If the lengths of the medians $A D, B E$ and $C F$ of the triangle $A B C$ , are $6, 8, 10$ respectively, then

A D

and

B E

are perpendicular

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

B E

and

C F

are perpendicular

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

area of

Δ A B C = 32

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

area of

Δ D E F = 8

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

Open in App

Solution

The correct option is B area of $Δ D E F = 8$
$\begin{matrix} A D = \sqrt{2 A B^{2} + 2 A C^{2} - B C^{2}} = 6 2 A B^{2} + 2 A C^{2} - B C^{2} = 36 B E = \sqrt{2 A B^{2} + 2 B C^{2} - A C^{2}} = 8 2 A B^{2} + 2 B C^{2} - A C^{2} = 64 C F = \sqrt{2 A C^{2} + 2 B C^{2} - A B^{2}} = 10 2 A C^{2} + 2 B C^{2} - A B^{2} = 100 A B^{2} = x A C^{2} = y B C^{2} = z 2 x + 2 y - z = 36 2 x + 2 z - y = 64 2 y + 2 z - x = 100 x = A B^{2} = \frac{100}{9} y = A C^{2} = \frac{208}{9} z = B C^{2} = \frac{292}{9} A D = 6, B E = 8, C F = 10 i n, Δ A B E A D ⊥ B E a r e a, Δ B E C = 16 a r e a, Δ A B E = 16 + 16 = 32 \frac{a r e a, Δ A B E}{a r e a, Δ D E F} = 4 \frac{32}{a r e a, Δ D E F} = 4 a r e a, Δ D E F = 8 \end{matrix}$

Suggest Corrections

Similar questions

A D, B E a n d C F

Δ A B C

B E a n d C F

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})

If AD, BE and CF are the medians of an equilateral triangle ABC, then the true statement is -

A D = 15 c m

A G

Join BYJU'S Learning Program