Practice Set 15.2 8th Std Maths Answers Chapter 15 Area
Question 1
Maharashtra Board Solution
Lengths of the diagonals of a rhombus are 15 cm and 24 cm, find its area.
Solution & Step-by-Step Answer:
Lengths of the diagonals of a rhombus are 15 cm and 24 cm. Area of a rhombus = × product of lengths of diagonals = × 15 × 24 = 15 × 12 = 180 sq.cm ∴ The area of the rhombus is 180 sq. cm.
Question 2
Maharashtra Board Solution
Lengths of the diagonals of a rhombus are 16.5 cm and 14.2 cm, find its area.
Solution & Step-by-Step Answer:
Lengths of the diagonals of a rhombus are 16.5 cm and 14.2 cm. Area of a rhombus = × product of lengths of diagonals = × 16.5 × 14.2 = 16.5 × 7.1 = 117.15 sq cm ∴ The area of the rhombus is 117.15 sq. cm.
Question 3
Maharashtra Board Solution
If perimeter of a rhombus is 100 cm and length of one diagonal is 48 cm, what is the area of the quadrilateral?
Solution & Step-by-Step Answer:
Let ₹ABCD be the rhombus. Diagonals AC and BD intersect at point E. l(AC) = 48 cm …(i) l(AE) = …[Diagonals of a rhombus bisect each other] = × 48 …[From (i)] = 24 cm …(ii) Perimeter of rhombus = 100 cm …[Given] Perimeter of rhombus = 4 × side ∴ 100 = 4 × l(AD) ∴ l(AD) = = 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) = ….[Diagonals of a rhombus bisect each other] ∴ 7 = …[From (iv)] ∴ l(BD) = 7 × 2 = 14 cm …(v) Area of a rhombus = × product of lengths of diagonals = × l(AC) × l(BD) = × 48 × 14 … [From (i) and (v)] = 48 × 7 = 336 sq.cm ∴ The area of the quadrilateral is 336 sq.cm.
Question 4
Maharashtra Board Solution
If length of a diagonal of a rhombus is 30 cm and its area is 240 sq.cm, find its perimeter.
Solution & Step-by-Step Answer:
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 = × product of lengths of diagonal ∴ 240 = × l(AC) x l(BD) …[From (ii)] ∴ 240 = × 30 × l(BD) …[From (i)] ∴ l(BD) = ∴ l(BD) = 8 × 2 = 16 cm …(iii) Diagonals of a rhombus bisect each other. ∴ l(AE) = = × 30 … [From (i)] = 15 cm …(iv) and l(DE) = = × 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.