+0

# Rhombus

0
130
2

ABCD is a rhombus. If angle A is 129 degrees and the perimeter of ABCD is 60, then find the length of each diagonal.

Feb 20, 2022

#1
0

Draw the rhombus, its diagonals and label the points.

You know the hypotenuse of one of the smaller triangles formed by the diagonals 15.

The diagonals split the triangle into right triangles with one angle of 64.5 degrees, and a hypotenuse of 15.

To find the longer side, we do $$15\times \sin(64.5) \approx13.53877927$$

Multiply this by 2, and one diagonal is: $$30 \times \sin(64.5) \approx \color{brown}\boxed{27.07755853}$$

The other length of the original right triangle is: $$15 \times \cos(64.5) \approx {6.457666452}$$

Multiply this by 2, and the other diagonal is: $${30 \times \cos(64.5)} \approx \color{brown}\boxed{12.9153329}$$

Feb 20, 2022
#2
+1

Another way to get the same answer that BuilderBoi has:

Since the perimeter of the rhombus is 60, each side is 15.

Let's start by drawing the rhombus:

each side is 15;

angle(C) = angle(A) = 129o;

angle(B) = angle(D) = 180o - 129o = 51o.

To find the longer diagonal, DB, use the Law of Cosines on triangle(ABD):

DB2  =  DA+ AB2 - 2·DA·AB·cos(A)

DB2  =  152 + 152 - 2·15·15·cos(129)

DB2  =  733.194176

DB    =  27.08

To find the shorter diagonal, AC, use the Law of Cosines on triangle(CDA):

AC2  =  CD2 + DA2 - 2·CD·DA·cos(51)

Finish this to get the length of the shorter diagonal.

Feb 21, 2022
edited by geno3141  Feb 21, 2022