Area Class 8th Mathematics (new) MHB Solution

Area

Class 8th Mathematics (new) MHB Solution


Class 8th Mathematics (new) MHB Solution

Practice Set 15.1
  1. If the base of a parallelogram is 18 cm and its height is 11 cm, find its area.…
  2. If the area of a parallelogram is 29.6 sq cm and its base is 8 cm, find its height.…
  3. Area of a parallelogram is 83.2 sq cm. If its height is 6.4 cm, find the length of its…
Practice Set 15.2
  1. Lengths of the diagonals of a rhombus are 15cm and 24 cm, find its area.…
  2. Length of the diagonals of a rhombus are 16.5 cm and 14.2 cm, find its area.…
  3. If the perimeter of a rhombus is 100 cm and length of one diagonal is 48 cm, what is…
  4. If the length of a diagonal of a rhombus is 30 cm and its area is 240 sq cm, find its…
Practice Set 15.3
  1. In 􀁵 ABCD, l (AB) = 13 cm, l (DC) = 9 cm, l (AD) = 8 cm, find the area of 􀁵ABCD. .5…
  2. Length of the two parallel sides of a trapezium is 8.5 cm and 11.5 cm respectively and…
  3. 􀁵PQRS is an isosceles trapezium l (PQ) = 7 cm. seg PM ⊥ seg SR, l(SM) = 3 cm,Distance…
Practice Set 15.4
  1. Sides of a triangle are cm 45 cm, 39 cm, and 42 cm, find its area.…
  2. Look at the measures shown in the adjacent figure and find the area of 􀁵PQRS.…
  3. Some measures are given in the adjacent figure, find the area of􀁵 ABCD.…
Practice Set 15.5
  1. Find the areas of given plots. (All measures are in meters.)(1) (2)…
Practice Set 15.6
  1. Radii of the circles are given below, find their areas.(1) 28 cm(2) 10.5 cm(3) 17.5 cm…
  2. Areas of some circles are given below find their diameters.(1) 176 sq cm(2) 394.24 sq…
  3. The diameter of the circular garden is 42 m. There is a 3.5 m wide road around the…
  4. Find the area of the circle if its circumference is 88 cm.

Practice Set 15.1

Question 1.

If the base of a parallelogram is 18 cm and its height is 11 cm, find its area.


Answer:

we know that,

Area of parallelogram = base × height


Given that base of parallelogram = 18cm



And, the height of parallelogram = 11cm


Area of parallelogram = 18 × 11


198 sq cm



Question 2.

If the area of a parallelogram is 29.6 sq cm and its base is 8 cm, find its height.


Answer:

We know that,

Area of parallelogram = base × height




Given that area of parallelogram = 29.6cm


And, the base of parallelogram = 8cm


= 3.7 cm



Question 3.

Area of a parallelogram is 83.2 sq cm. If its height is 6.4 cm, find the length of its base.


Answer:

We know that,

Area of parallelogram = base × height




Given that area of parallelogram = 83.2cm


And, the height of parallelogram = 6.4cm



= 13 cm




Practice Set 15.2

Question 1.

Lengths of the diagonals of a rhombus are 15cm and 24 cm, find its area.


Answer:

We know that,

Area of rhombus = ×product of the length of diagonals


Given that length of one of the diagonals is 15cm


And the other is 24cm


⇒ Area of rhombus = 1/2×15×24


= 180 sq cm



Question 2.

Length of the diagonals of a rhombus are 16.5 cm and 14.2 cm, find its area.


Answer:

We know that,

Area of rhombus = ×product of the length of diagonals


Given that length of one of the diagonals is 16.5cm


And the other is 14.2cm



= 117.5 sq cm



Question 3.

If the perimeter of a rhombus is 100 cm and length of one diagonal is 48 cm, what is the area of the quadrilateral?


Answer:


We know that perimeter of rhombus = 4 × side of the rhombus


Given perimeter of rhombus = 100cm


Side AB of rhombus = 100/4 = 25cm


Let BD be the diagonal given = 48cm


We know that diagonals of a rhombus bisect each other


 E is the midpoint of BD


BE = 24 cm


Now, ∆ABE is the right angle triangle at E


∴using Pythagoras theorem,


AE2 + BE2= AB2




AE = 7cm


Area of rhombus = 4×area of ∆ABE



= 2 × 24 × 7


= 336 sq cm




Question 4.

If the length of a diagonal of a rhombus is 30 cm and its area is 240 sq cm, find its perimeter.


Answer:


we know that,



Given that area of rhombus = 240 sq cm


And diagonal BD = 30cm



⇒other diagonal, AC = 240 × 2 ÷ 30


AC = 16cm


We know that diagonals of a rhombus bisect each other,


So let E be the midpoint of their point of intersection.


Now, AE = 16/2 = 8cm


And BE = 30/2 = 15cm


Now, ∆ABE is right angle triangle


∴ using Pythagoras theorem,


AE2+ BE2 = AB2




⇒AB = 17cm


We know that perimeter of rhombus = 4 × side of rhombus


= 4 × 17


= 68 cm




Practice Set 15.3

Question 1.

In 􀁵 ABCD, l (AB) = 13 cm, l (DC) = 9 cm, l (AD) = 8 cm, find the area of 􀁵ABCD.




Answer:

We know that,



From the fig. it is clear that AB and CD are the 2 parallel sides


Given that AB = 13cm, CD = 9cm and AD = 8cm


Here sum of parallel sides, i.e., AB+CD = 13+9 = 22


Hence,


= 88 sq cm



Question 2.

Length of the two parallel sides of a trapezium is 8.5 cm and 11.5 cm respectively and its height is 4.2 cm, find its area.


