Try this beautiful Geometry problem from PRMO, 2018 based on Medians.
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 \(A\).
Geometry
Medians
Triangle
But try the problem first...
Answer:\(24\)
PRMO-2018, Problem 13
Pre College Mathematics
First hint
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 ..........
Second Hint
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........
Final Step
Now \(AF\) is the median , \(AF=\frac{a}{2}=24\)
Try this beautiful Geometry problem from PRMO, 2018 based on Medians.
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 \(A\).
Geometry
Medians
Triangle
But try the problem first...
Answer:\(24\)
PRMO-2018, Problem 13
Pre College Mathematics
First hint
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 ..........
Second Hint
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........
Final Step
Now \(AF\) is the median , \(AF=\frac{a}{2}=24\)
