NCERT Exercise 11.4-part-2

Perimeter and Area 7th Math

NCERT Exercise 11.4-part-2

Question (7) Through a rectangular field of length 90 m and breadth 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is 3 m, find

(i) the area covered by the roads.

(ii) the cost of constructing the roads at the rate of ₹ 110 per m²

Solution:

class 7th math Perimeter and area solution of ncert exercise 11.4 Q7

Let ABCD is the given rectangular field, in which according to question

Length (DC) = 90 m

Breadth (AD) = 60 m

Width of road EH and PQ = 3 m

Rate of construction of road = ₹ 110 per m²

Thus, area of roads and cost of construction of road =?

Calculation of area of road parallel to the length of the field

Length of the road parallel to the length of the field = length of the field = 90 m

Width of the road = 3 m

Now, we know that, Area of rectangle = Length × Width

Thus, area of the rectangular road = 90 m × 3 m

= 270 m²

Calculation of area of road parallel to the width of the field

Length of the road parallel to the width of the field = width of the field = 60 m

Width of the road = 3 m

Now, we know that, Area of rectangle = Length × Width

Thus, area of the rectangular road parallel to the width = 60 m × 3 m

= 180 m²

Calculation of area of the square where roads cross each other

Side of the square where roads cross each other = 3 cm

We know that, Area of square = (Side)²

= (3 m)²

= 9 m²

Thus; area of square at which both the roads cross each other

= 9 m²

(i) Area covered by roads

Area of covered by roads

= Area of road parallel to the length + Area of the road parallel to width – Area of square where roads crosses

= 270 m² + 180m² – 9 m²

= 450 m² – 9 m

= 441 m²

Thus, total areas covered by roads = 441 m²

(ii) Cost of construction of roads

∵ cost of construction of 1 m² of road = ₹ 110

∴ cost of construction of 441 m² of road = ₹ 110 × 441 m²

= ₹ 48510

Thus, area covered by road = 441 m² and cost of construction of road = ₹ 48510 Answer

Question (8) Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown). Did she have any cord left? (`pi` = 3.14)

class 7th math Perimeter and area solution of ncert exercise 11.4 Q8

Solution:

Given, radius of circular pipe = 4 cm

And, side of square box = 4 cm

We know that, circumference of a circle `= 2 pi r`

Thus, radius of given circular pipe `=2 pi 4 cm`

= 3.14 × 8 cm

= 25.12 cm

Thus, length of wire required to wrap the given circular pipe = 25.12 cm

Now,

We know that, perimeter of a square = side × 4

Thus, perimeter of given square box = 4 cm × 4

= 16 cm

Thus, wire required to wrap the given square box = 16 cm

Thus, difference in the circumference and perimeter of circle and square

= 25.12 cm – 16 cm

= 9.12 cm

Thus, if the piece of wire which is wrapped around the circular pipe is wrapped over the given square, then cord left = 9.12 cm

Thus, length of cord left = 9.12 cm. And answer is Yes

Question (9) The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find:

class 7th math Perimeter and area solution of ncert exercise 11.4 Q9

(i) the area of the whole land

(ii) the area of the flower bed

(iii) the area of the lawn excluding the area of the flower bed

(iv) the circumference of the flower bed.

Solution :

Given,

Length of the rectangular lawn = 10 m

And, width of the rectangular lawn = 5 m

Radius of flower bed = 2 m

Thus,

(i) the area of the whole land

We know that, area of a rectangle = Length × Breadht

Thus, area of given rectangular land = 10 m × 5 m

= 50 m²

Thus, are of the whole land = 50 m² Answer

(ii) the area of the flower bed

We know that, area of a circle = `pi r^2`

Thus, area of given circular flower bed `=pi (2 m)^2`

`=22/7 xx 4 m^2`

`= (88 m^2)/7`

= 12.57 m²

Thus, Area of given circular flower bed = = 12.57 m² Answer

(iii) the area of the lawn excluding the area of the flower bed

The area of the lawn excluding the area of flower bed = The area of lawn – Area of flower bed

= 50 m² – 12.57 m²

= 37.43 m²

Thus, the area of the lawn excluding the area of flower bed = 37.43 m² Answer

(iv) the circumference of the flower bed.

