Question

in $\triangle pqr$, $m\angle p = 60^\circ$, $m\angle q = 30^\circ$, and $m\angle r = 90^\circ$. which of the following statements about $\triangle pqr$ are true?
check all that apply.
a. $qr = 2 \cdot pr$
b. $qr = \frac{\sqrt{3}}{2} \cdot pq$
c. $qr = \sqrt{3} \cdot pr$
d. $pq = \sqrt{3} \cdot pr$
e. $pr = \frac{\sqrt{3}}{2} \cdot pq$
f. $pq = 2 \cdot pr$

Explanation:

Step1: Assign side labels

Let $PR = x$. In $\triangle PQR$, $\angle R=90^\circ$, $\angle Q=30^\circ$, $\angle P=60^\circ$. The hypotenuse $PQ$ is opposite the right angle.

Step2: Find hypotenuse $PQ$

In a 30-60-90 triangle, the side opposite $30^\circ$ (which is $PR$) is half the hypotenuse.
$$PQ = 2 \cdot PR$$

Step3: Find side $QR$

Use Pythagorean theorem: $QR = \sqrt{PQ^2 - PR^2}$. Substitute $PQ=2x$:
$$QR = \sqrt{(2x)^2 - x^2} = \sqrt{4x^2 - x^2} = \sqrt{3x^2} = x\sqrt{3} = \sqrt{3} \cdot PR$$
Also, substitute $PR = \frac{PQ}{2}$ into $QR$:
$$QR = \sqrt{3} \cdot \frac{PQ}{2} = \frac{\sqrt{3}}{2} \cdot PQ$$

Step4: Verify each option

• A: $QR=2\cdot PR$ → $\sqrt{3}x

eq 2x$ (false)

• B: $QR=\frac{\sqrt{3}}{2}\cdot PQ$ → $\sqrt{3}x = \frac{\sqrt{3}}{2}\cdot 2x$ (true)
• C: $QR=\sqrt{3}\cdot PR$ → $\sqrt{3}x = \sqrt{3}\cdot x$ (true)
• D: $PQ=\sqrt{3}\cdot PR$ → $2x

eq \sqrt{3}x$ (false)

• E: $PR=\frac{\sqrt{3}}{2}\cdot PQ$ → $x

eq \frac{\sqrt{3}}{2}\cdot 2x$ (false)

• F: $PQ=2\cdot PR$ → $2x = 2\cdot x$ (true)

Answer:

B. $QR=\frac{\sqrt{3}}{2} \cdot PQ$
C. $QR=\sqrt{3} \cdot PR$
F. $PQ=2 \cdot PR$