Find the area of the trapezium PQRS with height PQ given in figure - Mathematics

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Sum

Find the area of the trapezium PQRS with height PQ given in figure

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Solution

We have, trapezium PQRD, in which draw a line RT perpendicular to PS.

Where, side, ST = PS – TP = 12 – 7 = 5 m.  ......[∵ TP = PQ = 7 m]

In right angled ΔSTR, (SR)2 = (ST)2 + (TR)2   .....[By Pythagoras theorem]

⇒ (13)2 = (5)2 + (TR)2  .......[By Pythagoras theorem]

⇒ (TR)2 = 169 – 25

⇒ (TR)2 = 144

∴ TR = 12 m  .....[Taking positive square root because length is always positive]

Now, Area of ΔSTR = `1/2 xx TR xx TS`  .....[∵ Area of triangle = `1/2` (base × height)]

= `1/2 xx 12 xx 5` = 30 m2

Now, Area of rectangle PQRT = PQ x RQ = 12 × 7  ......[∵ Area of a rectangle = length × breadth]

= 84 m2   .....[∵ PQ = TR = 12 m]

∴ Area of trapezium = Area of ΔSTR + Area of rectangle PQRT

= 30 + 84

= 114 m2

Hence, the area of trapezium is 114 m2.

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Chapter 12: Heron's Formula - Exercise 12.3 [Page 117]

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NCERT Exemplar Mathematics Class 9
Chapter 12 Heron's Formula
Exercise 12.3 | Q 10 | Page 117

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