We know that, circumference of a circle = `2 pi r`

Thus, circumference of the given flower bed `=2 xx 22/7xx2 m`

`= 88/7 m `

= 12.57 m²

Thus, the circumference of the flower bed = 12.57 m² Answer

Question (10) In the following figures, find the area of the shaded portions:

(i)

class 7th math Perimeter and area solution of ncert exercise 11.4 Q10-i

Solution :

Given,

AB = DC = Length of the rectangle = 18 cm

AD= BC = width of the rectangle = 10 cm

Now, we know that, area of a rectangle = Length × Width

Thus, area of the given rectangle = 18 cm × 10 cm

= 180 cm²

Now, in the triangle AFE,

Base (AF) = 6 cm and Height (AE) = 10 cm

Now, we know that, area of a triangle = `1/2` × Base × Height

Thus, area of the given triangle AFE `=1/2` × 6 cm × 10 cm

= 3 cm × 10 cm

= 30 cm²

Now, in the triangle ECB,

Base (EB) = 8 cm and Height (BC) = 10 cm

Now, Area of the triangle ECB `=1/2` × Base(EB) × Height (BC)

`= 1/2xx 8 cm xx 10 cm`

= 4 cm × 10 cm

= 40 cm²

Now, area of the shaded portion

= Area of the rectangle ADCB – (Area of `triangle` AFE + Area of `triangle` ECB)

= 180 cm² – (30 cm² + 40 cm²)

= 180 cm² – 70 cm²

= 110 cm²

Thus, Area of the shaded portion = 110 cm² Answer

(ii)

class 7th math Perimeter and area solution of ncert exercise 11.4 Q10-ii

Solution :

class 7th math Perimeter and area solution of ncert exercise 11.4 Q10-ii -answer

Given, PQ = QR = 20 cm

Since, both side is equal thus, given polygon is a square.

Now, we know that, Area of a square = side × side

Thus, area of the given square, PQRS = 20 cm × 20 cm

⇒ Area of square (PQRS) = 400 cm²

Now, in triangle, PTQ

Base (PT) = 10 cm and Height (QP) = 20 cm

We know that, area of a triangle = `1/2` × Base × Height

Thus, area of triangle PTQ `=1/2xx10 cmxx20 cm`

= 5 cm × 20 cm

⇒ Area of `triangle` PTQ = 100 cm²

Now, in triangle TSU,

Base (SU) = 10 cm and Height (TS) = 10 cm

Thus, area of `triangle` TSU = `1/2` Base (SU) × Height (TS)

`=1/2xx10 cm × 10 cm`

= 5 cm × 10 cm

= 50 cm²

Thus, Area of triangle TSU = 50 cm²

Now, in triangle URQ,

Base (UR) = 10 cm and Height (QR) = 20 cm

Thus, area of triangle URQ = `1/2` × Base (UR) × Height (QR)

= `1/2 xx 10 cmxx20 cm`

= 5 cm × 20 cm

= 100 cm²

Thus, area of triangle URQ = 100 cm²

Thus, Area of shaded portion = Area of square PQRS – (Area of triangle PQT + Area of triangle TSU + Area of triangle QUR)

= 400 cm² – (100 cm² + 50 cm² + 100 cm)

= 400 cm² – 250 cm²

= 150 cm²

Thus, Area of the shaded portion = 150 cm² Answer

Question (11) Find the area of the quadrilateral ABCD. Here, AC = 22 cm, BM = 3 cm, DN = 3 cm, and BM`_|_`AC, DN`_|_`AC

class 7th math Perimeter and area solution of ncert exercise 11.4 Q11

Solution

Given, in triangle BAC,

Base (AC) = 22 cm and Height (BM) = 3 cm

Now, we know that, Area of a triangle = `1/2` × Base × Height

Thus, area of triangle BAC = `1/2` × AC × BM

`=1/2` × 22 cm × 3 cm

= 11 cm × 3 cm

= 33 cm²

Thus, area of triangle BAC = 33 cm²

Now, in triangle ADC,

Base (AC) = 22 cm and Height (DN) = 3cm

Since, triangle ADC and triangle BAC has same height and base

Thus, area of triangle ADC = Area of triangle BAC = 33 cm²

Now, Area of quadrilateral ABCD = Area of triangle BAC + Area of triangle ADC

= 33 cm² + 33 cm²

= 66 cm²

Thus, area of given quadrilateral ABCD = 66 cm² Answer

Back to 7-math-home

❮ Previous

Reference: