The shorter diagonal of a rhombus is 8 cm long and the area of the rhombus is 68 sq. cm. Find the length of each side of the rhombus.
2007-07-30 00:07:44 · 4 answers · asked by double B 2 in Science & Mathematics ➔ Mathematics
area of a rhombus is 1/2 a x b
where a and b are the two diagonals
so 68 = 1/2 8 x b
so b = 68/4 = 17
2007-07-30 00:22:44 · answer #1 · answered by JAMES C 2 · 1⤊ 0⤋
The area of a rhombus is half the product of the lengths of its diagonals:
A=D1 * D2 / 2
with A = 68 and D1 = 8, so we can take D2
D2 = 68 *2 / 8 = 17
Half of the diagonals do a triangle rectangle with the side
(rhombus is a quadrilateral in which all of the sides are of equal length)
L^2 = (D1/2)^2 + (D2/2)^2
L = 9.4 cm
2007-07-30 00:42:17 · answer #2 · answered by GPC 3 · 0⤊ 0⤋
Area of rhombus = A = 1/2 x d1 x d2
68 = 1/2 x 8 x d2
d2 = 17 cm
Diagonals bisect each other to form 4 right angled triangles of side 8.5 , 4 and h, say.
Hypotenuse of each triangle is side of rhombus.
h² = 8.5² + 4²
h² = 88.25
h = 9.4 cm (to 1 dec. place)
Length of each side = 9.4 cm
2007-08-02 18:40:04 · answer #3 · answered by Como 7 · 0⤊ 0⤋
area = 68
diag1 = 8
area = 1/2 diag1 diag2
diag2 = 68(2)/8 = 17
side = sqr[ (17/2)^2 + (8/2)^2 ]
. . . = sqr(88.25)
side = 9.394 cm
2007-07-30 00:32:58 · answer #4 · answered by CPUcate 6 · 0⤊ 0⤋