In $Δ A B C$ , D E,F, are mid point of side AB, BC and AC reaspectively. If area( $C E D F$ )= $24 c m^{2}$ , find $a r e a (B E F D)$ :

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Solution

We have,

In $Δ A B C$

$D, E, F$ are mid point of the side $A B, B C, C A$ respectively.

Then,

$a r (C E D F) = 24 c m^{2}$

$a r (B E F D) = ?$

But According to given question

$a r (C E D F) = a r (B E F D) = 24 c m^{2}$
Hence, this is the answer.

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Similar questions

Q. In

Δ A B C

, D, E and F are the mid points of the sides BC, CA and AB respectively. The area of

Δ A B C

24 c m^{2}

, then the area of

Δ D E F

Q. In

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, D, E and F are mid-points of sides AB, BC and AC respectively.

D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC.

If D, E, F are respectively the mid-points of the sides BC, CA, and AB of $△$ ABC, and the area of $△$ ABC is 24 $c m^{2}$ , then find the area of $△$ DEF.

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are respectively the mid points of

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and

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, then

\to B E + \to A F

is equal to.

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InΔABC, D E,F, are mid point of side AB, BC and AC reaspectively. If area(CEDF)=24cm2, find area(BEFD):

We have, In ΔABC D,E,F are mid point of the side AB,BC,CA respectively. Then, ar(CEDF)=24cm2 ar(BEFD)=? But According to given question ar(CEDF)=ar(BEFD)=24cm2 Hence, this is the answer.

In $Δ A B C$ , D E,F, are mid point of side AB, BC and AC reaspectively. If area( $C E D F$ )= $24 c m^{2}$ , find $a r e a (B E F D)$ :

We have,

In $Δ A B C$

$D, E, F$ are mid point of the side $A B, B C, C A$ respectively.

Then,

$a r (C E D F) = 24 c m^{2}$

$a r (B E F D) = ?$

But According to given question

$a r (C E D F) = a r (B E F D) = 24 c m^{2}$
Hence, this is the answer.