The sides of a rhombus are 10 cm long. If the lengths of the diagonals differ by 4, what is the area of the rhombus? Explain to me how you got your answer.
2007-05-21 15:03:29 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
According to wikipedia:
"The area of any rhombus is one half the product of the lengths of its diagonals"
or:
1/2(D1*D2)
so...
D1 = 10
D2 = 10 (by definition see wikipedia)
sooo...
1/2(100) = 50 <-- thats my answer
If you didn't understand that or got lost, try the following site:
http://www.mathopenref.com/rhombusarea.html
Using number 2 - the diagnoals method.
2007-05-21 15:17:24 · answer #1 · answered by romeohsdrumline 3 · 0⤊ 1⤋
The diagonals of a rhombus bisect each other and are also perpendicular to each other. the side = 10 cm is the hypotenuse of a right triangle and its legs are 8 and 6 inches in length. Thus the two diagonals are 16cm and 12cm in length. The area of a rhombus is 1.2the product of the diagonals so the area is (1/2)(16)(12) = 96 cm^2.
2007-05-21 22:33:29 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