Find the altitude of a rhombus whose area is 420 sq cm and perimeter is 140 cm?

Answer 1

The rhombus in our problem has 4 sides with length of 35 cm, 140/4 (since the sides of the rhombus are congruent, and the perimeter is 140 cm ).

We know that the diagonals of the rhombus are perpendicular and bisect each other, so they form 4 right congruent triangles, so the area of one triangle is 105 cm^2.

Let the base of the triangle be x, then the altitude will be equal to √[(side)^2 - (base)^2] (from the pythagorean theorem). So the area of one triangle formed by drawing the diagonals of the rhombus

is:

(1/2)[x√(35^2 - x^2)] = 105 square both sides (1/4)[x^2(1,225 - x^2) = 11,025 multiply by 4 to both sides x^2(1,225 - x^2) = 44,100 1,225x^2 - (x^2)(x^2) = 44,100 let x^2 = t, and substitute t for x^2 into the equation 1,225t - (t)(t) = 44,100 1,225t - t^2 = 44,100 subtract 44,100 to both sides 1,225t - t^2 - 44,100 = 0 multiply by -1 to both sides and rearrange the terms of the equation

t^2 - 1,225t + 44,100 = 0 where a = 1, b = 1,225, and c = 44,100

Use the quadratic formula, x = [-b ± √(b^2 - 4ac)]/2, to solve for t

t = [1,225 ± √[1,225^2 - 4(1)(44,100)]]/2

t = (1,225 ± 1,150.75)/2 ignore +, because if you add it is a big number, and the side of rombus is only 35, and as the hypotenuse it is bigger than sides

t = (1,225 - 1,150.75)/2

t = 74.25/2

t = 37.125

Since t = x^2, then x^2 = 37.125, and x = √37.125 = 6.09 approximately (as length is always positive).

Thus the base is 6 cm, and the altitude is

√(35^2 - x^2) = √(1225 - 37.125) = 34.46 approximately. Thus the diagonals of the rhombus are d1 = 2 x 6.09 = 12.18, (approximately), and d2 = 2 x 34.46 = 68.92 We know that the area of the rhombus is equal to one halph of the product of its diagonals. Check:

(1/2)(12.18 x 68.92) = 419.7228 approximately to 420.