Question

If ∆ABC ~ ∆PQR, A (∆ABC) = 80, A (∆PQR) = 125, then fill in the blanks.
$\frac{\mathrm{A}\left(∆\mathrm{ABC}\right)}{\mathrm{A}\left(∆....\right)}=\frac{80}{125}\therefore \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\boxed{}}{\boxed{}}$

Answer 1

Given:
∆ABC ~ ∆PQR
A (∆ABC) = 80
A (∆PQR) = 125
According to theorem of areas of similar triangles "When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides".
$$\begin{gathered} \therefore \frac{\mathrm{A}\left(∆\mathrm{ABC}\right)}{\mathrm{A}\left(∆\mathrm{PQR}\right)}=\frac{{\mathrm{AB}}^2}{{\mathrm{PQ}}^2} \\ \Rightarrow \frac{80}{125}=\frac{{\mathrm{AB}}^2}{{\mathrm{PQ}}^2} \\ \Rightarrow \frac{16}{25}=\frac{{\mathrm{AB}}^2}{{\mathrm{PQ}}^2} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{4^2}{5^2}=\frac{{\mathrm{AB}}^2}{{\mathrm{PQ}}^2} \\ \Rightarrow \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\boxed{4}}{\boxed{5}} \end{gathered}$$
Therefore, $\frac{\mathrm{A}\left(∆\mathrm{ABC}\right)}{\boxed{\mathrm{A}\left(∆\mathrm{PQR}\right)}}=\frac{80}{125}\mathrm{and}\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\boxed{4}}{\boxed{5}}$