Question

A
1. In AABC, ZABC = 90°, AD=DC, AB = 12cm
and BC=6.5 cm. Find the area of AADB.
12 cm​

Answer 1

ANSWER

In triangle ABC, using Pythagoras theorem

AC

2

=AB

2

+BC

2

AC

2

=12

2

+(6.5)

2

AC

2

=144+42.5

AC

2

=186.25

AC=13.6cms

Also,

AD=DC=

2

AC

=

2

13.6

=6.8cms

Falling a perpendicular line from D to AB, such that ED is parallel to BC. ∠AED=∠ABC=90

o

Since D is the mid point of AC and ED is parallel to BC, so applying mid pint theorem,

ED=

2

BC

=

2

6.5

=3.25cms

Now in triangle ABD, ED is the altitude and AB is the base

So the area of triangle ABD=

2

(AB×ED)

Ar(ABD)=

2

12×3.25

=

2

39

=19.5sq.cms