Name: Areas of Similar Triangles
Uploaded: 2019-07-19
Duration: 6 min 54 s
Description: The corresponding sides of similar triangles are proportional. Are the ratios of areas of similar triangles the same as that of their sides? Watch this video...

Question

$B L$ and $C M$ are medians on sides $A C$ and $A B$ of triangle $A B C$ , right angled at $A$ . Prove that $4 (B L^{2} + C M^{2}) = 5 B C^{2}$ .

Accepted Answer

Given: $B L$ and $C M$ are medians on sides $A C$ and $A B$ of $△ A B C$ .

To Prove: $4 (B L^{2} + C M^{2}) = 5 B C^{2}$

Proof: $B L$ is median

$\Rightarrow A L = C L = \frac{1 }{2} A C$ $. . [1]$

$C M$ is median

$\Rightarrow A M = M B = \frac{1}{2} A B$ $. . [2]$

In $△ B A C$ , By Pythagoras theorem

$(B C)^{2} = (A B)^{2} + (A C)^{2}$ $. . [3]$

In $△ B A L$ , By Pythagoras theorem

$\Rightarrow (B L)^{2} = (A B)^{2} + {(A L)}^{2} \Rightarrow (B L)^{2} = (A B)^{2} + {(\frac{A C}{2})}^{2} . . [using (1)] \Rightarrow 4 B L^{2} = 4 A B^{2} + A C^{2} . . [4]$

In $△ M A C,$ By Pythagoras theorem

$C M^{2} = A M^{2} + A C^{2}$

$\Rightarrow C M^{2} = {(\frac{A B}{2})}^{2} + A C^{2} . . [using (2)] \Rightarrow 4 C M^{2} = A B^{2} + 4 A C^{2} . . [5]$

Now adding $[4]$ and $[5]$ we get

$4 B L^{2} + 4 C M^{2} = 5 A B^{2} + 5 A C^{2}$

$\Rightarrow 4 (B L^{2} + C M^{2}) = 5 (A B^{2} + A C^{2}) \Rightarrow 4 (B L^{2} + C M^{2}) = 5 B C^{2} . . [using [3]]$

Hence Proved.

$B L$ and $C M$ are medians on sides $A C$ and $A B$ of triangle $A B C$ , right angled at $A$ . Prove that $4 (B L^{2} + C M^{2}) = 5 B C^{2}$ .

Important Questions on Triangles

Learn and Prepare for any exam you want !

BL and CM are medians on sides AC and AB of triangle ABC, right angled at A. Prove that 4(BL2+CM2)=5BC2.

Important Questions on Triangles

Learn and Prepare for any exam you want !

$B L$ and $C M$ are medians on sides $A C$ and $A B$ of triangle $A B C$ , right angled at $A$ . Prove that $4 (B L^{2} + C M^{2}) = 5 B C^{2}$ .