Question

In △𝐀𝐁𝐂, ∠𝐀𝐁𝐂=𝟗𝟎°,𝐀𝐃=𝐃𝐂, 𝐀𝐁=𝟏𝟐 𝐜𝐦 and 𝐁𝐂=𝟔.𝟓𝐜𝐦. Find the area of △𝐀𝐃𝐁.

Answer 1

In triangle ABC, using Pythagoras theorem
AC${}^2$ = AB${}^2$ + BC${}^2$
AC${}^2$ = 12${}^2$ + (6.5)${}^2$
AC <!--td {border: 1px solid #cccccc;}br {mso-data-placement:same-cell;}--> ${}^2$ = 144 + 42.5
AC${}^2$ = 186.25
AC = 13.6cms
Also,
AD = DC = $\frac{AC}{2}$ = <!--td {border: 1px solid #cccccc;}br {mso-data-placement:same-cell;}--> ​​​​​​​$\frac{13.6}{2}$
=6.8cms

Falling a perpendicular line from D to AB, such that ED is parallel to BC. ∠AED = ∠ABC = 90${}^{\circ }$

Since D is the mid point of AC and ED is parallel to BC, so applying mid pint theorem,
ED = ​​​​​​​$\frac{BC}{2}$ = ​​​​​​​$\frac{6.5}{2}$ = 3.25 cm

​Now in triangle ABD, ED is the altitude and AB is the base
So the area of triangle ABD= ​​​​​​​$\frac{1}{2}$× (AB×ED)
​
ar(ABD) = <!--td {border: 1px solid #cccccc;}br {mso-data-placement:same-cell;}--> ​​​​​​​$\frac{1}{2}$ × 12 × 3.25
ar(ABD) = ​​​​​​​$\frac{39}{2}$ = 19.5 sq.cm

​​​​​​​Answer = 19.5 sq.cm