0 like 0 dislike
91 views

Given $Q U=24 \mathrm{~cm}, Q C=16 \mathrm{~cm}, D A=6 \mathrm{~cm}$ and $C D=x$.

1. Prove $\triangle Q U C \| \triangle A C D$

2. Calculate $x$

| 91 views

0 like 0 dislike
\begin{aligned}
&\text { Solutions: }\\
&\text { 1. In } \triangle Q U C \text { and } \triangle A C D\\
&\hat{C}_{2}=\hat{C}_{4} \text { (vert. opp. } \angle \text { 's) }\\
&\hat{Q}_{2}=\hat{D}_{2} \text { ( } \angle \text { 's in same seg.) }\\
&\hat{U}_{1}=\hat{A}_{1} \text { (sum } \angle \text { 's of } \triangle \text { or } \angle \text { 's in same seg.) }\\
&\therefore \triangle C Q U \| \triangle \triangle C D A \text { (AAA) }\\
&\therefore \frac{C Q}{C D}=\frac{Q U}{D A}=\frac{C U}{C A}(\|\| \triangle \text { 's })\\
&\text { 2. } \frac{C Q}{C D}=\frac{Q U}{D A}\\
&\frac{16}{x}=\frac{24}{6}\\
&96=24 x\\
&\therefore x=4 \mathrm{~cm}
\end{aligned}
by Diamond (58,513 points)

1 like 0 dislike
0 like 0 dislike
1 like 0 dislike
0 like 0 dislike
0 like 0 dislike
1 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike