Question

word problems on congruent triangles worksheet
1 for each of these given sets of triangles, state the rule that tells you that they are congruent. find the unknown values.
a) b)
x = _, y = _ ∠oqr = _, ∠qro = _

Explanation:

Step1: Analyze part a)

• Congruence Rule:
• In \(\triangle ABC\) and \(\triangle DEF\), \(BC = DF=15\) (given), \(\angle B=\angle D = 90^{\circ}\), and \(\angle ACB=\angle EFD\) (vertically - opposite angles). So, by \(AAS\) (Angle - Angle - Side) congruence rule \(\triangle ABC\cong\triangle DEF\).
• Finding \(x\):
• Since \(\triangle ABC\cong\triangle DEF\), the corresponding sides are equal. So, \(x = CE = 20\) (by CPCTC - Corresponding Parts of Congruent Triangles are Congruent).
• Finding \(y\):
• Also, \(y=DE = 13\) (by CPCTC).

Step2: Analyze part b)

• Congruence Rule:
• In \(\triangle MNO\) and \(\triangle RQO\), \(MN = RQ = 3\), \(MO=RQ = 5\), \(NO = OQ = 2\). So, by \(SSS\) (Side - Side - Side) congruence rule \(\triangle MNO\cong\triangle RQO\).
• Finding \(\angle OQR\):
• By CPCTC, \(\angle OQR=\angle ONM = 62^{\circ}\).
• Finding \(\angle QRO\):
• We know that the sum of angles in a triangle is \(180^{\circ}\). In \(\triangle RQO\), if \(\angle RQO = 62^{\circ}\) and \(\angle QOR=30^{\circ}\), then \(\angle QRO=180^{\circ}-(62^{\circ}+30^{\circ})\).
• \(\angle QRO = 88^{\circ}\).

Answer:

a) \(x = 20\), \(y = 13\) (congruence rule \(AAS\))
b) \(\angle OQR = 62^{\circ}\), \(\angle QRO=88^{\circ}\) (congruence rule \(SSS\))