Maharashtra Board Class 8 Maths Chapter 15 Area Set 15.2 Solutions

Question 1. Lengths of the diagonals of a rhombus are 15 cm and 24 cm, find its area.
Answer:
Solution:
Lengths of the diagonals of a rhombus are 15 cm and 24 cm.
Area of a rhombus
= \( \frac{1}{2} \) \( \times \) product of lengths of diagonals
= \( \frac{1}{2} \) \( \times \) 15 \( \times \) 24
= 15 \( \times \) 12
= 180 sq.cm
.. The area of the rhombus is 180 sq. cm.
In simple words: The area of a rhombus is calculated by taking half of the product of its diagonal lengths. Here, with diagonals of 15 cm and 24 cm, the area is 180 sq. cm.

🎯 Exam Tip: Remember the formula for the area of a rhombus is crucial for direct application in such problems. Ensure units are correctly stated in the final answer.

Question 2. Lengths of the diagonals of a rhombus are 16.5 cm and 14.2 cm, find its area.
Answer:
Solution:
Lengths of the diagonals of a rhombus are 16.5 cm and 14.2 cm.
Area of a rhombus
= \( \frac{1}{2} \) \( \times \) product of lengths of diagonals
= \( \frac{1}{2} \) \( \times \) 16.5 \( \times \) 14.2
= 16.5 x 7.1
= 117.15 sq cm
.. The area of the rhombus is 117.15 sq. cm.
In simple words: Using the formula that area equals half the product of diagonals, for diagonals 16.5 cm and 14.2 cm, the rhombus area is 117.15 sq. cm.

🎯 Exam Tip: Pay close attention to decimal calculations for accuracy. Clearly write down each step of the formula application.

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?
Answer:
Solution:
Let ABCD be the rhombus. Diagonals AC and BD intersect at point E.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समचतुर्भुज ABCD दिखाया गया है, जिसके विकर्ण AC और BD एक दूसरे को बिंदु E पर काटते हैं। यह समचतुर्भुज के गुणों को समझने में मदद करता है।
I(AC) = 48 cm ...(i)
I(AE) = \( \frac{1}{2} \) I(AC) ...[Diagonals of a rhombus bisect each other]
= \( \frac{1}{2} \) \( \times \) 48 ...[From (i)]
= 24 cm ...(ii)
Perimeter of rhombus = 100 cm ...[Given]
Perimeter of rhombus = 4 \( \times \) side
.. 100 = 4 \( \times \) I(AD)
.. I(AD) = \( \frac{100}{4} \) = 25 cm ...(iii)
In \( \triangle \)ADE,
m\( \angle \)AED = 90° ...[Diagonals of a rhombus are perpendicular to each other]
.. [I(AD)]\( ^2 \) = [I(AE)]\( ^2 \) + [I(DE)]\( ^2 \) ... [Pythagoras theorem]
.. (25)\( ^2 \) = (24)\( ^2 \) + I(DE)\( ^2 \) ... [From (ii) and (iii)]
.. 625 = 576 + I(DE)\( ^2 \)
.. I(DE)\( ^2 \) = 625 - 576
.. I(DE)\( ^2 \) = 49
.. I(DE) = \( \sqrt{49} \)
...[Taking square root of both sides]
I(DE) = 7 cm ...(iv)
I(DE) = \( \frac{1}{2} \) I(BD)....[Diagonals of a rhombus bisect each other]
.. 7 = \( \frac{1}{2} \) I(BD) ...[From (iv)]
.. I(BD) = 7 \( \times \) 2
= 14 cm ...(v)
Area of a rhombus
= \( \frac{1}{2} \) \( \times \) product of lengths of diagonals
= \( \frac{1}{2} \) \( \times \) I(AC) \( \times \) I(BD)
= \( \frac{1}{2} \) \( \times \) 48 \( \times \) 14 ... [From (i) and (v)]
= 48 \( \times \) 7
= 336 sq.cm
.. The area of the quadrilateral is 336 sq.cm.
In simple words: Given the perimeter, we find the side length. Using the property that diagonals bisect each other perpendicularly, we apply the Pythagorean theorem to find the length of the other diagonal. Finally, we calculate the area using the lengths of both diagonals.

🎯 Exam Tip: This question combines perimeter, properties of diagonals, and the Pythagorean theorem. Ensure each step is clearly linked to a geometric property or formula for full marks.

Question 4. If length of a diagonal of a rhombus is 30 cm and its area is 240 sq.cm, find its perimeter.
Answer:
Solution:
Let ABCD be the rhombus.
Diagonals AC and BD intersect at point E.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समचतुर्भुज ABCD दिखाया गया है, जिसके विकर्ण AC और BD एक दूसरे को बिंदु E पर काटते हैं। यह समचतुर्भुज के गुणों को समझने में मदद करता है।
I(AC) = 30 cm ...(i)
and A(ABCD) = 240 sq. cm .(ii)
Area of the rhombus = \( \frac{1}{2} \) \( \times \) product of lengths of diagonal
.. 240 = \( \frac{1}{2} \) \( \times \) I(AC) x I(BD) ...[From (ii)]
.. 240 = \( \frac{1}{2} \) \( \times \) 30 \( \times \) I(BD) ...[From (i)]
.. I(BD) = \( \frac{240 \times 2}{30} \)
.. I(BD) = 8 \( \times \) 2 = 16 cm ...(iii)
Diagonals of a rhombus bisect each other.
.. I(AE) = \( \frac{1}{2} \) I(AC)
= \( \frac{1}{2} \) \( \times \) 30 ... [From (i)]
= 15 cm ...(iv)
and I(DE) = \( \frac{1}{2} \) I(BD)
= \( \frac{1}{2} \) \( \times \) 16
= 8 cm
In \( \triangle \)ADE,
m\( \angle \)AED = 90°
...[Diagonals of a rhombus are perpendicular to each other]
..[I(AD)]\( ^2 \) = [I(AE)]\( ^2 \) + [I(DE)]\( ^2 \)
..[Pythagoras theorem]
..I(AD)\( ^2 \) = (15)\( ^2 \) + (8)\( ^2 \) ... [From (iv) and (v)]
= 225 + 64
..I(AD)\( ^2 \) = 289
..I(AD) = \( \sqrt{289} \)
...[Taking square root of both sides]
..I(AD) = 17 cm
Perimeter of rhombus = 4 \( \times \) side
= 4 \( \times \) I(AD)
= 4 x 17
= 68 cm
..The perimeter of the rhombus is 68 cm.
In simple words: Given the area and one diagonal, we first find the length of the second diagonal using the area formula. Then, applying properties of rhombus diagonals and the Pythagorean theorem to one of the right-angled triangles formed, we find the side length. Finally, the perimeter is calculated as four times the side length.

🎯 Exam Tip: This problem requires a multi-step approach. Clearly state each formula used and the property of the rhombus applied at each step to demonstrate a thorough understanding.