Question

Find the perimeter and area of the quadrilateral ABCD in which AB=17cm,AD=9cm,CD=12cm,angle ACB=90degree ,AC=15cm.

Answer 1

hello friend!!


by Pythagoras theorem, 
BC = √AB²−√AC²
=√17² - √15²
=√289 - 225BC
= √64
= 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  =S√(S−a)(S−b)(S−c)  (Heron's formula)
=18√(18−15)(18−12)(18−9)
=√18×3×6×9 
=√18×18×3×3
= 18×3
=54cm²

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

hope it helps!! 

Answer 2

Hiiiiiii friends........
here is your answer..........
in triangle acb. area of triangle=1/2×b×h

1/2×8×15=60cm
and now

in triangle ACD let,a=15,b=12,c=9
then s=1/2(a+b+c)

1/2(15+12+9)=36/2=18
s=18

area of ACD=under root√s(s-a)(s-b)(s-c)
√18(18-15)(18-12)(18-9)
=54cm

so the total area =54+60=114cm

now the perimeter of quadrilateral
sum of all out line side=46cm

hope it help you