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