Question 55

Let $$\triangle$$ABC $$\sim$$ $$\triangle$$RPQ and $$\frac{ar(\triangle \text{ABC})}{ar(\triangle \text{PQR})}=\frac{16}{25}$$. If PQ = 4 cm, QR = 6 cm and PR = 7 cm, then AC (in cm) is equal to:

Solution

$$\triangle$$ABC $$\sim$$ $$\triangle$$RPQ

$$\Rightarrow$$ $$\left(\frac{AC}{QR}\right)^2=\frac{ar(\triangle\text{ABC})}{ar(\triangle\text{PQR})}$$

$$\Rightarrow$$ $$\left(\frac{AC}{6}\right)^2=\frac{16}{25}$$

$$\Rightarrow$$ $$\frac{AC}{6}=\frac{4}{5}$$

$$\Rightarrow$$ AC = 4.8 cm

Hence, the correct answer is Option B

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