Maharashtra Board Class 8 Maths Chapter 15 Area Set 15.3 Solutions

 Question 1. In the given figure, ABCD is a trapezium, side AB || side DC, l(AB) = 13 cm, l(DC) = 9 cm, l(AD) = 8 cm, find the area ABCD.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक समलम्ब चतुर्भुज ABCD को दर्शाता है, जिसमें भुजा AB, भुजा DC के समानांतर है। इसमें AD भुजा समलम्ब चतुर्भुज की ऊँचाई को दर्शाती है।
Answer:
ABCD is a trapezium, side AB || side DC,
l(AB) = 13 cm, l(DC) = 9 cm, l(AD) = 8 cm,
Area of a trapezium = \( \frac{1}{2} \times \) sum of lengths of parallel sides \( \times \) height
\(\therefore A(\square ABCD) = \frac{1}{2} \times [l(AB) + l(DC)] \times l(AD)\)
\( = \frac{1}{2} \times (13 + 9) \times 8 \)
\( = \frac{1}{2} \times 22 \times 8 \)
= 11 \( \times \) 8
= 88 sq.cm
\(\therefore\) The area of \(\square ABCD\) is 88 sq. cm.
[Note: The question is modified.]
In simple words: To find the area of the trapezium, we sum the lengths of its parallel sides (AB and DC), multiply by the height (AD), and then divide by two. This gives a total area of 88 square centimeters for the given dimensions.

🎯 Exam Tip: Remember the formula for the area of a trapezium: half the sum of parallel sides multiplied by the height. Clearly label each given dimension in your solution.

Question 2. Length of the two parallel sides of a trapezium are 8.5 cm and 11.5 cm respectively and its height is 4.2 cm, find its area.
Answer:
Length of the two parallel sides of a trapezium are 8.5 cm and 11.5 cm and its height is 4.2 cm.
Area of a trapezium
\( = \frac{1}{2} \times \) sum of lengths of parallel sides \( \times \) height
\( = \frac{1}{2} \times (8.5 + 11.5) \times 4.2 \)
\( = \frac{1}{2} \times 20 \times 4.2 \)
= 10 \( \times \) 4.2
= 42 sq. cm
\(\therefore\) The area of the trapezium is 42 sq. cm.
In simple words: The area of the trapezium is calculated by taking half the sum of the lengths of its two parallel sides (8.5 cm and 11.5 cm) and multiplying it by its height (4.2 cm), resulting in 42 square centimeters.

🎯 Exam Tip: Accurately substitute the given lengths of parallel sides and height into the trapezium area formula. Double-check your arithmetic, especially with decimal numbers, for precision.

Question 3. PQRS is an isosceles trapezium. l(PQ) = 7 cm, seg PM ⊥ seg SR, l(SM) = 3 cm. Distance between two parallel sides is 4 cm, find the area of PQRS.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक समद्विबाहु समलम्ब चतुर्भुज PQRS को दर्शाता है। इसमें PQ भुजा SR के समानांतर है। PM, SR पर लंबवत है, जिसकी लंबाई 4 इकाई है, SM की लंबाई 3 इकाई है, और PQ की लंबाई 7 इकाई है।
Answer:
PQRS is an isosceles trapezium.
l(PQ) = 7 cm, seg PM \( \perp \) seg SR,
l(SM) = 3 cm, l(PM) = 4cm
Draw seg QN \( \perp \) seg SR.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र समद्विबाहु समलम्ब चतुर्भुज PQRS को दर्शाता है जिसमें QN भी SR पर लंबवत है। PMNQ एक आयत के रूप में दिखाया गया है, जिसमें PM की लंबाई 4, SM की लंबाई 3, और PQ की लंबाई 7 इकाई है।
In \(\square\)PMNQ,
seg PQ || seg MN
\(\angle PMN = \angle QNM = 90^\circ\)
\(\therefore\) PMNQ is a rectangle.
Opposite sides of a rectangle are congruent.
\(\therefore\) l(PM) = l(QN) = 4 cm and
l(PQ) = l(MN) = 7 cm
In \(\triangle\)PMS, \(m\angle PMS = 90^\circ\)
\(\therefore [l(PS)]^2 = [l(PM)]^2 + [l(SM)]^2\) ... [Pythagoras theorem]
\(\therefore [l(PS)]^2 = (4)^2 + (3)^2\)
\(\therefore [l(PS)]^2 = 16 + 9 = 25\)
\(\therefore l(PS) = \sqrt{25} = 5\) cm
...[Taking square root of both sides]
PQRS is an isosceles trapezium.
\(\therefore\) l(PS) = l(QR) = 5 cm
In \(\triangle\)QNR, \(m\angle QNR = 90^\circ\)
\(\therefore [l(QR)]^2 = [l(QN)]^2 + [l(NR)]^2\)
...[Pythagoras theorem]
\(\therefore (5)^2 = (4)^2 + [l(NR)]^2\)
\(\therefore 25 = 16 + [l(NR)]^2\)
\(\therefore [l(NR)]^2 = 25 - 16 = 9\)
\(\therefore l(NR) = \sqrt{9} = 3\) cm
...[Taking square root of both sides]
l(SR) = l(SM) + l(MN) + l(NR)
= 3 + 7 + 3
= 13 cm
Area of a trapezium
\( = \frac{1}{2} \times \) sum of lengths of parallel sides \( \times \) height
\(\therefore A(\square PQRS) = \frac{1}{2} \times [l(PQ) + l(SR)] \times l(PM)\)
\( = \frac{1}{2} \times (7 + 13) \times 4 \)
\( = \frac{1}{2} \times 20 \times 4 \)
= 40 sq.cm
\(\therefore\) The area of \(\square PQRS\) is 40 sq. cm.
In simple words: For the isosceles trapezium PQRS, we first determine the lengths of all sides, including the non-parallel side SR, by using the properties of rectangles and the Pythagorean theorem. Once all side lengths and the height are known, the area is calculated using the standard trapezium formula, yielding 40 square centimeters.

🎯 Exam Tip: For complex geometry problems involving trapeziums and right-angled triangles, carefully break down the figure into simpler shapes. Use the Pythagorean theorem to find unknown lengths and properties of rectangles to deduce congruent sides, which are crucial steps before calculating the area.