In the triangle ABC, if AD = 16 and DC = 9 then what is the area of

Solution:

We are asked the area of \(\triangle ABC = \frac{1}{2}\times BC (base) \times AB (height)\).

But we have just the value of \(AD\), \(DC\) and thus \(AC\).

Thie given triangle has 2 right-angle triangles with a common angle \(\angle C\)



Thus we can conclude that \(\triangle ABC\) and \(\triangle BDC\) are similar by \(AAA\) condition of similarity.

After proving them similar, we can say that their ratio of corresponding sides will be the same i.e., \(\frac{AC}{BC} = \frac{BC}{DC}\)

\(⇒ BC^2=AC\times DC\)

\(⇒ BC^2=25\times 9\)

\(⇒ BC=\sqrt{25\times 9}\)

\(⇒ BC=5\times 3=15\)

Now we have the value of \(BC\). But we still want the value of \(AB\), which we can find very easily using the Pythagoras theorem on \(\triangle ABC\). We can say:

\(AC^2=AB^2+BC^2\)

\(⇒ 25^2=AB^2+15^2\)

\(⇒ AB^2=625-225\)

\(⇒ AB^2=400\)

\(⇒ AB=\sqrt{400}\)

\(⇒ AB=20\)

Now area of \(\triangle ABC = \frac{1}{2}\times BC \times AB=\frac{1}{2}\times 15\times 20 = 150\).

Hence the right answer is Option D.