Maharashtra Board Class 8 Maths Chapter 15 Area Set 15.4 Solutions

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
= \( \sqrt{s(s-a)(s-b)(s-c)} \)
= \( \sqrt{63(63-45)(63-39)(63-42)} \)
= \( \sqrt{63 \times 18 \times 24 \times 21} \)
= \( \sqrt{7 \times 9 \times 2 \times 9 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7} \)
= \( \sqrt{7^2 \times 9^2 \times 2^2 \times 2^2 \times 3^2} \)
= \( 7 \times 9 \times 2 \times 2 \times 3 \)
= 756 sq. cm
∴ The area of the triangle is 756 sq.cm.
In simple words: To find the area of a triangle given its sides, first calculate the semi-perimeter using (a+b+c)/2. Then apply Heron's formula, which is the square root of s(s-a)(s-b)(s-c), to get the area.

🎯 Exam Tip: Remember Heron's formula for finding the area of a triangle when only the side lengths are known, as it is a fundamental concept for such problems.

Question 2. Look at the measures shown in the given figure and find the area of □PQRS.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक चतुर्भुज PQRS है जिसे दो त्रिभुजों, ΔPSR और ΔPQR, में विभाजित किया गया है। ΔPSR एक समकोण त्रिभुज है जिसमें ∠S 90 डिग्री है, जिसकी भुजाएँ PS = 36m और SR = 15m हैं। ΔPQR की भुजाएँ PQ = 56m, QR = 25m हैं, और इसकी तीसरी भुजा PR है जिसे ΔPSR से गणना करके ज्ञात किया जा सकता है।
Solution:
A(PQRS) = Α(ΔPSR) + Α(ΔPQR)
In ΔPSR, l(PS) = 36 m, l(SR) = 15 m
Α(ΔPSR)
= \( \frac{1}{2} \) 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 ΔPSR, 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 ΔPQR, a = 56m, b = 25m, c = 39m
Semiperimeter of ΔPQR = s = \( \frac{1}{2}(a + b + c) \)
= \( \frac{56 + 25 + 39}{2} \)
= \( \frac{120}{2} \)
= 60
∴ A(ΔPQR) = \( \sqrt{s(s-a)(s-b)(s-c)} \)
= \( \sqrt{60(60-56)(60-25)(60-39)} \)
= \( \sqrt{60 \times 4 \times 35 \times 21} \)
= \( \sqrt{3 \times 4 \times 5 \times 4 \times 5 \times 7 \times 3 \times 7} \)
= \( \sqrt{3^2 \times 4^2 \times 5^2 \times 7^2} \)
= \( 3 \times 4 \times 5 \times 7 \)
= 420 sq. m
A(PQRS) = Α(ΔPSR) + Α(ΔPQR)
= 270 + 420
= 690 sq. m
∴ The area of PQRS is 690 sq.m
In simple words: To find the area of the quadrilateral, divide it into two triangles. Calculate the area of the right-angled triangle using ½ × base × height, and find its hypotenuse using the Pythagorean theorem. Use this hypotenuse as a side for the second triangle and apply Heron's formula to find its area. Finally, sum the areas of both triangles.

🎯 Exam Tip: Decomposing complex shapes into simpler geometric figures (like triangles) is a common strategy. Ensure accurate application of Pythagorean theorem and Heron's formula.

Question 3. Some measures are given in the figure, find the area of ABCD.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक चतुर्भुज ABCD है जिसे एक विकर्ण BD द्वारा दो त्रिभुजों, ΔBAD और ΔBCD, में विभाजित किया गया है। ΔBAD एक समकोण त्रिभुज है जिसमें ∠A 90 डिग्री है, जिसकी भुजाएँ AB = 40m और AD = 9m हैं। ΔBCD की भुजा CD = 60m है और BT = 13m इसकी ऊँचाई है, जहाँ BT भुजा CD पर लंब है।
Solution:
A(ABCD) = Α(ΔΒAD) + Α(ΔBCD)
In ΔBAD, m∠BAD = 90°, l(AB) = 40m, l(AD) = 9m
A(ΔBAD) = \( \frac{1}{2} \) 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 ΔBDC, l(BT) = 13m, l(CD) = 60m
A(ΔBDC) = \( \frac{1}{2} \) 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) = Α(ΔΒAD) + A(ΔBDC)
= 180 + 390
= 570 sq. m
∴ The area of ABCD is 570 sq.m.
In simple words: To find the area of the quadrilateral, split it into two triangles. Calculate the area of the right-angled triangle using ½ × base × height. Then, calculate the area of the second triangle using ½ × base × height (where the height is perpendicular to the base). Add both areas to get the total area of the quadrilateral.

🎯 Exam Tip: For quadrilaterals, dividing them into triangles simplifies area calculation. Identify right angles or perpendicular heights to use the basic area formula for triangles efficiently.