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Class 8 MAHARASHTRA STATE TEXTBOOK BUREAU Solutions Maths Chapter 15: Area

Area Exercise Ex. 15.1

Solution 1

Given that base of parallelogram is 18 cm and height is 11 cm.

 

Area of parallelogram = base of parallelogram × height of parallelogram

 = 18 × 11

 = 198 cm2

The area of a parallelogram is 198 cm2.

Solution 2

Given that area of parallelogram is 29.6 sq cm and base of parallelogram is 8 cm.

Area of parallelogram = 29.6

Base × height = 29.6

8 × height = 29.6

Height =   

The height of a parallelogram is 3.7 cm.

Solution 3

Given that area of parallelogram is 83.2 sq cm and height of parallelogram is 6.4 cm.

Since, Area of parallelogram = 83.2

Base × height = 83.2

Base × 6.4 = 83.2

Base =   

The base of a parallelogram is 13 cm.

Area Exercise Ex. 15.2

Solution 1

The lengths of diagonals of a rhombus are 15 cm and 24 cm.

Area of a rhombus =   

=   

= 15  12

= 180 cm2

∴ Area of rhombus is 180 cm2.

Solution 2

The lengths of diagonals of a rhombus are 16.5 cm and 14.2 cm.

Area of rhombus =   

=   

= 7.1  16.5

= 117.15 cm2

∴ Area of rhombus is 117.15 cm2.

Solution 3

Here, perimeter of a rhombus = 100 cm and length of one diagonal = 48 cm.

Since, the four sides of a rhombus are equal.

∴ The length of a rhombus is   .

Consider, a rhombus PQRS where diagonals interest at O.

  

Here, point O bisects both the diagonals PR and QS at right angles.

i.e. OS = OQ and OP = OR

∴ PO =   =   

Now, in right angled ∆POQ,

(PO)2 + (OQ)2 = (PQ)2 … Pythagoras theorem

∴ (24)2 + (OQ)2 = (25)2

∴ OQ2 = (25)2 - (24)2

∴ OQ2 = 625 - 576

∴ OQ2 = 49 cm2

∴ OQ = 7 cm

∴ QS = 2 × OQ = 2 × 7 = 14 cm

∴ The lengths of diagonals of a rhombus are 48 cm and 14 cm.

Area of rhombus =   

=   

=   

= 336 cm2

∴ Area of rhombus is 336 cm2.

Solution 4

Given: Length of a diagonal = 30 cm and area of a rhombus = 240 sq. cm

Let the length of the other diagonal of a rhombus be x cm.

Area of a rhombus =   

∴ 240 = 15x

∴ 15x = 240

∴ x = 16 cm

Consider, a rhombus ABCD where diagonals interest at O.

The diagonals of a rhombus intersect at O where point O bisects both the diagonals at right angles.

i.e. OD = OB and OA = OC

Now, AC = 30 cm and BD = 16 cm

OA =   

∴ OA =   

Similarly, OB =   

∴ OB =   

In right angled ∆AOB,

AB2  = AO2 + OB2

∴ AB2 = 152 + 82

∴ AB2 = 225 + 64

∴ AB2 = 289

∴ AB = 17 cm

The length of the side of a rhombus is 17 cm.

∴ Perimeter of a rhombus is 4 × side = 4 × 17 = 68 cm

Area Exercise Ex. 15.3

Solution 1

In □ABCD, l (AB) = 13 cm, l (DC) = 9 cm, l (AD) = 8 cm

Here, AB and DC are parallel sides and height = AD = 8 cm

Area of trapezium =   

=   

 

=   

 

= 11   

= 88 cm2

∴ Area of trapezium is 88 cm2.

Solution 2

Lengths of the two parallel sides of a trapezium are 8.5 cm and 11.5 cm and its height is 4.2 cm.

Area of trapezium =    

=   

 

=   

 

= 10   

 

= 42 cm2

 

∴ Area of trapezium is 42 cm2. 

Solution 3

  

Let QN ⟂ NR.

From the given figure, seg PM ⟂ seg SR and seg QN ⟂ seg SR

Thus, side PQ || side MN

∴ □PQNM is a rectangle.

 

PQ = MN = 7 cm, PM = QN = 4 cm … (Opposite sides of a rectangle)

In right angled triangle ∆SMP and ∆RNQ,

PS = QR  … (Opposite sides of an isosceles trapezium)

 

PM = QN = 4 cm and SM = NR = 3 cm

 

SR = SM + MN + NR (S - M - N - R)

= 3 + 7 + 3 = 13 cm

 

Area of trapezium =     

=   

 

=   

 

=   

 

= 10   

 

= 40 cm2

 

∴ Area of trapezium is 40 cm2. 

