Question

If \[ABC \sim PQR\] and \[Area\left( {\vartriangle ABC} \right) = 80,Area\left( {\vartriangle PQR} \right) = 245\] then find the value of \[\dfrac{{AB}}{{PQ}}\& \dfrac{{A{B^2}}}{{P{Q^2}}}\].

Answer 1

Hint: Here we will use the theorem that If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides. By using this property of a similar triangle we will find the required values.

Complete step by step solution:
According to the theorem we can say that
\[\dfrac{{Area\left( {\vartriangle ABC} \right)}}{{Area\left( {\vartriangle PQR} \right)}} = \dfrac{{A{B^2}}}{{P{Q^2}}} = \dfrac{{B{C^2}}}{{Q{R^2}}} = \dfrac{{C{A^2}}}{{R{P^2}}}\]
Where AB,BC,QR are sides of the triangle ABC and PQ,QR,RP are sides of the triangle PQR corresponding to the sides of the triangle ABC.
Now,
\[\therefore \dfrac{{Area\left( {\vartriangle ABC} \right)}}{{Area\left( {\vartriangle PQR} \right)}} = \dfrac{{80}}{{245}}\]
Which means,
\[ \Rightarrow \dfrac{{Area\left( {\vartriangle ABC} \right)}}{{Area\left( {\vartriangle PQR} \right)}} = \dfrac{{16}}{{49}}\]
So we can say that
\[ \Rightarrow \dfrac{{A{B^2}}}{{P{Q^2}}} = \dfrac{{16}}{{49}}\]
By square rooting both the sides we are getting
\[\begin{array}{l}
 \Rightarrow \dfrac{{AB}}{{PQ}} = \sqrt {\dfrac{{16}}{{49}}} \\
 \Rightarrow \dfrac{{AB}}{{PQ}} = \dfrac{4}{7}
\end{array}\]
So from here We can clearly see that
\[\dfrac{{A{B^2}}}{{P{Q^2}}} = \dfrac{{16}}{{49}}\& \dfrac{{AB}}{{PQ}} = \dfrac{4}{7}\]

Note:
The Theorem is only true for corresponding sides for example \[\dfrac{{Area\left( {\vartriangle ABC} \right)}}{{Area\left( {\vartriangle PQR} \right)}} \ne \dfrac{{A{B^2}}}{{Q{R^2}}}\] . One must keep this in mind while doing these types of Questions.