Question

The diagonals of rhombus abcd intersect at point

Answer 1

The area of a rhombus is 1/2 times the lengths of the diagonals

A = (1/2)(AC)(BD)

Since the diagonals bisect each other, and ed = 8, BD must be 16. Substituting in this for BD and the given area (168) we get the following:

168 = (1/2)(AC)16
168 = 8(AC)
21 = AC

Answer 2

Answer:

ac= 21

Step-by-step explanation:

Rhombus is a parallelogram in which diagonals bisect each other .

Given: Area of rhombus is 168 .

The diagonals of rhombus abcd intersect at point e. It's diagonals are ac and bd . We know that diagonals of rhombus bisect each other ,so, bd=2de . So, $bd=2(8)=16$ .

Also, area of rhombus is $A=\frac{1}{2}d_1d_2$ where $d_1,d_2$ are the diagonals of rhombus .

As area of rhombus is 168 , we get ,

$168=\frac{1}{2}d_1d_2$

Let $d_1=bd=16$

So, we get ,

$168=\frac{1}{2}d_1d_2\\168=\frac{1}{2}\left ( 16 \right )d_2\\168=8d_2\\d_2=\frac{168}{8}\\=21$

Therefore, $ac=21$