Let $\triangle \mathrm{ABC} \sim \triangle \mathrm{RPQ}$ and $\frac{{ar}(\triangle A B C)}{{ar}(\triangle P Q R)}=\frac{16}{25}$. If $\mathrm{PQ}=4 \mathrm{~cm}, \mathrm{QR}=6 \mathrm{~cm}$ and $\mathrm{PR}=7 \mathrm{~cm}$, then $\mathrm{AC}($ in $\mathrm{cm})$ is equal to:
Answer & explanation
Correct answer: option 2
\(\frac{ae(ABC)}{ar(RPQ)}\) = \(\frac{16}{25}\)
Now,
According to the question,
\(\frac{ae(ABC)}{ar(RPQ)}\) = \( {AC }^{ 2} \)/\( {QR }^{ 2} \)
= \(\frac{16}{25}\) = \( {AC }^{ 2} \)/\( {6 }^{ 2} \)
= 25\( {AC }^{ 2} \) = 36 x 16
= AC = \(\frac{24}{5}\)
= AC = 4.8 cm
Therefore, AC Is 4.8 cm.