Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 15.2 8th Std Maths Answers Solutions Chapter 15 Area.

## Practice Set 15.2 8th Std Maths Answers Chapter 15 Area

Question 1.

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

Solution:

Lengths of the diagonals of a rhombus are 15 cm and 24 cm.

Area of a rhombus

= \(\frac { 1 }{ 2 }\) × product of lengths of diagonals

= \(\frac { 1 }{ 2 }\) × 15 × 24

= 15 × 12

= 180 sq.cm

∴ The area of the rhombus is 180 sq. cm.

Question 2.

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

Solution:

Lengths of the diagonals of a rhombus are 16.5 cm and 14.2 cm.

Area of a rhombus

= \(\frac { 1 }{ 2 }\) × product of lengths of diagonals

= \(\frac { 1 }{ 2 }\) × 16.5 × 14.2

= 16.5 × 7.1

= 117.15 sq cm

∴ The area of the rhombus is 117.15 sq. cm.

Question 3.

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

Solution:

Let ₹ABCD be the rhombus. Diagonals AC and BD intersect at point E.

l(AC) = 48 cm …(i)

l(AE) = \(\frac { 1 }{ 2 }l(AC)\) …[Diagonals of a rhombus bisect each other]

= \(\frac { 1 }{ 2 }\) × 48 …[From (i)]

= 24 cm …(ii)

Perimeter of rhombus = 100 cm …[Given]

Perimeter of rhombus = 4 × side

∴ 100 = 4 × l(AD)

∴ l(AD) = \(\frac { 100 }{ 4 }\) = 25 cm …(iii)

In ∆ADE,

m∠AED = 90° …[Diagonals of a rhombus are perpendicular to each other]

∴ [l(AD)]² = [l(AE)]² + [l(DE)]² … [Pythagoras theorem]

∴ (25)² = (24)² + l(DE)² … [From (ii) and (iii)]

∴ 625 = 576 + l(DE)²

∴ l(DE)² = 625 – 576

∴ l(DE)² = 49

∴ l(DE) = √49

… [Taking square root of both sides]

l(DE) = 7 cm …(iv)

l(DE) = \(\frac { 1 }{ 2 } l(BD)\) ….[Diagonals of a rhombus bisect each other]

∴ 7 = \(\frac { 1 }{ 2 } l(BD)\) …[From (iv)]

∴ l(BD) = 7 × 2

= 14 cm …(v)

Area of a rhombus

= \(\frac { 1 }{ 2 }\) × product of lengths of diagonals

= \(\frac { 1 }{ 2 }\) × l(AC) × l(BD)

= \(\frac { 1 }{ 2 }\) × 48 × 14 … [From (i) and (v)]

= 48 × 7

= 336 sq.cm

∴ The area of the quadrilateral is 336 sq.cm.

Question 4.

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

Solution:

Let ₹ABCD be the rhombus.

Diagonals AC and BD intersect at point E.

l(AC) = 30 cm …(i)

and A(₹ABCD) = 240 sq. cm .. .(ii)

Area of the rhombus = \(\frac { 1 }{ 2 }\) × product of lengths of diagonal

∴ 240 = \(\frac { 1 }{ 2 }\) × l(AC) x l(BD) …[From (ii)]

∴ 240 = \(\frac { 1 }{ 2 }\) × 30 × l(BD) …[From (i)]

∴ l(BD) = \(\frac { 240\times 2 }{ 30 }\)

∴ l(BD) = 8 × 2 = 16 cm …(iii)

Diagonals of a rhombus bisect each other.

∴ l(AE) = \(\frac { 1 }{ 2 }l(AC)\)

= \(\frac { 1 }{ 2 }\) × 30 … [From (i)]

= 15 cm …(iv)

and l(DE) = \(\frac { 1 }{ 2 }l(BD)\)

= \(\frac { 1 }{ 2 }\) × 16

= 8 cm

In ∆ADE,

m∠AED = 90°

…[Diagonals of a rhombus are perpendicular to each other]

∴[l(AD)]² = [l(AE)]² + [l(DE)]²

…[Pythagoras theorem]

∴l(AD)² = (15)² + (8)² … [From (iv) and (v)]

= 225 + 64

∴l(AD)² = 289

∴l(AD) = √289

…[Taking square root of both sides]

∴l(AD) = 17 cm

Perimeter of rhombus = 4 × side

= 4 × l(AD)

= 4 × 17

= 68 cm

∴The perimeter of the rhombus is 68 cm.