directions: compare the sides and angles by filling in the blank with a symbol. 21. qr____rs 22. ab____cd 23. m∠fjg____m∠hjg 24. m∠qsp____m∠qsr

Question

directions: compare the sides and angles by filling in the blank with a < or > symbol.
21. qr____rs
22. ab____cd
23. m∠fjg____m∠hjg
24. m∠qsp____m∠qsr

Accepted Answer

### Problem 21: Compare $ QR $ and $ RS $ # Explanation: ## Step 1: Analyze the angles opposite the sides In triangle $ QTR $ and $ STR $, we know that $ \angle QTR = 121^\circ $ and $ \angle STR = 123^\circ $. In a triangle, the larger angle is opposite the longer side. Also, consider the sides opposite these angles in the context of the two triangles sharing the side $ TR $. The side opposite $ \angle STR $ is $ QR $ and the side opposite $ \angle QTR $ is $ RS $? Wait, no, let's correct. Wait, actually, in triangle $ QRS $, we can look at the angles at $ T $. Wait, $ \angle QTR = 121^\circ $, $ \angle STR = 123^\circ $. Since $ \angle STR > \angle QTR $, and in the triangles, the sides opposite these angles: the side opposite $ \angle STR $ is $ QR $ and the side opposite $ \angle QTR $ is $ RS $? Wait, no, let's think again. The angle at $ T $ for $ \triangle QTR $ is $ 121^\circ $, and for $ \triangle STR $ is $ 123^\circ $. Since $ \angle STR > \angle QTR $, and the side opposite $ \angle STR $ is $ QR $, and the side opposite $ \angle QTR $ is $ RS $? Wait, no, actually, in triangle $ QRS $, the angles at $ T $: $ \angle QTR = 121^\circ $, $ \angle STR = 123^\circ $. Since $ \angle STR > \angle QTR $, then the side opposite $ \angle STR $ (which is $ QR $) and the side opposite $ \angle QTR $ (which is $ RS $)? Wait, no, maybe better: in a triangle, larger angle implies longer opposite side. So if $ \angle STR = 123^\circ $ and $ \angle QTR = 121^\circ $, then the side opposite $ \angle STR $ (which is $ QR $) and the side opposite $ \angle QTR $ (which is $ RS $)? Wait, no, let's see: $ \triangle QTR $ has angle $ 121^\circ $ at $ T $, so the side opposite is $ QR $? Wait, no, $ \triangle STR $ has angle $ 123^\circ $ at $ T $, side opposite is $ QR $? Wait, maybe I mixed up. Wait, actually, $ TR $ is a common side. The angle at $ T $ for $ QR $ is $ 121^\circ $, and for $ RS $ is $ 123^\circ $. Since $ 123^\circ > 121^\circ $, the side opposite the larger angle is longer. So the side opposite $ 123^\circ $ is $ QR $, and the side opposite $ 121^\circ $ is $ RS $? Wait, no, that can't be. Wait, no, actually, in $ \triangle QTR $, angle at $ T $ is $ 121^\circ $, so the side opposite is $ QR $. In $ \triangle STR $, angle at $ T $ is $ 123^\circ $, side opposite is $ RS $? No, that's wrong. Wait, maybe the other way: the angle at $ T $ for $ RS $ is $ 123^\circ $, so the side $ RS $ is opposite $ 121^\circ $? No, I think I made a mistake. Wait, let's recall: in a triangle, the larger angle is opposite the longer side. So if we have two angles, one $ 121^\circ $ and one $ 123^\circ $, the side opposite the $ 123^\circ $ angle is longer. So if $ \angle STR = 123^\circ $, then the side opposite to it (which is $ QR $) and the side opposite $ \angle QTR = 121^\circ $ (which is $ RS $)? Wait, no, maybe the sides: $ QR $ is opposite $ \angle STR = 123^\circ $, and $ RS $ is opposite $ \angle QTR = 121^\circ $. Since $ 123^\circ > 121^\circ $, then $ QR > RS $? Wait, no, that would mean $ QR > RS $, but wait, no, let's check again. Wait, $ \angle STR = 123^\circ $, so the side opposite is $ QR $, and $ \angle QTR = 121^\circ $, side opposite is $ RS $. So since $ 123 > 121 $, then $ QR > RS $? Wait, no, that's not right. Wait, maybe the other way: the angle at $ T $ for $ QR $ is $ 121^\circ $, so $ QR $ is opposite $ 121^\circ $, and $ RS $ is opposite $ 123^\circ $. Then since $ 123 > 121 $, $ RS > QR $. Ah, there we go. I had it reversed. So the side opposite the larger angle is longer. So $ \angle STR = 123^\circ $ is opposite $ QR $? No, wait, $ \triangle STR $: angles are at $ S $, $ T $, $ R $. Wait, maybe the diagram is such that $ QT = ST $ (marked with ticks). So $ QT = ST $, and $ TR $ is common. Then in $ \triangle QTR $ and $ \triangle STR $, we have $ QT = ST $, $ TR = TR $, and $ \angle QTR = 121^\circ $, $ \angle STR = 123^\circ $. By the Hinge Theorem (SAS Inequality), 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. So here, $ QT = ST $, $ TR = TR $, and $ \angle STR > \angle QTR $, so the side opposite $ \angle STR $ (which is $ QR $) and the side opposite $ \angle QTR $ (which is $ RS $)? Wait, no, the Hinge Theorem: in $ \triangle QTR $ and $ \triangle STR $, $ QT = ST $, $ TR = TR $, and $ \angle QTR = 121^\circ $, $ \angle STR = 123^\circ $. So the included angle for $ \triangle STR $ is larger, so the side opposite, which is $ QR $, is longer? Wait, no, the Hinge Theorem says that if two sides are equal, and the included angle is larger, the third side is longer. So $ QT = ST $, $ TR = TR $, included angle at $ T $ for $ \triangle STR $ is $ 123^\circ $, for $ \triangle QTR $ is $ 121^\circ $. So the third side of $ \triangle STR $ (which is $ QR $) is longer than the third side of $ \triangle QTR $ (which is $ RS $)? Wait, no, $ \triangle STR $'s third side is $ QR $? No, $ \triangle STR $ has sides $ ST $, $ TR $, $ SR $. $ \triangle QTR $ has sides $ QT $, $ TR $, $ QR $. So since $ ST = QT $, $ TR = TR $, and $ \angle STR > \angle QTR $, then $ SR > QR $ (because $ SR $ is the third side of $ \triangle STR $, $ QR $ is the third side of $ \triangle QTR $). So $ QR < RS $. ## Step 2: Conclusion Since $ \angle STR = 123^\circ > \angle QTR = 121^\circ $, by the Hinge Theorem, $ RS > QR $, so $ QR < RS $. # Answer: $ QR < RS $ ### Problem 22: Compare $ AB $ and $ CD $ # Explanation: ## Step 1: Analyze the triangles We have a quadrilateral $ ABCD $ with $ AB $ and $ CD $ sides, and triangles $ ABD $ and $ CDB $. Wait, the diagram shows $ AD = BC $ (ticks) and $ AB = CD $? Wait, no, the angles: $ \angle ADB = 58^\circ $, $ \angle CBD = 65^\circ $. Wait, in triangle $ ABD $ and $ CDB $, we have $ AD = BC $ (marked), $ BD = BD $ (common side). Now, look at the angles opposite $ AB $ and $ CD $. In $ \triangle ABD $, angle at $ D $ is $ 58^\circ $, in $ \triangle CDB $, angle at $ B $ is $ 65^\circ $. Wait, maybe better: in triangle $ ABD $, angle at $ D $ is $ 58^\circ $, in triangle $ CDB $, angle at $ B $ is $ 65^\circ $. Wait, no, let's see the sides. Wait, the quadrilateral has $ AD = BC $ (ticks) and $ AB $ and $ CD $ with angles at $ D $ and $ B $. Wait, actually, in triangle $ ABD $ and $ CDB $, $ AD = BC $, $ BD = BD $, and the angles at $ D $ and $ B $: $ \angle ADB = 58^\circ $, $ \angle CBD = 65^\circ $. Wait, no, maybe