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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.

LetBC = 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

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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.

LetBC = 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

Subscribe to Cheenta at Youtube


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