Rhombus problem.?

Question

In rhombus ABCD, AB=13 and AC=20. Find BD to the nearest tenth. Step by step and show work.

Answer 1

Draw rhombus ABCD on a piece of paper in front of you. Remember that in a rhombus, the sides are congruent and the diagonals are perpendicular. Write in that AB = 13 and AC = 20 (you'll need to draw in a line form A to C). Then draw in a line from B to D. Let's call the middle point X. Because the diagonals form perpendicular liens, you know that triangle AXB is a right triangle, right? So lets use the Pythagorean theorem: 10^2 x XB^2 = 13^2 --> 100 x XB^2 = 169. So XB equals the square root of 69, which is about 8.306623...
You're supposed to find BD to the nearest tenth, so let's round the number above to 8.3. Therefore, since 8.3 = XB, and XB = 1/2 BD, BD = 8.3 (2) = 16.6 units.

Answer 2

Let x and y be the diagonals,
and a = side (all are equal)

x^2+y^2 = 4a^2 (Rhombus formula)

x = AC = 20
y = BD
a = 13

20^2 + y^2 = 4(13^2)

Simplify

y^2 = 176
y = 16.6 which is BD
.