# Mathsight Class 7 Solutions Chapter 15

## Mathsight Class 7 Solutions Chapter 15 Perimeter and Area

Welcome to NCTB Solutions. Here with this post we are going to help 7th class students to solve MathSight Class 7 Mathematics Book, Chapter 15, Perimeter and Area. Here students can easily get all the exercise questions solution for Chapter 15, Perimeter and Area Exercise 15.1, 15.2 and 15.3

Perimeter and Area Exercise 15.1 Solution :

Question no – (1)

Solution :

Area of triangle,

= 1/2 × b × 2h

bh = b × h

(a) b × h

Question no – (2)

Solution :

Area  = 50 × 40

= 2000 m2

Question no – (3)

Solution :

Area = 1/2 × b × h

= 1/2 × 5 × 8

= 60 cm2.

Question no – (4)

Solution :

Let, OX = H

We know that △XOY and △XOZ –

YO/XY = XO/XZ = XY/YZ

XO/XZ = XY/YZ

Or, H = XZ × XY/YZ = 12 × 9/15

= 36/5 = 7.2

OX = 7.2 cm.

Question no – (5)

Solution :

Let CF = X

base = 2x

Height = 3x

According to question,

1/2 × 2x × 3x = 108

Or, 3x2 = 108

Or, x2 = 108/3 = 36

Or, x2 = 62

Or, x = 6

Question no – (6)

Solution :

Area of △ACB = 1/2 × base × height

= 1/2 × 25 × 8 = 100 cm2

Area of △BCD = 1/2 × base × height

= 1/2 × 16 × 5 = 40 cm2

Area of quadrilateral ABCD = △ACB + △BCD

= 100 + 40

= 140 cm2

Question no – (8)

Solution :

Figure, Area of ABCD = 30 × 20 = 600 cm2

OQ, PS, R is midpoint

AQ = QB = DS = SC

= 30/2

= 15 cm.

AP = PD = BR = RC

= 20/2

= 10 cm.

Area of △APQ = 1/2 × base × height

= 1/2 × 10 × 15 = 75 cm2

∴ Area △APQ = Area of △BQR

= Area of △CRS = Area of △PDS

= 75 cm2

Area of △APS × △BQR × △CRS × △PDS

= 75 × 4

= 300 cm2

Area of PQRS

= 600 – 300

= 300 cm2

Question no – (9)

Solution :

Required, Total area of ABCD = (16 + 10) × (18 × 8)

= 26 × 26 = 676 cm2

Area of △AFE = 1/2 × 8 × 16 = 64 cm2

Area of △EGB = 1/2 × 10 × 18 = 90 cm2

△AFE + △EGB = 64 + 90 = 154 cm2

shaded region = 678 – 154 = 522 cm2

Question no – (10)

Solution :

Length of court yard = 19.5 m

Breath of court yard = 17 m

Area of court yard

= 19.5 × 17

= 331.5 m2

Side of square slab = 0.5 m

Area of square slab

= 0.5 × 0.5

= 0.25 m2

Number of square slab,

= 331.5/0.25

= 1326

Question no – (11)

Solution :

Aman’s rooms length = 9 m

Aman’s rooms Breath = 7 m

Aman’s rooms Area

= 9 × 7

= 63 m2

Riddhi’s rooms side length = 8 m

Riddhi’s rooms Area

= 8 × 8

= 64 m2

64 > 63

Riddhi’s study room is greater than Aman.

Perimeter and Area Exercise 15.2 Solution :

Question no – (1)

Solution : Area of ABCD

= 100 × 80

= 8000 m2

Path wide = 10 m.

Length of PQRS,

= 100 – (10 × 2)

= 100 – 20

Breath of PQRS,

= 80 – (10 × 2)

= 80 – 20

= 40 m

Area of PQRS,

= 80 × 60

= 4800 m2

Area of path,

= 8000 – 4800

= 3200 m2

Question no – (2)

Solution : Area of ABCD

= 200 × 100

= 20000 m2

Path wide = 20 m

Length of PQRS,

= 200 – (20 × 2)

= 200 – 40

= 160 m.

= 100 – (20 × 2)

= 100 – 40

= 60 m.

Area of PQRS,

= 160 × 60

= 9600 m2

Question no – (3)

Solution : Path wide = 5 m

Area of lawn (PQRS) = 400 m2

Side of lawn (PQRS)

= √400 = 20 m

Side of lawn (ABCD),

= 20 + (2 × 5)

= 20 + 10

= 30 m

Area of (ABCD),

= 30 × 30

= 900 m2

Area of path,

= 900 – 400

= 500 m2

Question no – (4)

Solution :

Area of room (ABCD)

= 120 × 15

= 300 m2

Margin wide = 2 m

Length of carpet area (PQRS)

= 20 – (2 × 2)

= 20 – 4

= 16 m.

Breadth of carpet area (PQRS) ,

= 15 – (2 × 2)

= 15 – 4

= 11 m.

Area of carpet area (PQRS),

= 16 × 11

= 176 m2.

Cost of carpeting,

= 176 × 800

= 140,800

Question no – (5)

Solution : Border width,

= 50 cm.

= 0.50 m

Length of PQ,

= 20 + (0.50 × 2)

= 20 + 1

= 21 m.

Length of DD,

= 15 + (0.50 × 2)

= 15 + 1

= 16 m.

