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# Problem based on Triangles | PRMO-2018 | Problem 12

Try this beautiful problem from PRMO, 2018 based on Triangles.

## Problem based on Triangles | PRMO | Problem 12

In the triangle, ABC, the right angle at A. The altitude through A and the internal bisector of $\angle A$ have lengths $3$ and $4$, respectively. Find the length of the medium through A?

• $24$
• $30$
• $22$
• $18$

Geometry

Triangle

Pythagoras

## Check the Answer

Answer:$24$

PRMO-2018, Problem 12

Pre College Mathematics

## Try with Hints

We have to find out the value of $AM$ .Now we can find out the area of Triangle dividing two parts , area of $\triangle AMC$ + area of $\triangle ABM$ ( as M is the mid point of BC)

Can you now finish the problem ..........

Now $AN$ is the internal bisector...Therefore $\angle NAB=\angle NAC= 45^{\circ}$.Let $AC=b$ ,$AB=c$ and $BC=c$.using this values find out the area of triangles AMC and Triangle ABM

Area of $\triangle ABC$=$\frac{1}{2} bc=\frac{1}{2} \times a \times 3$

$\Rightarrow bc=3a$......................(1)

Now Area of $\triangle ABN$ + Area of $\triangle ANC$=Area of $\triangle ABC$

$\Rightarrow \frac{1}{2} c 4 sin 45^{\circ} +\frac{1}{2} b 4 sin 45^{\circ}=\frac{1}{2} bc$

$\Rightarrow b+c =\frac{1}{2\sqrt 2} bc$

(squarring both sides we get..........)

$\Rightarrow b^2 +c^2 +2bc=\frac{1}{8} b^2 c^2$

$\Rightarrow a^2 +6a=\frac{9}{8} a^2$ (from 1)

$\Rightarrow a +6 =\frac{9}{8} a$ $(as a \neq 0)$

$\Rightarrow a=48$

$\Rightarrow AM=BM=MC=\frac{a}{2}=24$

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