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Theorem 1: Parallelograms

X and Y are p...

Question

X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (See figure). Then:

A

A r e a (Δ L Z Y) = A r e a (M Z Y X)

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B

A r e a (Δ L Z N) = A r e a (M Z Y X)

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C

A r e a (Δ N M X) = A r e a (M Z Y X)

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D

N o n e o f t h e s e

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Solution

The correct option is A $A r e a (Δ L Z Y) = A r e a (M Z Y X)$
Since $Δ L X Z a n d Δ X M Z$ are on the same base and between the same parallel LM and XZ, we have
$a r (Δ L X Z) = a r (Δ X M Z)$ .....(1)
Adding $a r (Δ X Y Z)$ to both sides of (1), we get
$a r (Δ L X Z) + a r (Δ X Y Z) = a r (Δ X M Z) + a r (Δ X Y Z)$
$\Rightarrow a r (Δ L Z Y) = a r (M Z Y X)$

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Similar questions

Q. Question 2
X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (see figure). Prove that ar (\(\Delta\) LZY) = ar (MZYX).

Q. Question 2
X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (see figure). Prove that ar (

Δ

LZY) = ar (MZYX).

Q. X and Y are points on the side LN of the triangle LMN such that LX=XY=YN.Through X, a line is drawn parallel to LM to meet MN at Z. Prove that ar(LZY)=ar(MZYX).

Q. In the figure, XY is a line parallel to BC is drawn through A. If BE || CA and CF || BA are drawn to meet XY at E and F respectively. Show that

a r e a (△ A B E) = a r e a (△ A C F)

.

State true or false:

A B C

D

E

are points on side

A B

A D = D E = E B

D

E

lines are drawn parallel to

B C

which meet side

A C

F

G

respectively. Through

F

G

lines are drawn parallel to

A B

which meet side

B C

M

N

respectively, then

B M = M N = N C

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Theorem 1: Parallelograms

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