rhombus

I don't know if this is the easiest way, but it's the way that I saw.

I'm going to find the area of the rhombus and then use the formula:  Area  =  ½ · d₁ · d₂ 

Since a rhombus has 4 equal sides, each side has value:  68 / 4  =  17.

The area of the triangle formed by two sides of the rhombus and the diagonal can be found by using Heron's

formula:  Area  =  sqrt[ s(s - a)(s - b)(s - c) ]  where a, b, and c are the lengths of the three sides and 

s is the semiperimeter:  s  =  (a + b + c) / 2.

s  =  (17 + 17 + 16) / 2  =  25

Area(triangle)  =  sqrt[ 25(25 - 17)(25 - 17)(25 - 16) ]  =  120

Therefore, the area of the rhombus  =  2 x 120  =  240

Now using the formula:  Area  =  ½ · d₁ · d₂    --->   240  =  ½ · 16 · d₂   --->   d₂  =  30

Since all four sides in a rhombus are equal, each side should be equal to $\frac{68}{4}=17.$

The key idea in this problem is that the two diagonals in the rhombus are perpendicular to each other, and most importantly, the diagonals bisect each other...

We have a right triangle, with one leg as eight(8) units and the hypotenuse has length seventeen(17) units.

By the Pythagorean theorem, the other side length has length fifteen(15) units.

Thus, the second diagonal has length $2*15=30$ units.

If the perimeter is 68.....one side  =  68 / 4   =  17

1/2 of the  known diagonal length  =  8

So  ...using the Pythagorean Theorem 1/2  the length of the other diagonal  = sqrt (17^2 - 8^2)  = sqrt (289 - 64) = sqrt (225) = 15

So....the length of the other diagonal  =  2 * 15  =  30

perimeter of a rhombus is 68 one diagonal is 16 what is the lenth of the secound diagonal

Diagonal    D = {sqrt [(68/4)² - (16/2)²]} * 2 = 30