question 13 of 15 (1 point) | question attempt: 1 of unlimited
m∠deg

Question

question 13 of 15 (1 point) | question attempt: 1 of unlimited
m∠deg < m∠fge
cannot be determined
pr select xy
cannot be determined

Accepted Answer

# Explanation: ## Step1: Analyze Triangle QPR In triangle $ QPR $, two sides are marked equal (isosceles triangle). The given angle is $ 81^\circ $ at $ Q $. So the base angles at $ P $ and $ R $ can be found. The sum of angles in a triangle is $ 180^\circ $, so the other two angles sum to $ 180 - 81 = 99^\circ $. Since it's isosceles, each base angle is $ \frac{99}{2} = 49.5^\circ $. ## Step2: Analyze Triangle XZY In triangle $ XZY $, two sides are marked equal (isosceles triangle). The given angle is $ 96^\circ $ at $ Z $. The other two angles sum to $ 180 - 96 = 84^\circ $, so each base angle is $ \frac{84}{2} = 42^\circ $. ## Step3: Compare Sides PR and XY In a triangle, the larger angle is opposite the longer side. In triangle $ QPR $, the angle opposite $ PR $ is $ 81^\circ $. In triangle $ XZY $, the angle opposite $ XY $ is $ 96^\circ $? Wait, no. Wait, in triangle $ QPR $, sides $ QP = QR $ (marked), so angles at $ P $ and $ R $ are equal (49.5° each). In triangle $ XZY $, sides $ XZ = YZ $ (marked), so angles at $ X $ and $ Y $ are equal (42° each). Wait, maybe I mixed up. Wait, in triangle $ QPR $, the angle at $ Q $ is $ 81^\circ $, so sides $ QP = QR $, so $ PR $ is opposite $ 81^\circ $. In triangle $ XZY $, angle at $ Z $ is $ 96^\circ $, sides $ XZ = YZ $, so $ XY $ is opposite $ 96^\circ $. Wait, no, in an isosceles triangle, the equal sides are the legs, and the unequal side is the base, opposite the vertex angle. So in $ \triangle QPR $, $ QP = QR $ (legs), so base is $ PR $, opposite $ \angle Q = 81^\circ $. In $ \triangle XZY $, $ XZ = YZ $ (legs), so base is $ XY $, opposite $ \angle Z = 96^\circ $. Wait, but the other sides: Wait, maybe the marked sides are different. Wait, the problem is about comparing $ PR $ and $ XY $. Wait, maybe the triangles have two sides equal (the marked ones) and we can use the Hinge Theorem? Wait, no, the Hinge Theorem (SAS Inequality) states that if two sides of one triangle are congruent to two sides of another triangle, but the included angle is larger, then the third side is longer. But here, the triangles have two sides marked as equal (maybe $ QP = XZ $ and $ QR = YZ $? Wait, the diagram shows in $ \triangle QPR $, two sides are marked (maybe $ QP $ and $ QR $), and in $ \triangle XZY $, two sides are marked (maybe $ XZ $ and $ YZ $). Wait, maybe the marked sides are equal (i.e., $ QP = XZ $ and $ QR = YZ $). Then, in $ \triangle QPR $, the included angle is $ 81^\circ $, and in $ \triangle XZY $, the included angle is $ 96^\circ $. By the Hinge Theorem, if two sides of one triangle are congruent to two sides of another triangle, and the included angle is larger, then the third side is longer. So since $ 81^\circ < 96^\circ $, the side opposite the larger angle (in the triangle with the larger included angle) would be longer. Wait, no: Hinge Theorem: If $ AB = DE $, $ AC = DF $, and $ \angle A > \angle D $, then $ BC > EF $. So here, if $ QP = XZ $, $ QR = YZ $, and $ \angle Q = 81^\circ $, $ \angle Z = 96^\circ $, then since $ \angle Z > \angle Q $, the side