the triangles are isoceles? Wait, the diagram shows $ AD = AB $ (ticks) and $ BC = CD $ (ticks)? No, the ticks: $ AD $ and $ AB $ have ticks? Wait, no, the top side $ AD $ and bottom side $ BC $ have ticks, and $ AB $ and $ CD $ have ticks? Wait, the diagram: $ A $ connected to $ D $ and $ B $, $ D $ connected to $ C $, $ C $ connected to $ B $. $ AD $ has a tick, $ BC $ has a tick, $ AB $ has a tick, $ CD $ has a tick? Wait, no, the angle at $ D $ is $ 58^\circ $, angle at $ B $ is $ 65^\circ $. Wait, in triangle $ ABD $, angle at $ D $ is $ 58^\circ $, in triangle $ CDB $, angle at $ B $ is $ 65^\circ $. Wait, maybe we can compare the sides $ AB $ and $ CD $ by looking at the angles opposite them in their respective triangles. Wait, no, let's think again. The two triangles $ ABD $ and $ CDB $ have $ AD = BC $ (ticks), $ BD = BD $ (common side). The angle between $ AD $ and $ BD $ is $ 58^\circ $, and the angle between $ BC $ and $ BD $ is $ 65^\circ $. By the Hinge Theorem, since $ 65^\circ > 58^\circ $, the side opposite the larger angle (which is $ CD $) is longer than the side opposite the smaller angle (which is $ AB $). Wait, no: in $ \triangle ABD $, sides $ AD $, $ BD $, $ AB $; angle at $ D $ is $ 58^\circ $. In $ \triangle CDB $, sides $ BC $, $ BD $, $ CD $; angle at $ B $ is $ 65^\circ $. Since $ AD = BC $, $ BD = BD $, and $ \angle CBD = 65^\circ > \angle ADB = 58^\circ $, then by Hinge Theorem, $ CD > AB $, so $ AB < CD $. ## Step 2: Conclusion By the Hinge Theorem, since the included angle $ 65^\circ > 58^\circ $, $ AB < CD $. # Answer: $ AB < CD $ ### Problem 23: Compare $ m\angle FJG $ and $ m\angle HJG $ # Explanation: ## Step 1: Analyze the sides We have triangle $ FJG $ and $ HJG $, with $ FJ = HJ $ (ticks), $ GJ $ is common. The sides $ FG = 12 $ and $ HG = 13 $. So in $ \triangle FJG $ and $ \triangle HJG $, $ FJ = HJ $, $ GJ = GJ $, and $ HG > FG $ (13 > 12). By the Hinge Theorem (SAS Inequality), the larger side is opposite the larger angle. So the angle opposite $ HG $ is $ \angle HJG $, and the angle opposite $ FG $ is $ \angle FJG $. Since $ HG > FG $, then $ \angle HJG > \angle FJG $, so $ m\angle FJG < m\angle HJG $. ## Step 2: Conclusion Since $ HG > FG $ and the triangles share $ FJ = HJ $ and $ GJ $, by Hinge Theorem, $ m\angle FJG < m\angle HJG $. # Answer: $ m\angle FJG < m\angle HJG $ ### Problem 24: Compare $ m\angle QSP $ and $ m\angle QSR $ # Explanation: ## Step 1: Analyze the sides We have triangle $ QSP $ and $ QSR $, with $ SP = SR $ (ticks), $ QS $ is common. The sides $ QP = 6 $ and $ QR = 5 $. So $ QP > QR $ (6 > 5). By the Hinge Theorem (SAS Inequality), the larger side is opposite the larger angle. The angle opposite $ QP $ is $ \angle QSP $, and the angle opposite $ QR $ is $ \angle QSR $. Since $ QP > QR $, then $ \angle QSP > \angle QSR $, so $ m\angle QSP > m\angle QSR $. ## Step 2: Conclusion Since $ QP > QR $ and the triangles share $ SP = SR $ and $ QS $, by Hinge Theorem, $ m\angle QSP > m\angle QSR $. # Answer: $ m\angle QSP > m\angle QSR $

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 21: Compare \( QR \) and \( RS \)

Step 1: Analyze the angles opposite the sides

Step 1: Analyze the triangles

Step 2: Conclusion

Step 1: Analyze the sides

Step 2: Conclusion

Answer:

Problem 22: Compare \( AB \) and \( CD \)