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☐ 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 - SSC (English Medium) Class 8 - Mathematics

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Question

☐ 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

Solution

Draw a perpendicular from Q to line MR. Where it meets the line MR name it point N. 
MN = PQ = 7 cm
In Δ PMS, 

PM² + SM² = PS²

⇒ 4² + 3² = PS²

⇒ PS² = 16 + 9 = 25

⇒ PS = 5cm 

PQRS is an isosceles trapezium so, PS = QR = 5 cm
PM = QN = 4 cm
So, NR = SM = 3 cm
SR = SM + MN + NR = 3 + 7 + 3 = 13 cm
Area of trapezium PQRS = `1/2 xx` (sum of parallel sides) x height 

=`1/2 xx` (7 + 13) x 4

= 40 cm²

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APPEARS IN

 Balbharati Solution for Balbharati Class 8 Mathematics (2019 to Current)
Chapter 15: Area
Practice Set 15.3 | Q: 3 | Page no. 99
Solution ☐ 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 Concept: Area of a Trapezium.
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