{"id":585650,"date":"2022-03-16T10:50:45","date_gmt":"2022-03-16T10:50:45","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=585650"},"modified":"2022-03-19T04:11:34","modified_gmt":"2022-03-19T04:11:34","slug":"maharashtra-board-solutions-class-8-maths-practice-set-15-3-chapter-15-area","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-8-maths-practice-set-15-3-chapter-15-area\/","title":{"rendered":"Maharashtra Board Solutions Class 8-Maths (Practice Set 15.3): Chapter 15- Area"},"content":{"rendered":"\n

Class 8: Maths Chapter 15 solutions. Complete Class 8 Maths Chapter 15 Notes.<\/p>\n\n\n\n

Maharashtra Board Solutions Class 8-Maths (Practice Set 15.3): Chapter 15- Area<\/h2>\n\n\n\n

Maharashtra Board 8th Maths Chapter 15, Class 8 Maths Chapter 15 solutions<\/p>\n\n\n\n

Question 1.
In the given figure, \u2610ABCD is a trapezium, side AB || side DC, l(AB) = 13 cm, l(DC) = 9 cm, l(AD) = 8 cm, find the area \u2610ABCD.
\"Maharashtra
Solution:<\/strong>
\u2610ABCD is a trapezium, side AB || side DC,
l(AB) = 13 cm, l(DC) = 9 cm, l(AD) = 8 cm,
Area of a trapezium = 12 x sum of lengths of parallel sides x height
\u2234 A (\u2610ABCD) = 12 x [l(AB) + l(DC)] x l(AD)
= 12 x (13 + 9) x 8
= 12 x 22 x 8
= 11 x 8
= 88 sq.cm
\u2234 The area of \u2610ABCD is 88 sq. cm.
[Note: The question is modified.]<\/p>\n\n\n\n

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.
Solution:<\/strong>
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
= 12 x sum of lengths of parallel sides x height
= 12 x (8.5 + 11.5) x 4.2
= 12 x 20 x 4.2
= 10 x 4.2
= 42 sq. cm
\u2234 The area of the trapezium is 42 sq. cm.<\/p>\n\n\n\n

Question 3.
\u2610PQRS is an isosceles trapezium. l(PQ) = 7 cm, seg PM \u22a5 seg SR, l(SM) = 3 cm. Distance between two parallel sides is 4 cm, find the area of \u2610PQRS.
\"Maharashtra
Solution:<\/strong>
\u2610PQRS is an isosceles trapezium.
l(PQ) = 7 cm, seg PM \u22a5 seg SR,
l(SM) = 3 cm, l(PM) = 4cm
Draw seg QN \u22a5 seg SR.
In \u2610PMNQ,
seg PQ || seg MN
\u2220PMN = \u2220QNM = 90\u00b0
\u2234 \u2610PMNQ is a rectangle.
\"Maharashtra
Opposite sides of a rectangle are congruent.
\u2234 l(PM) = l(QN) = 4 cm and
l(PQ) = l(MN) = 7 cm
In \u2206PMS, m\u2220PMS = 90\u00b0
\u2234 [l(PS)]\u00b2 = [l(PM)]\u00b2 + [l(SM)]\u00b2 \u2026 [Pythagoras theorem]
\u2234 [l(PS)]\u00b2 = (4)\u00b2 + (3)\u00b2
\u2234 [l(PS)]\u00b2 = 16 + 9 = 25
\u2234 l(PS) = \u221a25 = 5 cm
\u2026[Taking square root of both sides]
\u2610PQRS is an isosceles trapezium.
\u2234 l(PS) = l(QR) = 5 cm
In \u2206QNR, m \u2220QNR = 90\u00b0
\u2234 [l(QR)]\u00b2 = [l(QN)]\u00b2 + [l(NR)]\u00b2
\u2026 [Pythagoras theorem]
\u2234 (5)\u00b2 = (4)\u00b2 + [l(NR)]\u00b2
\u2234 25 = 16 + [l(NR)]\u00b2
\u2234 [l(NR)]\u00b2 = 25 \u2013 16 = 9
\u2234 l(NR) = \u221a9 = 3 cm
\u2026[Taking square root of both sides]
l(SR) = l(SM) + l(MN) + l(NR)
= 3 + 7 + 3
= 13 cm
Area of a trapezium
= 12 x sum of lengths of parallel sides x height
\u2234 A(\u2610PQRS) = 12 x [l(PQ) + l(SR)] x l(PM)
= 12 x (7+ 13) x 4
= 12 x 20 x 4
= 40 sq.cm
\u2234 The area of \u2610PQRS is 40 sq. cm.<\/p>\n\n\n\n

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Maharashtra Board Solutions Class 8-Maths (Practice Set 15.3): Chapter 15- Area<\/p>\n\n\n\n

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Chapterwise Maharashtra Board Solutions Class 8 Maths :<\/strong><\/h2>\n\n\n\n