1600}{20}$

 Base $=80 \mathrm{~m}$

Question 6

The area of a rhombus is $42 \mathrm{~m}^{2}$. If its perimeter is $24 \mathrm{~m}$, find its altitude.

Sol: Perimeter = 24 m 

$4 \times $ side $=24 \Rightarrow$ side $=6 \mathrm{~m}$

Area = Base $ \times $ Height= 42

Height = $\frac{42}{6}$

H= 7m answer

SChand Composite Mathematics Class 7 Chapter 14 Perimeter and Area Exercise 14B

 Exercise 14 B

Question 1 

Find the area of the shaded portion in each case. 

(i) (Image to be added)

Sol: Area of shaded portion = Area of outer square -Area of Inner square 

$=8^{2}-4^{2}$

$=64-16$

$=48 m^{2}$

(ii)  (Image to be added) 

Sol:  : Area of shaded portion = Area of outer square -Area of Inner square

$=(24 \times 20)-(22 \times 18)$

$=480-396$

$=84 \mathrm{~m}^{2}$

Question 2

A photograph of sides 35 cm by 22 cm is mounted into a frame of external dimension 45 cm by 30 cm . find the area of the border surrounding the photograph . 

Sol: Area of border = External area of frame - Area of photograph 

$=(45 \times 30)-(35 \times 22)$

$=1350-770$

$=580 \mathrm{~cm}^{2}$

Question 3

A Verandah 1.25 m wide is constructed all along the outside of  a room 5.5m long and 4m wide. find the cost of cementing the floor of this verandah at the rate of Rs 15 per sqm.

Sol: (IMAGE TO BE ADDED)

To find the cost cementing the floor of verandah we have to find the area. 

Outer width = $\begin{aligned} & 5.5+1.25+1.21 \\=& 8 m \end{aligned}$

Outer length = $4+1.25+1.25=6.5 \mathrm{~m}$ 

Area of verandah = Area of outer - Area of inner 

$=(8 \times 6.5)-(5.5 \times 4)$

$=52-22$

$=30 \mathrm{~m}^{2}$

Cost of cementing floor = $15 \times 30$

= Rs 450 Answer 

Question 4

A sheet of paper measuring 30 cm by 20 cm. A strip 4cm wide is cut from it all around . Find the area of remaining sheet and also the area of the strip cut out . 

Sol: Width after cutting 4cm= 30-4+4

$=30-8=22 \mathrm{~cm}$

Length = 20 - 4+4= $20-8=12 \mathrm{~cm}$

Area of remaining sheet =  L$ \times $ B

$=22 \times 12=264 \mathrm{~cm}^{2}$

Area of strip = $(30 \times 20)-(264)$

$600-264=336 \mathrm{~cm}^{2}$

Question 5

Find the area of the crossroads at right angles to each other through the center of the field . 

Sol: (IMAGE TO BE ADDED)

Area of cross roads 

=Area of ABCD+ EFGH -IJKL 

$75 \times 2+62 \times 2-2 \times 2$

$=150+124-4$

$274-4=270 \mathrm{~m}^{2}$ Answer 

 

SChand Composite Mathematics Class 7 Chapter 14 Perimeter and Area Exercise 14A

 Exercise 14 A

Question 1 

Copy and complete the following table: 

$\begin{array}{c|c|c|c|}\hline \text { Length } & \text { Breadth } & \text { Area } & \text { Perimeter } \\\hline \text { (P) } 5 \mathrm{~m} & (3 \mathrm{~m}) & 15 \mathrm{~m}^{2} & 116 \mathrm{~m})\\\text { (ii) } 5 \mathrm{~cm} & (7 \mathrm{~m}) & \left(35 \mathrm{~m}^{2}\right) & 24\mathrm{~cm} \\\text { (iii) }(15 \mathrm{~cm}) & 9.5 \mathrm{~cm} & 142.5 \mathrm{~cm}^{2} & 49\mathrm{~cm} \\\text { (iv) }(16 \mathrm{~cm}) & 30 \mathrm{~cm} & 480 \mathrm{~cm}^{2} & (92\mathrm{~cm}) \\\hline\end{array}$

Question 2

Find (a) the area (b) the perimeter of the square PQRS if. 

 (i) $P Q=Q R=20 \mathrm{~cm} \quad$ Area $=P Q \times Q R=20 \times 20=400 \mathrm{~cm}^{2}$

Perimeter $=4 \times$ side $=4 \times 20=80 \mathrm{~cm}$

(ii) $P Q=Q R=1.2 \mathrm{~m} \quad$ Area $=P Q \times Q R=1.2 \times 1.2=1.14 \mathrm{~m}^{2}$

Perimeter $=4 \times$ side $=4 \times 1.2=4.8 \mathrm{~m}$

Question 3

Find the perimeter of the following squares: 

(i) Area = $=64 \mathrm{~cm}^{2}$

Sol:

$\begin{aligned} S^{2} &=64 \\ \text { Side } &=8 \mathrm{~cm} \end{aligned}$

Perimeter $=4 \times$ side

$=4 \times 8$

$=32 \mathrm{~cm}$

(ii) area $=196 \mathrm{~cm}^{2}$

Sol: $s^{2}=196$

Taking square root 

Side = 14 cm

Perimeter = $4 \times$ side

$=4 \times 14$

$=56 \text { Cm}$

(iii) Area = $2.25 \mathrm{m}^{2}$

Sol: $s^{2}=$ 2.25 

Taking square root side = 1.5m

P= $4 \times 1.5$

=6m

Question 4

(i) The perimeter of the square is 20cm . find its area . 

Sol: Side of square = $\frac{1}{4} \times $ perimeter of square

$=\frac{1}{4} \times 20=5 \mathrm{~cm}$

Area of square = Side $\times$ side 

$=5 \times 5=25$ $\operatorname{cm}^{2}$

(ii) The perimeter of a square field in 3km. Find its area in hectares. 

Sol: Side = $\frac{1}{4} \times $ perimeter

$=\frac{1}{4} \times 8=2 \mathrm{~km}$ or 2000m

Area of square = $side \times side$

$=2,000 \times 2000=40,00,000 \mathrm{~m}^{2}$

1 hectare = 10000 $m^{2}$

$1 m^{2}=\frac{1}{10000}$ Hectare 

$\frac{4000000}{10000}$= 400 hectare