The sides of a quadrilateral are 5 m, 12 m, 14 mand 15 m and the angle

1.	The sides of a quadrilateral are 5 m, 12 m, 14 mand 15 m and the angle between the first two sidesis a right-angle. Find its area.
Answer» AB = 12CM, BC = 5CM, CD = 14CM, AD = 15CM Now, angle formed by BC on AB is 900 By pythagoras theorem, AB2+ BC2= AC2 (12)2+ (5)2= AC2 144 + 25 = AC2 169 = AC2 13 = AC a = 12cm b = 5cm c = 13cm s = 12 + 13 + 5 / 2 = 30 / 2 = 15cm a = √15(15-12)(15-13)(15-5) = √1532*10 = 30 cm2 Area of ADC = 84cm2 Total area = 84cm2+ 30cm2= 114cm2

The sides of a quadrilateral are 5 m, 12 m, 14 mand 15 m and the angle between the first two sidesis a right-angle. Find its area.

Answer»

AB = 12CM, BC = 5CM, CD = 14CM, AD = 15CM

Now, angle formed by BC on AB is 900

By pythagoras theorem,

AB2+ BC2= AC2

(12)2+ (5)2= AC2

144 + 25 = AC2

169 = AC2

13 = AC

a = 12cm

b = 5cm

c = 13cm

s = 12 + 13 + 5 / 2

= 30 / 2

= 15cm

a = √15(15-12)(15-13)(15-5)

= √15*3*2*10

= 30 cm2

Area of ADC = 84cm2

Total area = 84cm2+ 30cm2= 114cm2