Question

3. In quadrilateral ABCD,
Angle B = Angle D = 90°
AB = 7 cm
AC = 25 cm
CD = 20 cm
Find the area of ABCD​

Answer 1

Answer:

Step-by-step explanation:

in Triangle ACD angle D is 90 degrees  so AC2 = AD2 + DC2

AC = 25,  DC = 20 cm

AD2 = AC2 - DC2

= 252 - 202 = 625 - 400 = 225

AD = 15 cm

So area is of ACD is (1/2)AD*DC  which is (1/2)*15*20 = 150 sq cm

Similarly in triangle ABC angle B is 90 degrees

so AC2 = AB2 + BC2

AC =25 cm,  AB = 7 cm

thus

252 = 72 + BC2

BC2 =625 - 49 = 576 sq cm

BC = 24 cm

Area of ABC is (1/2)*AB*BC = (1/2)*7*24 sq cm = 84 sq cm

Area of ABCD is area of ABC + DBC = 84  +  150 sq cm = 234 sq cm