Question

Find the area of the quadrilateral ABCD in which AB = 9 cm, BC = 40 cm, DC = 28 cm,

AD = 15 cm, diagonal AC = 41 cm. and ∆ABC = 90°​

Answer 1

Step-by-step explanation:

Area of quadrilateral=Ar∆ABC+Ar∆ACD

In ∆ABC, using pythagoras theorem

${ac}^{2} = {ab}^{2} + {bc}^{2} \\ ac = \sqrt{ {9 }^{2} } + \sqrt{40 {}^{2} } \\ ac = \sqrt{1681 } \\ ac = 41m$

Ar∆ACD=1/2×AB×BC

=1/2×9×40

=1800m sq.

Ar∆ACD=√s(s-a)(s-b)(s-c)

[s=a+b+c/2]

[s=15+28+41/2=42m]

Ar∆ACD=√42(42-15)(42-28)(42-51)

=√42×27×14×1

=126m sq.

therefore,Area of quadrilateral

=180+126

=306m sq.