Question

name the quadrilateral abcd, where ab= 5cm and cd=8cm, where angle a + angle d= 108 degrees? pls answer fast! thanks!​

Answer 1

Join AC

Since AB =6, AC =8 and Angle ABC is 90 degrees, ABC is a right angle triangle with base 8 and height 6

Area of Triangle ABC will be 1/2 x 8 x 6 = 24————-(1)

By theorem of Pythagorus,Hypotenuse AC will be square root of 6^2 + 8^2 =square root of (36 + 64 = 100) which will be 10

Now see the triangle ACD,

AC =10, AD =CD =13 so triangleACD is isosceles Triangle with base AC of 10

Perpendicular from D will bisect the base AC at say E.

Now if we find DE, which will be height of Triangle ACD, we will be able to calculate area of Triangle ACD.

Since the perpendicular drawn from vertex of Isosceles triangle bisects the base, AE =5

Now in Triangle AED, AD is 13 and AE is 5 so DE will be root of 13^2 - 5^2 = root of (169–25=144) which is 12. So in Triangle ACD, AC=10 and height DE =12 so area of triangle will be 1/2 x 10 x 12 = 60 ————(2)

Area of quadrilateral ABCD = Area of Triangle ABC + Area of Triangle ACD = (1 ) + (2) = 24 + 60 = 84 square cm.