Area of PQRS,

= 21 × 16

= 336 m2

cost of printing of border

= 336 × 200

= 67,200

Question no – (6)

Solution : Area of ABCD,

= 100 × 7000 m2

Area of QRST

= 70 × 5

= 350 m2

Area of MNOP

= 100 × 2

= 200 m2

Area of WXYZ

= 5 × 2

= 10

∴ Area of not covered by cross road,

= 7000 – [350 + (200 – 10)]

= 7000 – [350 + 190]

= 7000 – 540

= 6460 m2

Question no – (8)

Solution : Area of ABCD,

= 30 × 20

= 600 cm2

Frame wide = 2 cm

Length of PQ,

= 30 + (2 × 2)

= 30 + 4

= 34 cm

= 20 + (2 × 2)

= 24 cm.

Area of PQRS,

= 34 × 24

= 816 cm2

Cost of training,

= 816 × 20

= 16,320

Question no – (9)

Solution : Area of ABCD,

= 18.5 × 14

= 259 m2

Length of PQ,

= 18.5 + (4.5 × 2)

= 27.5 m

= 14 + (4.5 × 2)

= 14 + 9

= 23 m

Area of PQRS

= 27.5 × 23

= 632.5 m2

(i) Area of verandah,

= 632.5 – 259

= 373.5 m2

(ii) cost of paring tile in verandah,

= 373.5 × 20

= 7,470

Perimeter and Area Exercise 15.3 Solution :

Question no – (1)

Solution :

(a) 2 π r = 2 × 22/7 × 42/2

= 132 cm.

(b) 2 π r = 2 × 22/7 × 9.1/2

= 28.6 cm

(c) 2 π r = 2 × 22/7 × 105/2

= 330 cm

(d) 2 π r = 2 × 22/7 × 14.7/2

= 46.2 cm

Question no – (2)

Solution :

(a) r = c/2 π

= 44 × 7/2 × 22

= 7 cm.

(b) r = c/2 π = 15.4 × 7/2 × 22

= 2.45 cm

(c) r = c/2 π

= 242 × 7/2 × 22

= 38.5 cm.

(d) r = c/2 π

= 30.8 × 7/2 × 22

= 4.9 cm.

Question no – (3)

Solution :

r = c/2 π = 88 × 7/2 × 22

= 14 cm.

Area of circle = π r2

= 22/7 × 14 × 14

= 161 cm2

Question no – (4)

Solution :

C = 2 π r

= 2 × 22/7 × 35

= 220 cm.

Distance covered in 100 revolution,

= 220 × 100

= 22000 cm.

Question no – (5)

Solution :

Inner radi = r = c/2 π

= 264 × 7/2 × 22

= 42 cm.

Outer radi = R = c/2 π

= 352 × 7/2 × 22

= 56 cm.

Width of racing track,

= 56 – 42

= 14 cm.

Question no – (6)

Solution :

Perimeter of wire,

= 22 × 4

= 88 cm.

Circumference of circle = 88 cm.

radi = r = c/2 π

= 88 × 7/2 × 22

= 14 cm.

Area = π r2

= 22/7 × 14 × 14

= 616 cm2

Question no – (7)

Solution :

Perimeter of square

= 28 × 4

= 112 cm.

Area of square

= 28 × 28

= 784 cm2

Side f square = dia of circle = 28 cm.

= 28/2

= 14 cm

Area of circle = π r2

= 22/7 × 14 × 14

= 616 cm2

reaming area

= 784 – 616

= 168 cm2

Question no – (8)

Solution :

Radi of total lawn = R

= 28/2

= 14 cm

Area of total lawn = π × 14 × 14

= 21/7 × 14 × 14

= 616 cm2

Radi of cow graining lawn = 7 cm.

Area of cow graining lawn = π × (7)2

= 22/7 × 7 × 7

= 154 cm2

Area of cow not graining,

= 616 – 154

= 462 cm2

Question no – (9)

Solution :

Perimeter square = 44

Side of square = 44/4 = 11 cm

Area of square = (11)2 = 121 cm2

Perimeter square = circumference = 44  cm.

radi = r = 44 × 7/2 × 22

= 7 cm.

Area  of circle,

= π (7)2 = 22/7 × 7

= 154 cm2

Circle has more area,

= 154 – 121

= 33 m2

Question no – (10)

Solution :

= 4 × 2

= 8 cm

Area of semi circle = π R2

= 22/7 × 8 × 8

= 201.14 cm2

Area of small circle = π r2

= 22/7 × 4 × 4

= 50.28 cm2

= 201.14 – 50 – 28

= 150.86 cm2

(b) Area of triangle

= 1/2 × 15 × 20

= 150 cm2

Area of rectangle

= 25 × 40

= 1000 cm2

= 1000 – 150 = 850

(c) ∴ Area square

= 20 × 20 = 400 cm2

R = 20/2 = 10 cm.

Area of circle = π r2

= 22/7 × 10 × 10

= 314.28 cm2

Area of no of 1 half circle

= 314.28/2

= 157.14 cm2

Area of no of 4 half circle

= 157.14 × 4

= 628.56 cm2

= 628.56 + 400

= 1028.56 cm2

Question no – (11)

Solution :

Long a minute had = 15 cm

1 hr = 60 min

Distance covered,

= 60 × 15

= 900 cm

Question no – (12)

Solution :

Perimeter = 2 π r

= 2 × 22/7 × 18

= 113.14 cm

Perimeter of semi circular silence = p/2 + 2r

= 113.14/2 + (2 × 18)

= 56.57 + 36

= 92.57

Previous Chapter Solution :

Updated: May 27, 2023 — 9:29 am