find the area of of a quadrilateral whose sides measure 9 cm,40cm,28cmand 15cm. the angle between the first twosides of the quadrilateeral is a right angle
Answers
we get two triangles
for right angle triangle area=1/2*base*height
and 2nd part area of triangle=√s(s-a)(s-b)(s-c)
a,b,c are sides of triangle
s=(a+b+c)/2
I hope this understand
Answer:
306
Step - by - step explanation:
AB= 9
BC= 40
CD= 28
DA= 15
join A and C to form a diagonal AC in quadrilateral ABCD
using pythagorean theorem,
AB ^(2) + BC ^(2) = AC ^(2)
9*9 + 40*40 = AC ^(2)
81 + 1600 = AC ^(2)
1681 = AC^(2)
= AC
41 = AC
now, using herons formula, u can find the area of triangle ABC an d triangle ADC.
in triangle ABC
=
=
=
=
= 5*3*2*3*2
= 5*36
= 180
in triangle ADC
=
=
=
=
=2*7*3*3*
= 14*9
= 126
area of ABC + area of ADC = area of ABCD
180 + 126 = 306