Question

A rectangle is formed by joining the mid points of sides of a rhombus. Show that area of rectangle is half the area of the thombus.

Answer 1

here is ur answer....

Answer 2

$ar(PQRS) = \frac{1}{2} ar( ABCD )$

Step-by-step explanation:

In rhombus ABCD by joining the mid points of the the rhombus rectangle  PQRS  is  formed

since rectangle is also a parallelogram

then,

AB║ CD

In parallelogram QCDS ,

$\bigtriangleup SRQ$ and parallelogram SQCD are on the same base SQ and between the same parallel lines DC and  SQ

then ,

$ar(\bigtriangleup SRQ ) = \frac{1}{2}  ar(QSDC)$ .....(1)

similarly

in parallelogram ASBQ

$ar(\bigtriangleup PSQ ) = \frac{1}{2} ar(ABQS )$...(2)

add eq (1) and (2)

we get

$ar(\bigtriangleup SRQ )+ar(\bigtriangleup PSQ ) = [tex]\frac{1}{2} ar( QSDC) +  \frac{1}{2} ar(ABQS)$

$ar(PQRS) = \frac{1}{2} ar( ABCD )$

hence proved .

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