Question

# If $$\frac{2a}{3}=\frac{4b}{5}=\frac{3c}{4}$$, then what is the value of $$\frac{18}{a}\sqrt{a^2+c^2-b^2}$$?

Answer

Let $$\frac{2a}{3}=\frac{4b}{5}=\frac{3c}{4}=k$$

$$a=\frac{3k}{2}, b=\frac{5k}{4},c=\frac{4k}{3}$$

⇒$$\frac{18\times2}{3k}\sqrt{\frac{9k^2}{4}+\frac{16k^2}{9}-\frac{25k^2}{16}}$$

⇒$$\frac{18\times2}{3k}\sqrt{\frac{324k^2+256k^2-225k^2}{144}}$$

⇒$$\sqrt{355}$$

Answer (B) – $$\sqrt{355}$$