Question
Construct a $\triangle\text{PQR},$ in which $PQ = 6cm, QR = 7cm$ and $PR = 8cm.$ Then, construct another triangle whose sides are $\frac{4}{5}\text{ times}$ the corresponding sides of $\triangle\text{PQR}.$
✓
Answer
Steps of construction:
Step 1. Draw a line segment $QR = 7cm.$
Step 2. With $Q$ as centre and radius 6cm, draw an arc.
Step 3. With $R$ as centre and radius 8cm, draw an arc cutting the previous arc at $P$.
Step 4. Join $PQ$ and $PR$. Thus, $\triangle\text{PQR}$ is the required triangle.
Step 5. Below $QR$, draw an acute angle $\angle\text{RQX}.$
Step 6. Along $QX$, mark five points $R_1, R_2, R_3, R_4$ and $R_5$ such that $QR_1 = R_1R_2 = R_2R_3 = R_3R_4= R_4R_5.$
Step 7. Join $RR_5.$
Step 8. From $R_4$, draw $R_4R' || RR_5$ meeting $QR$ at $R'.$
Step 9. From $R'$, draw $P'R' || PR$ meeting $PQ$ in $P'$. Here, $\triangle\text{P}'\text{QR}'$ is the required triangle, each of whose sides are $\frac{4}{5}\text{ times}$ the corresponding sides of $\triangle\text{PQR}.$
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Solve for x and y:
$\frac{5}{\text{x}+\text{1}}-\frac{2}{\text{x}-\text{1}}=\frac{1}{2}$
$\frac{10}{\text{x}+\text{1}}+\frac{2}{\text{y}-\text{1}}=\frac{5}{2}$
→$\frac{5}{\text{x}+\text{1}}-\frac{2}{\text{x}-\text{1}}=\frac{1}{2}$
$\frac{10}{\text{x}+\text{1}}+\frac{2}{\text{y}-\text{1}}=\frac{5}{2}$