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Byju's Answer

Standard IX

Mathematics

The Mid-Point Theorem

D,E,F are mid...

Question

$D, E, F$ are mid points of the sides $B C, C A & A B$ respectively of $△ A B C$ , then the area of the parallelogram $B D E F$ is equal to

\frac{1}{2} a r (△ A B C)

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\frac{1}{4} a r (△ A B C)

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\frac{1}{3} a r (△ A B C)

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\frac{1}{6} a r (△ A B C)

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Solution

The correct option is D $\frac{1}{2} a r (△ A B C)$
Using the mid point theorem for triangles show that ADEF is a parallelogram.
Draw $A Q ⊥ B C$ so that AQ is the height of the triangle ABC.
PQ will be the height of BDEF.

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Similar questions

D, E

and

F

are respectively the mid-points of the sides

B C, C A

and

A B

of a

△ A B C

. Show that
(i)

B D E F

is a parallelogram
(ii)

a r (D E F) = \frac{1}{4} a r (A B C)

(iii)

a r (B D E F) = \frac{1}{2} a r (A B C)

D

E

and

F

are respectively the mid-points of the sides

B C

C A

and

A B

of a

Δ A B C

.
Show that

(a)

B D E F

is a parallelogram.

(b)

ar (D E F) = \frac{1}{4} ar (A B C)

(c)

ar (B D E F) = \frac{1}{2} ar (A B C)

Q. If

D, E, F

are the mid-points of the sides

B C, C A

and

A B

respectively of

△ A B C

, prove that

B D E F

is a parallelogram. And also show that

a r (△ D E F) = \frac{1}{4} a r (△ A B C)

D, E and F are the mid point of sides BC , CA,AB of a triangle ABC
1. BDEF is a parallelogram
2. $a r (D E F) = \frac{1}{4} a r (A B C)$
3. $(B D E F) = \frac{1}{2} a r (A B C)$

Q. D, E and F are respectively the mid-points of the sides BC, CA, and AB of a ABC
(i) BDEF is a parallelogram
(ii) ar (DEF) =

\frac{1}{4}

ar (ABC)
(iii) ar BDEF)

\frac{1}{2}

ar (ABC)

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D,E,F are mid points of the sides BC,CA& AB respectively of △ABC, then the area of the parallelogram BDEF is equal to

The correct option is D 12ar(△ABC)Using the mid point theorem for triangles show that ADEF is a parallelogram.Draw AQ⊥BC so that AQ is the height of the triangle ABC.PQ will be the height of BDEF.

$D, E, F$ are mid points of the sides $B C, C A & A B$ respectively of $△ A B C$ , then the area of the parallelogram $B D E F$ is equal to

The correct option is D $\frac{1}{2} a r (△ A B C)$
Using the mid point theorem for triangles show that ADEF is a parallelogram.
Draw $A Q ⊥ B C$ so that AQ is the height of the triangle ABC.
PQ will be the height of BDEF.