Question

Find the perimeter and area of a quadrilateral ABCD in which BC = 12 cm, CD = 9 cm, BD = 15 cm, DA = 17 cm and $\angle ABD={90}^{\circ }.$

Answer 1

by Pythagoras theorem,
$BC=\sqrt{A{B}^2-A{C}^2}$
$BC=\sqrt{{17}^2-{15}^2}$
$BC=\sqrt{289-225}$
$BC=\sqrt{64}$

BC = 8 cm

Perimeter of quadrilateral ABCD = 17 + 9 + 12 + 8 = 46 cm

Area of triangle △ABC=1/2×base×height
△ABC=12×8 x 15
= 60 cm²

NOW, for the area of the triangle, ACD
we will take a = 15 cm, b = 12 cm and c = 9 cm

S = 12(a+b+c)

S = 12(15+12+9)=36/2=18cm

Area of one triangular =$\sqrt{S(S-a)(S-b)(S-c)}$ (Heron's formula)
=$\sqrt{18(18-15)(18-12)(18-9)}$
=$\sqrt{18\times 3\times 6\times 9}$
=$\sqrt{18\times 18\times 3\times 3}$
= 18×3=54cm²

area of quadrilateral ABCD= area of △ABC + area of △ACD
= 60 + 54 = 114 cm²