show that the area of a rhombus is half the product of the lengths of its diagonals
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Hey !!
Here is your answer.⬇⬇
Proff :- let ◻ABCD is rhombus. AC and BD are two diagonals.
So, Area of ◻ ABCD = Area of ∆ABD + Area of ∆CBD
=( BD × AO )/2 + ( BD × OC )/2
.....( As we know that diagonals of rhombus are perpendicular to each other )
= BD/2 (AO + OC)
= ( BD × AC )/2 ......( AO + OC = AC )
= 1/2 × length of its diagonals. ..... ( proved )
HOPE IT HELPS YOU....
THANKS ^-^
Here is your answer.⬇⬇
Proff :- let ◻ABCD is rhombus. AC and BD are two diagonals.
So, Area of ◻ ABCD = Area of ∆ABD + Area of ∆CBD
=( BD × AO )/2 + ( BD × OC )/2
.....( As we know that diagonals of rhombus are perpendicular to each other )
= BD/2 (AO + OC)
= ( BD × AC )/2 ......( AO + OC = AC )
= 1/2 × length of its diagonals. ..... ( proved )
HOPE IT HELPS YOU....
THANKS ^-^
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