Question

for exercises 5 and 6, find the value of x.
5.

6.

for exercises 7 and 8, state whether the triangles are congruent and by which theorem.
7.

8.

1. why is ( lmnocong pqrs )?

Explanation:

5.

Step1: Determine the congruence of triangles

Since in \(\triangle ABC\) and \(\triangle PQR\), \(\angle B=\angle Q = 87^{\circ}\), \(\angle C=\angle R=40^{\circ}\), and \(AC = PR=9\) ft. By the \(AAS\) (Angle - Angle - Side) congruence criterion, \(\triangle ABC\cong\triangle PQR\).

Step2: Use the property of congruent triangles

Corresponding sides of congruent triangles are equal. So \(AB = PQ\). Given \(AB=(x + 4)\) ft and \(PQ = 7\) ft. Then \(x+4=7\).

Step3: Solve for \(x\)

Subtract \(4\) from both sides of the equation \(x+4 = 7\). We get \(x=7 - 4\).

For the first pair of triangles, we do not have enough information (either \(ASA\), \(AAS\), \(SSS\), \(SAS\)) to prove congruence. The marked angles and side do not match the congruence theorems.

For the second pair of triangles, we do not have enough information (either \(ASA\), \(AAS\), \(SSS\), \(SAS\)) to prove congruence. The marked angles and side do not match the congruence theorems.

For the pentagons \(LMNO\) and \(PQRS\), we assume that all corresponding sides and angles are equal. If we consider the side - side - side - side - side (for sides) and angle - angle - angle - angle - angle (for angles) relationships (in a sense similar to polygon congruence where all corresponding parts are equal).

Answer:

\(x = 3\)

6.

Step1: Determine the congruence of triangles

In \(\triangle DEF\) and \(\triangle YZX\), \(\angle D=\angle X=23^{\circ}\), \(\angle E=\angle Y = 64^{\circ}\), and \(EF=YZ = 13.5\) ft. By the \(AAS\) (Angle - Angle - Side) congruence criterion, \(\triangle DEF\cong\triangle YZX\).

Step2: Use the property of congruent triangles

Corresponding sides of congruent triangles are equal. So \(DE=YX\). Given \(DE = 15\) ft and \(YX=(4x-1)\) ft. Then \(4x-1=15\).

Step3: Solve for \(x\)

Add \(1\) to both sides: \(4x=15 + 1=16\). Divide both sides by \(4\): \(x=\frac{16}{4}\).