Medians | Geometry | PRMO-2018 | Problem 13

Try this beautiful Geometry problem from PRMO, 2018 based on Medians.

Medians | PRMO | Problem-13

In a triangle ABC, right-angled at A, the altitude through A and the internal bisector of $\angle A$ have lengths $3$ and $4$ , respectively. Find the length of the median through $A$ .

$20$
$24$
$13$

Key Concepts

Geometry

Medians

Triangle

Check the Answer

Answer: $24$

PRMO-2018, Problem 13

Pre College Mathematics

Try with Hints

medians of a triangle

Now in the Right angle Triangle ABC , $\angle A=90$ . $AD=3$ is Altitude and $AE=4$ is the internal Bisector and $AF$ is the median.Now we have to find out the length of $AF$

Now $\angle CAE = 45^{\circ }= \angle BAE$ .

Let $BC = a$ , $CA = b$ , $AB = c$
so $\frac{bc}{2}=\frac{3a}{2}$ $\Rightarrow bc=3a$

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

a triangle with medians

Therefore,

$\frac{2bc}{b+c} cos\frac{A}{2}=4$

$\Rightarrow \frac{6a}{b+c} .\frac{1}{\sqrt 2}=4$

$\Rightarrow 2\sqrt 2 (b+c)=3a$

$\Rightarrow 8(b^2 + c^2 + 2bc) = 9a^2$

$\Rightarrow 8(a^2 + 6a) = 9a^2$

$\Rightarrow 48a=a^2$

$\Rightarrow a=48$

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

Now $AF$ is the median , $AF=\frac{a}{2}=24$

Other useful links

Related

Try this beautiful Geometry problem from PRMO, 2018 based on Medians.

Medians | PRMO | Problem-13

In a triangle ABC, right-angled at A, the altitude through A and the internal bisector of $\angle A$ have lengths $3$ and $4$ , respectively. Find the length of the median through $A$ .

$20$
$24$
$13$

Key Concepts

Geometry

Medians

Triangle

Check the Answer

Answer: $24$

PRMO-2018, Problem 13

Pre College Mathematics

Try with Hints

medians of a triangle

Now in the Right angle Triangle ABC , $\angle A=90$ . $AD=3$ is Altitude and $AE=4$ is the internal Bisector and $AF$ is the median.Now we have to find out the length of $AF$

Now $\angle CAE = 45^{\circ }= \angle BAE$ .

Let $BC = a$ , $CA = b$ , $AB = c$
so $\frac{bc}{2}=\frac{3a}{2}$ $\Rightarrow bc=3a$

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

a triangle with medians

Therefore,

$\frac{2bc}{b+c} cos\frac{A}{2}=4$

$\Rightarrow \frac{6a}{b+c} .\frac{1}{\sqrt 2}=4$

$\Rightarrow 2\sqrt 2 (b+c)=3a$

$\Rightarrow 8(b^2 + c^2 + 2bc) = 9a^2$

$\Rightarrow 8(a^2 + 6a) = 9a^2$

$\Rightarrow 48a=a^2$

$\Rightarrow a=48$

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

Now $AF$ is the median , $AF=\frac{a}{2}=24$

Other useful links

Related

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