SECTION-D 10. In the figure SABC and ADBC have same base BC such that AB=120m, AC 122m,
BC=22m, BD=25m and CD = 25m. Find the area of the shaded region

For the triangle having the sides 122m, 120m and 22m :

s=

2

122+120+22

=132m

Area of triangle =

s(s−a)(s−b)(s−c)

=

132(132−122)(132−120)(132−22)

=

132×10×12×110

=1320m

2

For the triangle having the sides 22m, 24m and 26m :

s=

2

22+24+26

=36m

=

36(36−22)(36−24)(36−26)

=

36×14×12×10

=24

105

=24×10.25m

2

(approx.)=246m

2

There, the area of the shaded portion

=(1320−246)m

2

=1074 m

2

50x5=$250

step-by-step explanation: