A rhombus has sides 10 cm long and an angle of 60°. Find the diagonals of the rhombus.

When you know any 1 angle’s measure in a rhombus, you know all 4 angles’ measures. Opposite angles are congruent. And adjacent angles are supplementary - in other words any two adjacent angles sum to 180°. So if we have an angle that measures 60°, the other angles measure 120°, 60°, and 120°.

Now, since the diagonals of a rhombus bisect the pairs of opposite angles, each 120° angle is cut into two 60° angles by the shorter diagonal, forming 2 equilateral triangles within the rhombus, each sharing that shorter diagonal as one of its sides. So the shared side/ diagonal of the rhombus must be the same length as the 4 sides of the rhombus. And since we know the length of the rhombus’s sides is 10 cm, we know that the shorter diagonal is 10 cm as well.

The longer diagonal will bisect the pair of 60° angles. Therefore it will bisect each of the two equilateral triangles into two 30° –60°– 90° triangles. In these smaller triangles, the shorter side is one half of the 10 cm diagonal, or 5 cm. The hypotenuse is 10 cm, and the third side is opposite a 60° angle so it is  53–√  cm. That side of the smaller triangle is only half of the entire diagonal on which it lies, so double its length and you get:

103–√  cm for your longer diagonal. ← got it.

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2