Question

smaller diagonal of a rhombus is equal to length of its side if length of its side is 6 CM then what is the area of an equilateral triangle whose side is equal to the bigger diagonals of the Rhombus

Answer 1

Hope u understand it

I'm sorry if my answer is incorrect

Answer 2

AREA is $27 \times \sqrt3.$

Step-by-step explanation:

In the figure DB is smaller diagonal and its length is equal to one of the sides AD=6 cm.

Also by property of rhombus :

1. Diagonal bisects perpendicularly.

2. Diagonal cuts each other in equal parts.

Therefore, $DO= \dfrac{DB}{2}=3 \ cm.$ ( by property 2.)

Applying Pythagoras theorem.

$AD^2=DO^2+AO^2$.

So,  $AO=3\times \sqrt3.$

Therefore, $AC=6\times \sqrt3.$  ( by property 2.)

Now, We know area of equilateral triangle of side a is $\dfrac{\sqrt3}{4}\times a^2.$

So, area of equilateral triangle with side AC , $\dfrac{\sqrt3}{4}\times(6\times \sqrt3)^2=27\times \sqrt3.$

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AREA

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