Question
In the given figure, $\triangle\text{ACB}\sim\triangle\text{APQ}.$ If BC = 10cm, PQ = 5cm, BA = 6.5cm and AP = 2.8cm, find CA and AQ. Also, find the area$ (\triangle\text{ACB)}$ : area $ (\triangle\text{APQ)}.$
✓
Answer
Given: $\triangle\text{ACB}$ is similar to $\triangle\text{APQ}$ BC = 10cm, PQ = 5cm, BA = 6.5cm and AP = 2.8cmTo find:
We know that for any two similar triangles the sides are proportional.
Hence, $\frac{\text{AB}}{\text{AQ}}=\frac{\text{BC}}{\text{PQ}}=\frac{\text{AC}}{\text{AP}}$
$\frac{\text{AB}}{\text{AQ}}=\frac{\text{BC}}{\text{PQ}}$
$\frac{6.5}{\text{AQ}}=\frac{10}{5}$
AQ = 3.25cm
Similarly,
$\frac{\text{BC}}{\text{PQ}}=\frac{\text{CA}}{\text{AP}}$
$\frac{\text{CA}}{2.8}=\frac{10}{5}$
CA = 5.6cm
- CA and AQ
- Area of $\triangle\text{ACB}$ : Area of $\triangle\text{APQ}$
- It is given that $\triangle\text{ACB}\sim\triangle\text{APQ}$
We know that for any two similar triangles the sides are proportional.
Hence, $\frac{\text{AB}}{\text{AQ}}=\frac{\text{BC}}{\text{PQ}}=\frac{\text{AC}}{\text{AP}}$
$\frac{\text{AB}}{\text{AQ}}=\frac{\text{BC}}{\text{PQ}}$
$\frac{6.5}{\text{AQ}}=\frac{10}{5}$
AQ = 3.25cm
Similarly,
$\frac{\text{BC}}{\text{PQ}}=\frac{\text{CA}}{\text{AP}}$
$\frac{\text{CA}}{2.8}=\frac{10}{5}$
CA = 5.6cm
- We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
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