Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 15.4 8th Std Maths Answers Solutions Chapter 15 Area.

## Practice Set 15.4 8th Std Maths Answers Chapter 15 Area

Question 1.

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

Solution:

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

Here, a = 45cm, b = 39cm, c = 42cm

Semi perimeter of triangle = s = \(\frac { 1 }{ 2 }(a+b+c)\)

= \(\frac { 1 }{ 2 }(45+39+42)\)

= \(\frac { 126 }{ 2 }\)

= 63

Area of a triangle

∴ The area of the triangle is 756 sq.cm.

Question 2.

Look at the measures shown in the given figure and find the area of ☐PQRS.

Solution:

A (☐PQRS) = A(∆PSR) + A(∆PQR)

In ∆PSR, l(PS) = 36 m, l(SR) = 15 m

A(∆PSR)

= \(\frac { 1 }{ 2 }\) x product of sides forming the right angle

= \(\frac { 1 }{ 2 }\) x l(SR) x l(PS)

= \(\frac { 1 }{ 2 }\) x 15 x 36

= 270 sq.m

In ∆PSR, m∠PSR = 90°

[l(PR)]² = [l(PS)]² + [l(SR)]²

…[Pythagoras theorem]

= (36)² + (15)²

= 1296 + 225

∴ l(PR)² = 1521

∴ l(PR) = 39m

…[Taking square root of both sides]

In ∆PQR, a = 56m, b = 25m, c = 39m

A(☐PQRS) = A(∆PSR) + A(∆PQR)

= 270 + 420

= 690 sq. m

∴ The area of ☐PQRS is 690 sq.m

Question 3.

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

Solution:

A(☐ABCD) = A(∆BAD) + A(∆BDC)

In ∆BAD, m∠BAD = 90°, l(AB) = 40m, l(AD) = 9m

A(∆BAD) = \(\frac { 1 }{ 2 }\) x product of sides forming the right angle

= \(\frac { 1 }{ 2 }\) x l(AB) x l(AD)

= \(\frac { 1 }{ 2 }\) x 40 x 9

= 180 sq. m

In ∆BDC, l(BT) = 13m, l(CD) = 60m

A(∆BDC) = \(\frac { 1 }{ 2 }\) x base x height

= \(\frac { 1 }{ 2 }\) x l(CD) x l(BT)

= \(\frac { 1 }{ 2 }\) x 60 x 13

= 390 sq. m

A (☐ABCD) = A(∆BAD) + A(∆BDC)

= 180 + 390

= 570 sq. m

∴ The area of ☐ABCD is 570 sq.m.