Question

find the area of a quadrilateral ABCD in which ab is equal to 3 cm BC is equal to 4 cm CD is equal to 4 cm is equal to 5 cm and ac is equal to 5 cm​

Answer 1

Answer:

Find the area of a quadrilateral ABCD in which AB = 3 cm,

BC = 4 cm,

CD = 4 cm,

DA = 5 cm and

AC = 5 cm.

∴ ΔABC is right angled with ∠B = 90°. = 2 x 4.6 cm2 (approx.) = 9.2 cm2 (approx.)

Answer 2

Answer:

For ΔABC

a = 4 cm

b = 5 cm

c = 3 cm

∵ a2 + c2 = b2

For ΔABCa = 4 cmb = 5 cmc = 3 cm∵ a2 + c2 = b2

∴ ΔABC is righ

∴ ΔABC is right angled with ∠B = 90°.

∴ Area of right triangle ABC

equals 1 half cross times space Base space cross times space Height

equals space 1 half cross times 3 cross times 4 equals 6 space cm squared

For ΔACD

a = 4 cm b = 5 cm

c = 5 cm

therefore space space space space space space space straight s space equals space fraction numerator straight a plus straight b plus straight c over denominator 2 end fraction

space space space space space space space space space space space equals space fraction numerator 4 plus 5 plus 5 over denominator 2 end fraction equals 14 over 2 equals 7 space cm

∴ Area of the ΔACD

equals space square root of straight s left parenthesis straight s minus straight a right parenthesis left parenthesis straight s minus straight b right parenthesis left parenthesis straight s minus straight c right parenthesis end root

equals space square root of 7 left parenthesis 7 minus 4 right parenthesis left parenthesis 7 minus 5 right parenthesis left parenthesis 7 minus 5 right parenthesis end root

equals space square root of 7 left parenthesis 3 right parenthesis left parenthesis 2 right parenthesis left parenthesis 2 right parenthesis end root equals 2 square root of 21 space cm squared

= 2 x 4.6 cm2 (approx.)

= 9.2 cm2 (approx.)

∴ Area of the quadrilateral ABCD

= Area of ΔABC + Area of ΔACD

= 6 cm2 + 9.2 cm2

= 15.2 cm2, (approx.)