Question

# Find the perimeter and area of the quadrilateral ABCD in which AB = 21 cm, ∠BAC=90∘, AC = 20 cm, CD = 42 cm and AD = 34 cm.

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Solution

## ANSWER: In right angled ∆ ABC, BC2=AB2+AC2 (Pythagoras Theorem) ⇒ BC2=212+202 ⇒ BC2= 441 + 400 ⇒ BC2 = 841 ⇒ BC = 29 cm Area of ∆ ABC = 12×AB×AC = 12×21×20 = 210 cm2 ....(1) In ∆ ACD, The sides of the triangle are of length 20 cm, 34 cm and 42 cm. ∴ Semi-perimeter of the triangle is s=20+34+422=962=48 cm ∴ By Heron's formula, Area of ∆ACD=√s(s−a)(s−b)(s−c) = √48(48−20)(48−34)(48−42) = √48∗28∗14∗6 =336 cm2 Thus, Area of quadrilateral ABCD = Area of ∆ ABC + Area of ∆ ACD = (210 + 336) cm2 = 546 cm2 Also, Perimeter of quadrilateral ABCD = (34 + 42 + 29 + 21) cm = 126 cm Hence, the perimeter and area of quadrilateral ABCD is 126 cm and 546 cm2 , respectively.

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