In the following figure, AD and EF are medians. What’s the length of FC?

4 cm

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

3 cm

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

2 cm

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

1 cm

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

Open in App

Solution

The correct option is C 2 cm

Since AD is the median, BD = DC.
So, CD = 4 cm

Also, since EF is the median, $D F = C F = \frac{C D}{2} = \frac{4}{2} = 2 cm$

Suggest Corrections

Similar questions

In the following figure, AD and EF are medians. What’s the length of FC?

In the given figure AD is median of $Δ A B C$ , DE is median of $Δ A B D$ and EF is median of $Δ B E D$ . If area of $Δ A B C$ is $128 c m^{2}$ , then area of $Δ B E F$ is

In the given figure, AB = 5 cm and CA= 7cm, and AD is the line of symmetry.

Find the length of EF.

Q. In the given figure,

A D = B C = 5 c m, A B = 4 c m, C D = 10 c m

. What is the length of CF if B is the midpoint of ED and EF

⊥

DF?

Q. In the given figure, ABC is a right triangle right-angled at B, EF ⊥ BC. If AF = 3 cm, AB =

\sqrt{21}

cm and FC = 4 cm, then EF is equal to

Join BYJU'S Learning Program

In the following figure, AD and EF are medians. What’s the length of FC?

The correct option is C 2 cm Since AD is the median, BD = DC. So, CD = 4 cm Also, since EF is the median, DF=CF=CD2=42=2 cm

The correct option is C 2 cm

Since AD is the median, BD = DC.
So, CD = 4 cm

Also, since EF is the median, $D F = C F = \frac{C D}{2} = \frac{4}{2} = 2 cm$