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**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 m^{2}

**Question no – (3)**

**Solution :**

Area = 1/2 × b × h

= 1/2 × 5 × 8

= 60 cm^{2}.

**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, 3x^{2} = 108

Or, x^{2} = 108/3 = 36

Or, x^{2} = 6^{2}

Or, x = 6

**Question no – (6)**

**Solution :**

Area of △ACB = 1/2 × base × height

= 1/2 × 25 × 8 = 100 cm^{2}

Area of △BCD = 1/2 × base × height

= 1/2 × 16 × 5 = 40 cm^{2}

**∴** Area of quadrilateral ABCD = △ACB + △BCD

= 100 + 40

= 140 cm^{2}

**Question no – (8)**

**Solution :**

Figure,

Area of ABCD = 30 × 20 = 600 cm^{2}

**∴** 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 cm^{2}

∴ Area △APQ = Area of △BQR

= Area of △CRS = Area of △PDS

= 75 cm^{2}

Area of △APS × △BQR × △CRS × △PDS

= 75 × 4

= 300 cm^{2}

**∴** Area of PQRS

= 600 – 300

= 300 cm^{2}

**Question no – (9)**

**Solution :**

Required,

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

= 26 × 26 = 676 cm^{2}

Area of △AFE = 1/2 × 8 × 16 = 64 cm^{2}

Area of △EGB = 1/2 × 10 × 18 = 90 cm^{2}

**∴** △AFE + △EGB = 64 + 90 = 154 cm^{2}

**∴** shaded region = 678 – 154 = 522 cm^{2}

**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 m^{2}

Side of square slab = 0.5 m

**∴** Area of square slab

= 0.5 × 0.5

= 0.25 m^{2}

**∴** 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 m^{2}

Riddhi’s rooms side length = 8 m

**∴** Riddhi’s rooms Area

= 8 × 8

= 64 m^{2}

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 m^{2}

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 m^{2}

**∴** Area of path,

= 8000 – 4800

= 3200 m^{2}

**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.

Breadth of PQRS,

= 100 – (20 × 2)

= 100 – 40

= 60 m.

**∴** Area of PQRS,

= 160 × 60

= 9600 m^{2}

**Question no – (3)**

**Solution :**

Path wide = 5 m

Area of lawn (PQRS) = 400 m^{2}

Side of lawn (PQRS)

= √400 = 20 m

Side of lawn (ABCD),

= 20 + (2 × 5)

= 20 + 10

= 30 m

**∴** Area of (ABCD),

= 30 × 30

= 900 m^{2}

**∴** Area of path,

= 900 – 400

= 500 m^{2}

**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 m^{2}.

**∴** 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 m^{2}

**∴** cost of printing of border

= 336 × 200

= 67,200

**Question no – (6)**

**Solution :**

Area of ABCD,

= 100 × 7000 m^{2}

Area of QRST

= 70 × 5

= 350 m^{2}

Area of MNOP

= 100 × 2

= 200 m^{2}

Area of WXYZ

= 5 × 2

= 10

∴ Area of not covered by cross road,

= 7000 – [350 + (200 – 10)]

= 7000 – [350 + 190]

= 7000 – 540

= 6460 m^{2}

**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

Breadth of PQ,

= 20 + (2 × 2)

= 24 cm.

**∴** Area of PQRS,

= 34 × 24

= 816 cm^{2}

Cost of training,

= 816 × 20

= 16,320

**Question no – (9)**

**Solution :**

Area of ABCD,

= 18.5 × 14

= 259 m^{2}

Length of PQ,

= 18.5 + (4.5 × 2)

= 27.5 m

Breadth of PQ

= 14 + (4.5 × 2)

= 14 + 9

= 23 m

Area of PQRS

= 27.5 × 23

= 632.5 m^{2}

**(i)** Area of verandah,

= 632.5 – 259

= 373.5 m^{2}

**(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 :**

Let, radi = r

**∴** r = c/2 π = 88 × 7/2 × 22

= 14 cm.

**∴** Area of circle = π r^{2}

= 22/7 × 14 × 14

= 161 cm^{2}

**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 r^{2}

= 22/7 × 14 × 14

= 616 cm^{2}

**Question no – (7)**

**Solution :**

Perimeter of square

= 28 × 4

= 112 cm.

Area of square

= 28 × 28

= 784 cm^{2}

Side f square = dia of circle = 28 cm.

**∴** radi of circle

= 28/2

= 14 cm

**∴** Area of circle = π r^{2}

= 22/7 × 14 × 14

= 616 cm^{2}

**∴** reaming area

= 784 – 616

= 168 cm^{2}

**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 cm^{2}

Radi of cow graining lawn = 7 cm.

**∴** Area of cow graining lawn = π × (7)^{2}

= 22/7 × 7 × 7

= 154 cm^{2}

**∴** Area of cow not graining,

= 616 – 154

= 462 cm^{2}

**Question no – (9)**

**Solution :**

Perimeter square = 44

Side of square = 44/4 = 11 cm

Area of square = (11)^{2} = 121 cm^{2}

Perimeter square = circumference = 44 cm.

**∴** radi = r = 44 × 7/2 × 22

= 7 cm.

**∴** Area of circle,

= π (7)^{2} = 22/7 × 7

= 154 cm^{2}

**∴** Circle has more area,

= 154 – 121

= 33 m^{2}

**Question no – (10)**

**Solution :**

**(a)** radi of semi circle

= 4 × 2

= 8 cm

Area of semi circle = π R^{2}

= 22/7 × 8 × 8

= 201.14 cm^{2}

Area of small circle = π r^{2}

= 22/7 × 4 × 4

= 50.28 cm^{2}

**∴** Shaded area

= 201.14 – 50 – 28

= 150.86 cm^{2}

**(b)** Area of triangle

= 1/2 × 15 × 20

= 150 cm^{2}

Area of rectangle

= 25 × 40

= 1000 cm^{2}

**∴** shaded Area

= 1000 – 150 = 850

**(c) ∴** Area square

= 20 × 20 = 400 cm^{2}

R = 20/2 = 10 cm.

Area of circle = π r^{2}

= 22/7 × 10 × 10

= 314.28 cm^{2}

**∴** Area of no of 1 half circle

= 314.28/2

= 157.14 cm^{2}

**∴** Area of no of 4 half circle

= 157.14 × 4

= 628.56 cm^{2}

Shaded Area,

= 628.56 + 400

= 1028.56 cm^{2}

**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 :**