## anonymous 4 years ago In Rhombus ABCD, AB= 15, and AC= 29. Find the area of the rhombus to the nearest tenth?

1. Mertsj

|dw:1328733982966:dw|

2. Mertsj

Do you know Heron's formula?

3. anonymous

no I dont?

4. Mertsj

How about the area = 1/2 the product of the diagonals? Is that what you're studying?

5. Mertsj

The diagonals bisect each other and are perpendicular. So let's use right triangle BEC and find BE using the :Pythagorean Theorem.|dw:1328734330510:dw|

6. Mertsj

CE= 29/2 and 15 is the hypotenuse. So $(\frac{29}{2})^2+(BE)^2=225$

7. Mertsj

$BE=3.84$

8. Mertsj

BD=7.68

9. Mertsj

$A=\frac{1}{2}BD \times AC=\frac{1}{2} \times 7.68 \times 29=111.36$

10. Mertsj

111.4 to the nearest tenth

11. anonymous

okay thanks.