Maharashtra Board Class 8 Maths Chapter 15 Area Set 15.5 Solutions

Question 1. Find the areas of given plots. (All measures are in meters.)
(i)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक भूखंड QPTSR को दर्शाता है जो एक समकोण त्रिभुज QAP, एक समलंब QACR, एक समकोण त्रिभुज ARCS और एक अन्य समकोण त्रिभुज APTS में विभाजित है। विभिन्न खंडों के लिए प्रमुख माप मीटर में दिए गए हैं।
(ii)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक भूखंड ABCDE को दर्शाता है जो एक त्रिभुज ABE, एक त्रिभुज BCE और एक त्रिभुज CED में विभाजित है। विभिन्न भुजाओं की लंबाई और BAE पर एक समकोण इंगित किए गए हैं।
Answer:
Solution:
i. Here, ΔQAP, ΔRCS are right angled triangles and □QACR is a trapezium.
In ΔQAP, l(AP) = 30 m, l(QA) = 50 m
A(ΔQAP)
\( = \frac{1}{2} \times \text{product of sides forming the right angle} \)
\( = \frac{1}{2} \times l(\text{AP}) \times l(\text{QA}) \)
\( = \frac{1}{2} \times 30 \times 50 \)
\( = 750 \text{ sq. m} \)
In □QACR, l(QA) = 50 m, l(RC) = 25 m,
l(AC) = l(AB) + l(BC)
\( = 30 + 30 = 60 \text{ m} \)
A(□QACR)
\( = \frac{1}{2} \times \text{sum of lengths of parallel sides} \times \text{height} \)
\( = \frac{1}{2} \times [l(\text{QA}) + l(\text{RC})] \times l(\text{AC}) \)
\( = \frac{1}{2} \times (50 + 25) \times 60 \)
\( = \frac{1}{2} \times 75 \times 60 \)
\( = 2250 \text{ sq.m} \)
In ΔRCS, l(CS) = 60 m, l(RC) = 25 m A(ΔRCS)
\( = \frac{1}{2} \times \text{product of sides forming the right angle} \)
\( = \frac{1}{2} \times l(\text{CS}) \times l(\text{RC}) \)
\( = \frac{1}{2} \times 60 \times 25 \)
\( = 750 \text{ sq. m} \)
In ΔPTS, l(TB) = 30 m,
l(PS) = l(PA) + l(AB) + l(BC) + l(CS)
\( = 30 + 30 + 30 + 60 \)
\( = 150 \text{m} \)
A(ΔPTS) \( = \frac{1}{2} \times \text{base} \times \text{height} \)
\( = \frac{1}{2} \times l(\text{PS}) \times l(\text{TB}) \)
\( = \frac{1}{2} \times 150 \times 30 \)
\( = 2250 \text{ sq. m} \)
∴ Area of plot QPTSR = A(ΔQAP) + A(□QACR) + A(ΔRCS) + A(ΔPTS)
\( = 750 + 2250 + 750 + 2250 \)
\( = 6000 \text{ sq. m} \)
∴ The area of the given plot is \( 6000 \text{ sq.m.} \)
ii. In ΔABE, m∠BAE = 90°, l(AB) = 24 m, l(BE) = 30 m
∴ \([l(\text{BE})]^2 = [l(\text{AB})]^2 + [l(\text{AE})]^2\)
...[Pythagoras theorem]
∴ \((30)^2 = (24)^2 + [l(\text{AE})]^2\)
∴ \(900 = 576 + [l(\text{AE})]^2\)
∴ \([l(\text{AE})]^2 = 900 - 576\)
∴ \([l(\text{AE})]^2 = 324\)
∴ \(l(\text{AE}) = \sqrt{324} = 18 \text{ m}\)
...[Taking square root of both sides]
A(ΔABE)
\( = \frac{1}{2} \times \text{product of sides forming the right angle} \)
\( = \frac{1}{2} \times l(\text{AE}) \times l(\text{AB}) \)
\( = \frac{1}{2} \times 18 \times 24 \)
\( = 216 \text{ sq. m} \)
In ΔBCE, a = 30m, b = 28m, c = 26m
Semiperimeter of ΔBCE = s = \(\frac{1}{2}\)(a+b+c)
\( = \frac{30+28+26}{2} \)
\( = \frac{84}{2} \)
\( = 42 \text{ m} \)
A(ΔBCE) = \(\sqrt{s(s-a)(s-b)(s-c)}\)
\( = \sqrt{42(42-30)(42-28)(42-26)} \)
\( = \sqrt{42 \times 12 \times 14 \times 16} \)
\( = \sqrt{2 \times 3 \times 7 \times 2 \times 2 \times 3 \times 2 \times 7 \times 4 \times 4} \)
\( = \sqrt{2^2 \times 2^2 \times 3^2 \times 4^2 \times 7^2} \)
\( = 2 \times 2 \times 3 \times 4 \times 7 \)
\( = 336 \text{ sq. m} \)
In ΔEDC, l(CE) = 28 m, l(DF) = 16 m
A(ΔEDC)
\( = \frac{1}{2} \times \text{base} \times \text{height} \)
\( = \frac{1}{2} \times l(\text{CE}) \times l(\text{DF}) \)
\( = \frac{1}{2} \times 28 \times 16 \)
\( = 224 \text{ sq. m} \)
∴ Area of plot ABCDE
\( = \text{A(ΔABE)} + \text{A(ΔBCE)} + \text{A(ΔEDC)} \)
\( = 216 + 336 + 224 \)
\( = 776 \text{ sq. m} \)
∴ The area of the given plot is \( 776 \text{ sq.m.} \)
[Note: In the given figure, we have taken l(DF) = 16 m]
In simple words: For the first plot (QPTSR), we divided it into three triangles and one trapezium, calculated the area of each component using relevant formulas, and summed them up. For the second plot (ABCDE), we split it into three triangles and used basic area formulas and Heron's formula for the irregular triangle, then added the areas.

🎯 Exam Tip: Accurately identifying the correct geometric shapes within a complex plot and applying the appropriate area formulas (e.g., triangle, trapezium, Heron's formula) is crucial for scoring. Ensure all units are consistent and calculations are precise.