opposite $ \angle Z $ (which is $ XY $) would be longer than the side opposite $ \angle Q $ (which is $ PR $). Wait, no, wait: in $ \triangle QPR $, sides $ QP $ and $ QR $ are equal, and included angle $ \angle Q = 81^\circ $, so side $ PR $ is opposite $ \angle Q $. In $ \triangle XZY $, sides $ XZ $ and $ YZ $ are equal, included angle $ \angle Z = 96^\circ $, so side $ XY $ is opposite $ \angle Z $. If $ QP = XZ $ and $ QR = YZ $ (marked as equal in the diagram, same tick marks), then by Hinge Theorem, since $ \angle Z > \angle Q $, $ XY > PR $, so $ PR < XY $. Wait, but the question is $ PR $ [Select] $ XY $. So the options are $ < $, $ > $, or cannot be determined. Wait, but maybe the marked sides are not equal? Wait, the diagram shows in $ \triangle QPR $, two sides with one tick, and in $ \triangle XZY $, two sides with one tick. So maybe $ QP = XZ $ and $ QR = YZ $ (same tick marks). So then, by Hinge Theorem, since $ \angle Q = 81^\circ $ and $ \angle Z = 96^\circ $, and $ 81 < 96 $, the side opposite the larger angle (XY) is longer than the side opposite the smaller angle (PR). So $ PR < XY $. Wait, but let's recheck. Wait, in $ \triangle QPR $, angles: $ \angle Q = 81^\circ $, $ \angle P = \angle R = (180 - 81)/2 = 49.5^\circ $. In $ \triangle XZY $, $ \angle Z = 96^\circ $, $ \angle X = \angle Y = (180 - 96)/2 = 42^\circ $. Now, if the triangles have two sides equal (e.g., $ QP = XZ $ and $ QR = YZ $), then we can compare the third sides. Alternatively, maybe the triangles are isosceles with two sides equal, and we can compare the sides opposite the given angles. Wait, the angle at $ Q $ is $ 81^\circ $, opposite $ PR $. The angle at $ Z $ is $ 96^\circ $, opposite $ XY $. But unless we know the lengths of the equal sides, we can't directly compare $ PR $ and $ XY $? Wait, no, the Hinge Theorem requires two sides to be congruent. If the two sides (the equal ones) are congruent, then we can apply Hinge Theorem. Since the diagram marks the equal sides with the same tick marks, we can assume $ QP = XZ $ and $ QR = YZ $. Then, included angle $ \angle Q = 81^\circ $, $ \angle Z = 96^\circ $. So by Hinge Theorem, $ XY > PR $, so $ PR < XY $. Wait, but the options are $ PR $ [Select] $ XY $, with options $ < $, $ > $, or cannot be determined. Wait, maybe I made a mistake. Wait, let's think again. In $ \triangle QPR $, sides $ QP = QR $, so it's isosceles with base $ PR $, vertex angle $ 81^\circ $. In $ \triangle XZY $, sides $ XZ = YZ $, isosceles with base $ XY $, vertex angle $ 96^\circ $. The length of the base in an isosceles triangle is given by $ 2a \sin(\theta/2) $, where $ a $ is the length of the equal sides, and $ \theta $ is the vertex angle. So $ PR = 2a \sin(81/2) = 2a \sin(40.5^\circ) $, and $ XY = 2b \sin(96/2) = 2b \sin(48^\circ) $. Now, if $ a = b $ (since the equal sides are marked with the same tick, so $ a = b $), then $ PR = 2a \sin(40.5^\circ) $, $ XY = 2a \sin(48^\circ) $. Since $ \sin(40.5^\circ) < \sin(48^\circ) $ (because $ 40.5 < 48 $ and sine is increasing in $ [0, 90^\circ] $), then $ PR < XY $. So $ PR < XY $. # Answer: $ PR < XY $ (so the correct symbol is $ < $)

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Analyze Triangle QPR

Step2: Analyze Triangle XZY

Step3: Compare Sides PR and XY

Answer: