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Rhombus ABCD has perimeter 148, and one of its diagonals has length 24. what is the area of ABCD?

 May 14, 2015

Best Answer 

 #1
avatar+23252 
+8

Since its a rhombus, it has four equal sides:  148 ÷ 4 = 37

The diagonal divides the rhombus into two congruent triangle, each 37 x 37 x 24.

If we can find the area of one of the triangles, we can multiply that by 2 and get the total area.

Hero's formula for the area of a triangle uses the three sides; it is: Area = √[ s(s-a)(s-b)(s-c) ] 

where a, b, and c are the lengths of the three sides:  a = 37, b = 37, and c = 24.

s is the semiperimeter:  s = (a + b + c) / 2  =  (37 + 37 + 24)/2  =  98/2  =  49.

Area  = √[ 49(49-37)(49-37)(49-24) ]  =  = √[ 49(12)(12)(25) ]  =  420

So, the whole area is 420 x 2 = 840.

 May 14, 2015
 #1
avatar+23252 
+8
Best Answer

Since its a rhombus, it has four equal sides:  148 ÷ 4 = 37

The diagonal divides the rhombus into two congruent triangle, each 37 x 37 x 24.

If we can find the area of one of the triangles, we can multiply that by 2 and get the total area.

Hero's formula for the area of a triangle uses the three sides; it is: Area = √[ s(s-a)(s-b)(s-c) ] 

where a, b, and c are the lengths of the three sides:  a = 37, b = 37, and c = 24.

s is the semiperimeter:  s = (a + b + c) / 2  =  (37 + 37 + 24)/2  =  98/2  =  49.

Area  = √[ 49(49-37)(49-37)(49-24) ]  =  = √[ 49(12)(12)(25) ]  =  420

So, the whole area is 420 x 2 = 840.

geno3141 May 14, 2015

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