Answer:

We know that,


Given that length of 2 parallel sides = 8.5cm and 11.5cm


⇒sum of parallel sides = 8.5 + 11.5 = 20


And, distance between them = 4.2cm



= 42 sq cm



Question 3.

􀁵PQRS is an isosceles trapezium l (PQ) = 7 cm. seg PM ⊥ seg SR, l(SM) = 3 cm,

Distance between two parallel sides is 4 cm, find the area of􀁵 PQRS




Answer:


Given that the trapezium is isosceles. Therefore from the fig. it is clear that SM = NR = 3cm


Also, PQ = MN = 7cm


Now, length of side SR = 3 + 7 + 3 = 13cm


Therefore, the sum of parallel sides of trapezium = 7 + 13 = 20


And the distance between them = 4 cm




= 40 sq cm




Practice Set 15.4

Question 1.

Sides of a triangle are cm 45 cm, 39 cm, and 42 cm, find its area.


Answer:

To find the area of a triangle whose three sides are given we have the Heron’s formula


Where, ∆ is an area of a triangle.




And a, b, c are the three sides of the triangle


In this question, we have the three sides of the triangle which are 45cm, 39cm, and 42cm



= 63m


S - a = 63 - 45 = 18


S - b = 63 - 39 = 24


S - c = 63 - 42 = 21


Hence 


= 756 sq m



Question 2.

Look at the measures shown in the adjacent figure and find the area of 􀁵PQRS.




Answer:

In the given fig. ∆ PRS is right angle triangle at S


Therefore, using Pythagoras theorem,


PS2 + SR2 = PR2




= 39m


Now,


Area of ∆PRS =× base × height


×PS×SR


×36×15


= 270 sq m


Now the area of triangle PQR, using heron’s formula


Here, sides are 56 cm, 25 cm, and 39 cm


Therefore,


S = 60


S - a = 60 - 56 = 4


S - b = 60 - 25 = 35


S - c = 60 - 39 = 21




= 420 sq m


Hence, the area of the quadrilateral PQRS = area of ∆PQR + ∆PSR


= 420 + 270


= 690 sq m



Question 3.

Some measures are given in the adjacent figure, find the area of􀁵 ABCD.




Answer:

In the given fig. ABD is right angled triangle at A,

Given that AB = 40cm, and AD = 9cm



Therefore, the area of triangle ABD






= 180 sqm


Now, the area of triangle, ∆BCD






Now area of quadrilateral ABCD,


= 180 + 390


= 570 sq m




Practice Set 15.5

Question 1.

Find the areas of given plots. (All measures are in meters.)

(1) 

(2)


Answer:

(1)



Given that,


PA = 30m, AC = 30m, and CT = 30m


PC = PA + AC = 30 + 30 = 60m


∆PCT is right angled triangle at C


Area of ∆PCT = 1/2 × PC × CT



= 900m…………(1)


In, ∆SCT is right angled triangle at C


SB = 60m, BC = 30m, and CT = 30m


Area of ∆SCT = 1/2× base × height



×30×90


= 1350m…………….(2)


In ∆SBR is right angled triangle at B


SB = 60m, BR = 25m


Area of ∆SBR = 1/2 × base × height


 × SB × BR


 × 60 × 25


= 750m…………..(3)


In ∆APQ is right angled triangle at A


AP = 30m, AQ = 50m


Area of ∆APQ = × base × height


×AP×AQ


×50×30


= 750m…………(4)


Now, in trapezium ABRQ


AQ and RB are the 2 parallel sides


Also, AQ = 50m and BR = 25m


⇒AQ + BR = 75m


The distance between AQ and BR = 60m


Hence,



= 2250 sq m………….(5)


Now area of quadrilateral PQRST = (1)+(2)+(3)+(4)+(5)


= 900+1350+750+750+2250


= 6000 sq m


(2) the data for this question is inadequate.




Practice Set 15.6

Question 1.

Radii of the circles are given below, find their areas.

(1) 28 cm

(2) 10.5 cm

(3) 17.5 cm


Answer:

(1) We know that

area of circle = πr2


Here given that radius of the circle is 28cm



= 784π sq cm


= 2464 sq cm


(2) Here the radius of the circle = 10.5 cm



= 110.25π sq cm


= 346.5 sq cm


(3) Here the radius of the circle is 17.5cm



= 306.25π sq cm


= 961.625 sq cm



Question 2.

Areas of some circles are given below find their diameters.

(1) 176 sq cm

(2) 394.24 sq cm

(3) 12474 sq cm


Answer:

(1) We know that area of circle = πr2


Here area of circle = 176cm






(2) Here area of circle = 394.24 sq cm




r = 11.2 cm


D = 2r = 2(11.20) = 22.4 cm


(3) Here area of circle = 12474 sq cm





D = 2r = 2(63) = 126cm



Question 3.

The diameter of the circular garden is 42 m. There is a 3.5 m wide road around the garden. Find the area of the road.




Answer:

Given that the diameter of the garden (inner circle) = 42m

Therefore, inner radius, r = 21m


Also, given that road surrounds the garden and is 3.5 m wide.


Therefore, the diameter of the road (outer circle) will be = 42+2(3.5) = 49m


And then outer radius, R = 24.5m


Now, the area of road = area of the outer circle – area of the inner circle


area of outer circle = πR2


= π(24.5)2


= 1885 sq m


area of inner circle = πr2


= π(21)2


= 1385 sq m


Hence, area of road = 1885-1385 = 500 sq m



Question 4.

Find the area of the circle if its circumference is 88 cm.


Answer:

we know that,

The Circumference of a circle = 2πr


Given circumference = 88cm


⇒2πr = 88


r = 14cm


Now, area of circle = πr2


= π(14)2


= 615.75 sq cm


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