Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 15.3 8th Std Maths Answers Solutions Chapter 15 Area.

## Practice Set 15.3 8th Std Maths Answers Chapter 15 Area

Question 1.

In the given figure, ☐ABCD is a trapezium, side AB || side DC, l(AB) = 13 cm, l(DC) = 9 cm, l(AD) = 8 cm, find the area ☐ABCD.

Solution:

☐ABCD is a trapezium, side AB || side DC,

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

Area of a trapezium = \(\frac { 1 }{ 2 }\) x sum of lengths of parallel sides x height

∴ A (☐ABCD) = \(\frac { 1 }{ 2 }\) x [l(AB) + l(DC)] x l(AD)

= \(\frac { 1 }{ 2 }\) x (13 + 9) x 8

= \(\frac { 1 }{ 2 }\) x 22 x 8

= 11 x 8

= 88 sq.cm

∴ The area of ☐ABCD is 88 sq. cm.

[Note: The question is modified.]

Question 2.

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

Solution:

Length 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 a trapezium

= \(\frac { 1 }{ 2 }\) x sum of lengths of parallel sides x height

= \(\frac { 1 }{ 2 }\) x (8.5 + 11.5) x 4.2

= \(\frac { 1 }{ 2 }\) x 20 x 4.2

= 10 x 4.2

= 42 sq. cm

∴ The area of the trapezium is 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.

Solution:

☐PQRS is an isosceles trapezium.

l(PQ) = 7 cm, seg PM ⊥ seg SR,

l(SM) = 3 cm, l(PM) = 4cm

Draw seg QN ⊥ seg SR.

In ☐PMNQ,

seg PQ || seg MN

∠PMN = ∠QNM = 90°

∴ ☐PMNQ is a rectangle.

Opposite sides of a rectangle are congruent.

∴ l(PM) = l(QN) = 4 cm and

l(PQ) = l(MN) = 7 cm

In ∆PMS, m∠PMS = 90°

∴ [l(PS)]² = [l(PM)]² + [l(SM)]² … [Pythagoras theorem]

∴ [l(PS)]² = (4)² + (3)²

∴ [l(PS)]² = 16 + 9 = 25

∴ l(PS) = √25 = 5 cm

…[Taking square root of both sides]

☐PQRS is an isosceles trapezium.

∴ l(PS) = l(QR) = 5 cm

In ∆QNR, m ∠QNR = 90°

∴ [l(QR)]² = [l(QN)]² + [l(NR)]²

… [Pythagoras theorem]

∴ (5)² = (4)² + [l(NR)]²

∴ 25 = 16 + [l(NR)]²

∴ [l(NR)]² = 25 – 16 = 9

∴ l(NR) = √9 = 3 cm

…[Taking square root of both sides]

l(SR) = l(SM) + l(MN) + l(NR)

= 3 + 7 + 3

= 13 cm

Area of a trapezium

= \(\frac { 1 }{ 2 }\) x sum of lengths of parallel sides x height

∴ A(☐PQRS) = \(\frac { 1 }{ 2 }\) x [l(PQ) + l(SR)] x l(PM)

= \(\frac { 1 }{ 2 }\) x (7+ 13) x 4

= \(\frac { 1 }{ 2 }\) x 20 x 4

= 40 sq.cm

∴ The area of ☐PQRS is 40 sq. cm.