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slavikrds [6]
3 years ago
6

The diagonals of rhombus abcd intersect at point

Mathematics
2 answers:
Nadya [2.5K]3 years ago
3 0
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
Reptile [31]3 years ago
3 0

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

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