if the area of a rhombus is 48 cm2 and one of its diagonals is 12 cm, find the side of the rhombus.
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area of rhombus= 1/2 × d1.d2
48÷12=1/2×d1
d1=4×2
=8 cm
We know that the diagonls of the rhombus bisect each other at 90' .
Let AB =BC= CD= AD be the sides of the rhombus.
So by Pythagoras theorem,
AB^2=(d1^2 ×d2^2 )1/4
AB SQ. = (64+144)×1/4
=208÷4
=52
AB= ROOT 52
=2_/13 cm
so the side of the rhombus is 2 root 13 cm.
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48÷12=1/2×d1
d1=4×2
=8 cm
We know that the diagonls of the rhombus bisect each other at 90' .
Let AB =BC= CD= AD be the sides of the rhombus.
So by Pythagoras theorem,
AB^2=(d1^2 ×d2^2 )1/4
AB SQ. = (64+144)×1/4
=208÷4
=52
AB= ROOT 52
=2_/13 cm
so the side of the rhombus is 2 root 13 cm.
I HOPE IT WILL HELP YA!!!
IF YES THEN MARK ME AS BRAINLIEST!!!
ALSO YOU CAN FOLLOW ME ON THE BRAINLY APP @cutie1312!!!
☺☺☺☺
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Answered by
1
Answer:
Step-by-step explanation:
Area of rhombus= 1/2 (product of its diagonals)
48cm² = 1/2 (12 × second diagonal)
48×2/12= IInd diagonal
IInd diagonal= 8
perimeter of rhombus= 2(8² + 12²)
= 2(64+ 144)
= 2(208)
= 416
Also, perimeter= 4× side (all the sides are equal)
416/4= side
side= 104cm
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