Area Exercise Ex. 15.4

Solution 1

Let sides of a triangle be a = 45 cm, b = 39 cm, c = 42 cm.

 

Semi-perimeter (s) =   

 

Area of triangle =   

=   

=   

=   

=   

= 9 × 3 × 7 × 4

= 27 × 28

= 756 cm2

∴ Area of triangle is 756 cm2.

Solution 2

In ∆PSR, PS ⟂ SR, using Pythagoras theorem,

PR2 = PS2 + SR2

 = 362 + 152

 = 1296 + 225

 = 1521

∴ PR2 = 1521 cm2

∴ PR = 39 m … (i)

 

In ∆PSR,

Area of ∆PSR =   

  

 =   

 = 270 m2

 

In ∆PQR,

PQ = a = 56 m, QR = b = 25 m, PR = c = 39 m

 

∴ Semi-perimeter (s) =   

 

Area of triangle =   

=   

=   

=   

=   

= 5 × 4 × 7 × 3

= 20 × 21

= 420 cm2

 

∴ Area of □ ABCD = area of ∆PSR + area of ∆PQR

 = 270 + 420

 = 690 cm2

Solution 3

In right angled triangle ∆BAD,

Base = AD = 9 m, height = AB = 40 m

 

∴ Area of ∆BAD =   

 = 9 × 20

 = 180 m2

 

In ∆BDC,

Base = DC = 60 m, height = BT = 13 m

 

∴ Area of ∆BDC =   

 = 30 × 13

 = 390 m2

 

Area of □ABCD = Area of ∆BAD + Area of ∆BDC

 = 180 + 390

 = 570 m2

Area Exercise Ex. 15.5

Solution 1(1)

In ∆ PAQ,

Base = PA = 30 m, height = QA = 50

 

Area of ∆ PAQ =  

= 15 × 50

= 750 sq. m

 

In trapezium QACR, AQ and CR are parallel sides where AQ = 50 m, CR = 25 m and height = h = AC = AB + BC = 30 + 30 = 60 m

 

Area of trapezium QACR =  

=  

= 75 = = = 24

= 216 sq. m

 

In ∆ DEC,

Area of ∆ DEC =  

= 14 = = = = 2 4 2

= 336 sq. m

 

Area of plots

= A(∆ ABE) + A(∆ BCE) + A(∆ EDC)

= 216 + 336 + 224

= 776 sq. m

Area Exercise Ex. 15.6

Solution 1(1)

Given that radius of a circle = r = 28 cm

Area of circle   

   =   

   = 22 × 4 × 28

   = 2464 cm2

Solution 1(2)

Given that radius of a circle = r = 10.5 cm

 

Area of circle   

 =   

 = 22 × 1.5 × 28

 = 346.5 cm2

Solution 1(3)

Given that radius of a circle = r = 17.5 cm

Area of circle   

 =   

 = 22 × 2.5 × 17.5

 = 962.5 cm2

Solution 2(1)

Given that area of a circle = 176 sq cm

Since, Area of circle   

  

  

 

∴ r2 = 176 ×   

 

∴ r2 = 56

∴ r =   

∴ Diameter = 2r = 2  cm

Solution 2(2)

Given that area of a circle = 394.24 sq. cm

Since, Area of circle   

  

  

 

∴ r2 = 394.24 ×   

 

∴ r2 = 125.44

 

∴ r =   = 11.2

 

∴ Diameter = 2r = 2   = 22.4 cm

Solution 2(3)

Given that area of a circle = 12474 sq. cm

Since, Area of circle   

  

  

 

∴ r2 = 12474 ×   

 

∴ r2 = 3969

 

∴ r =   = 63

 

∴ Diameter = 2r = 2  = 126 cm

Solution 3

Diameter of a garden = d = 42 m

∴ Radius of inner circle (r)  =   

 

Radius of outer circle (R) = 21 + width of a road

 = 21 + 3.5

 = 24.5 m

 

Area of a road = area of outer circle - area of inner circle

 =  -   

 =  -  

 

 =   

 

 =   

 

 =   

 = 500.50 sq. m

Solution 4

Circumference of a circle = 88 cm … (Given)

 

∴ 2  = 88

 

  = 88

 

∴ r =   = 14

 

∴ r = 14 cm

 

∴ Area of a circle =   =   = 616 sq. cm