Question

PQRS is a Rhombus. m<PQS = (3x=10) and m<SQR = (x=40) Find m<QRS

Answer 1

Answer on Question #57556 – Math – Geometry

Question

PQRS is a Rhombus. $m < PQS = (3x + 10)$ and $m < SQR = (x + 40)$. Find $m < QRS$.

Given:

$$\angle P Q S = 3 x + 1 0$$ $$\angle S Q R = x + 4 0$$ $$\angle Q R S - ?$$

Solution

1) QS is a diagonal of the rhombus, therefore $\angle PQS = \angle SQR$ (the diagonals bisect the angels)

$$3 x + 1 0 = x + 4 0$$ $$2 x = 3 0$$ $$x = 1 5$$

2) Then $\angle SQR = x + 40 = 15 + 40 = 55$, hence $\angle PQR = 2\angle SQR = 2 \cdot 55 = 110$

3) Then $\angle QRS = 180 - 110 = 70$.

Recall rhombus' property: adjacent sides (ones next to each other) of a rhombus are supplementary. This means that their measures add up to 180 degrees.

Answer: $\angle QRS = 70$.

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