if the area of a rhombus are 24cm square and one of a style men before cm find the perimeter of rhombus. Give step by step explanation please
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The area of a regular rhombus is [math]A = pq/2 [/math], where p and q are the diagonals of the shape.
given A = 24 and p = 4
[math]24 = 4q/2[/math]
[math]48 = 4q[/math]
[math]q = 12[/math]
so the other diagonal is 12 cm long.
now, by Pythagorean theorem, each side of the rhombus (annotate as s) is given by:
[math]s^2 = (p/2)^2 + (q/2)^2[/math]
[math]s^2 = (4/2)^2 + (12/2)^2[/math]
[math]s^2 = 4+36[/math]
[math]s^2 = 40[/math]
[math]s = 2 \sqrt{10}[/math]
since there are 4 sides in a rhombus, so the perimeter of it is:
[math]4*2 \sqrt{10}[/math]
[math]= 8 \sqrt{10}[/math]
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