You visited us 1 times! Enjoying our articles? Unlock Full Access!

Apollonius's Theorem

If AD, BE a...

Question

If $A D, B E$ and $C F$ are the medians of a $Δ A B C$ then $(A D^{2} + B E^{2} + C F^{2}) : (B C^{2} + C A^{2} + A B^{2})$ is equal to:

4 : 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3 : 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3 : 4

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2 : 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $3 : 4$
In $Δ A B C$
$A D, B E, C F$ are medians
By Apollonius's theorem
$A B^{2} + A C^{2} = 2 (A D^{2} + B D^{2})$
$\Rightarrow A B^{2} + A C^{2} = 2 (A D^{2} + {(\frac{B C}{2})}^{2})$
$\Rightarrow A B^{2} + A C^{2} - \frac{{B C}^{2}}{2} = 2 A D^{2}$ ...(1)
similarly $B C^{2} + B A^{2} - \frac{{A C}^{2}}{2} = 2 B E^{2}$ ...(2)
and $B C^{2} + A C^{2} - \frac{{A B}^{2}}{2} = 2 C F^{2}$ ...(3)
Adding (1), (2) & (3), we get
$\Rightarrow \frac{3 (A B^{2} + A C^{2} + B C^{2})}{2} = 2 (A D^{2} + B E^{2} + C F^{2})$
Therefore, $(A D^{2} + B E^{2} + C F^{2}) : (B C^{2} + C A^{2} + A B^{2}) = 3 : 4$
Ans: C

Suggest Corrections

Similar questions

If AD, BE and CF are the medians of a $△ A B C t h e n (A D^{2} + B E^{2} + C F^{2}) : (B C^{2} + C A^{2} + A B^{2})$ is equal to

A D, B E

C F

Δ A B C,

(A D^{2} + B E^{2} + C F^{2}) : (B C^{2} + C A^{2} + A B^{2})

A D, B E

C F

Δ A B C,

(A D^{2} + B E^{2} + C F^{2}) : (B C^{2} + C A^{2} + A B^{2}) =

A D, B E

C F

△ A B C

(A D^{2} + B E^{2} + C F^{2}) : (B C^{2} + C A^{2} + A B^{2})

△ A B C

⊥

A C^{2} = A D^{2} + B C \times D M + \frac{B C^{2}}{4}

A B^{2} = A D^{2} - B C \times D M + \frac{B C^{2}}{4}

Join BYJU'S Learning Program