Question

State and prove area theorem’s?

Answer 1

Hint- In this question, we have to prove the ratio of areas of two similar triangles is always equal to the square of its corresponding sides. Now to solve this question one should know about the basic properties of similarity of triangles. The basic is that the angles in similar triangles are equal. Then second is that the ratio of corresponding sides of a similar triangle is also equal. These are the properties used in the question also one will notice that later on the triangle is divided into two parts to be able to prove the theorem.

Complete step-by-step answer:
Area theorem, the ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

Given,$\vartriangle ABD \sim \vartriangle PQR$
$
  \therefore \angle A = \angle P,\angle B = \angle Q,\angle C = \angle R \\
  \dfrac{{AB}}{{PQ}} = \dfrac{{BC}}{{QR}} = \dfrac{{CA}}{{PR}}{\text{ }}......\left( 1 \right) \\
 $
The ratio of $\vartriangle ABD$and $\vartriangle PQR$ is given as,
$
  \dfrac{{ar\left( {\vartriangle ABC} \right)}}{{ar\left( {\vartriangle PQR} \right)}} = \dfrac{{\dfrac{1}{2} \times BC \times AD}}{{\dfrac{1}{2} \times QR \times PS}} \\
  \dfrac{{ar\left( {\vartriangle ABC} \right)}}{{ar\left( {PQR} \right)}} = \dfrac{{BC}}{{QR}} = \dfrac{{AD}}{{PS}}{\text{ }}......\left( 2 \right) \\
 $
In $\vartriangle ABD$ and $\vartriangle PQS$
$
  \angle B = \angle Q{\text{ }}\left[ {Given} \right] \\
  \angle ABD = \angle PSQ{\text{ }}\left[ {Each{\text{ }}{{90}^ \circ }} \right] \\
  \therefore \vartriangle ABD \sim \vartriangle PQS{\text{ }}\left[ {AA{\text{ similarity}}} \right] \\
 $
$ \Rightarrow \dfrac{{AD}}{{PS}} = \dfrac{{AB}}{{PQ}} = \dfrac{{BD}}{{QS}}{\text{ }}.....\left( 3 \right){\text{ }}\left[ {Corresponding{\text{ sides are propotional}}} \right]$
From (1) and (3), we get
$\dfrac{{AB}}{{PQ}} = \dfrac{{BC}}{{QR}} = \dfrac{{CA}}{{PR}} = \dfrac{{AD}}{{PS}}$
$\dfrac{{BC}}{{QR}} = \dfrac{{AD}}{{PS}}$
From (2) and (4), we get
$
  \dfrac{{ar\left( {\vartriangle ABC} \right)}}{{ar\left( {\vartriangle PQR} \right)}} = \dfrac{{BC}}{{QR}} \times \dfrac{{BC}}{{QR}} = \dfrac{{B{C^2}}}{{Q{R^2}}} \\
  \therefore \dfrac{{ar\left( {ABC} \right)}}{{ar\left( {PQR} \right)}} = \dfrac{{A{B^2}}}{{P{Q^2}}} = \dfrac{{B{C^2}}}{{Q{R^2}}} = \dfrac{{C{A^2}}}{{P{R^2}}} \\
    \\
 $
Using (1), Hence proved.

Note- In this question it should be noted that we use the basic properties of similarity in the triangles in the question. But we should know that we have to prove that the ratio of areas of the triangle is equal to the square of its corresponding sides. By keeping this in mind we should know the basic properties of similarity in triangles like angles in similar triangles are equal, then the ratio of corresponding sides is equal. Now to make the task easier we should know that the triangle is divided so this will make the question easier to solve. Thus by these steps it will become easy to solve the question.