Question

i.
In a trapezium ABCD,
seg AB || seg DC, seg BD I seg AD, seg AC I seg BC. If
AD=15, BC = 15, AB=25 and CD = 7, then find A (OABCD).​

Answer 1

Answer:

192 sq. units

Step-by-step explanation:

According to Pythagoras theorem,

In ∆ABD

AB²= AD²+DB²

(25)²=(15)²+DB²

625=225+BD²

BD²=625-225

BD²=400

BD=√400

BD=20

Now, Area of Triangle ABD=√s(s-a) (s-b) (s-c)

s= a+b+c/2

=20+25+15/2

=60/2

=30

Area of triangle=√30(30-20) (30-25) (30-15)

=√30×5×10×15

=150 sq. units

Also,

Area of triangle=½×base×height

150=½×25×DP

DP=300/25

DP=12

Therefore, area of trapezium=12

Now,

According to Pythagoras theorem

In ∆ADP

AD²=AP²+DP²

(15)²=AP²+(12)²

225=AP²+144

AP²=225-144

AP²=81

AP=√81

AP=9

AP=QB=9

CD=PQ=25-(9+9)=7

Area of trapezium=½ ×sum of parallel sides×height

=½×(25+7)×12

=½×32×12

=32×6

=192 sq. units

Hence A( trapezium ABCD) = 192sq.units