Question

State which pairs of triangles in figure are similar? Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
          

Answer 1

Hint: To solve this variety of triangle, given to use a similar triangle concept. Similar triangles, two figures having the same shape (but not necessarily the same size) are called similar figures.
Two triangles are said to be similar, if
Their corresponding angles are equal and
Their corresponding sides are in the same ratio proportion.
Check the two conditions for all parts $$\left[\left(1\right)\text{ to }\left(6\right)\right]$$whether we get similar or not.

Complete step-by-step answer:
1. Given in $$\mathrm{\Delta }ABC$$ and $$\mathrm{\Delta }PQR$$
The corresponding angles are
$$\mathrm{\angle }A=\mathrm{\angle }P={60}^{\circ }$$
$$\mathrm{\angle }B=\mathrm{\angle }Q={80}^{\circ }$$
$$\mathrm{\angle }C=\mathrm{\angle }R={40}^{\circ }$$ $$$$
By $$AAA$$ similarity, $$\mathrm{\Delta }ABC$$ and $$\mathrm{\Delta }PQR$$ is similar
This is $$\mathrm{\Delta }ABC\sim \mathrm{\Delta }PQR$$
2. In $$\mathrm{\Delta }ABC$$and $$\mathrm{\Delta }QRP$$
The sides are
 $$AB=2$$, $$BC=2.5$$,$$CA=3$$
$$QR=4$$, $$RP=5$$, $$PQ=6$$
Thus, $$\mathrm{\Delta }ABC$$ and $$\mathrm{\Delta }QRP$$ Corresponding sides are in the same ratio, that is,
$${\displaystyle \frac{AB}{QR}}={\displaystyle \frac{2}{4}}={\displaystyle \frac{1}{2}}$$
$${\displaystyle \frac{AC}{QP}}={\displaystyle \frac{3}{6}}={\displaystyle \frac{1}{2}}$$
$${\displaystyle \frac{BC}{RP}}={\displaystyle \frac{2.5}{5}}={\displaystyle \frac{1}{2}}$$
Hence, $${\displaystyle \frac{AB}{QR}}={\displaystyle \frac{AC}{QP}}={\displaystyle \frac{BC}{RP}}={\displaystyle \frac{1}{2}}$$
By $$SSS$$ similarity, $$\mathrm{\Delta }ABC$$ and $$\mathrm{\Delta }QRP$$ is similar
This is $$\mathrm{\Delta }ABC\sim \mathrm{\Delta }QRP$$
3. In $$\mathrm{\Delta }MPL$$ and $$\mathrm{\Delta }EDF$$
The sides are
$$MP=2,PL=3,LM=2.7$$
$$ED=4,DF=6,FE=5$$
Thus,
$$\mathrm{\Delta }MPL$$ and $$\mathrm{\Delta }EDF$$ corresponding sides are not in the same ratio, that is,
$${\displaystyle \frac{MP}{ED}}={\displaystyle \frac{2}{4}}={\displaystyle \frac{1}{2}}$$
$${\displaystyle \frac{PL}{DF}}={\displaystyle \frac{3}{6}}={\displaystyle \frac{1}{2}}$$
$${\displaystyle \frac{LM}{FE}}={\displaystyle \frac{2.7}{5}}$$
Hence, the ratio is not here, the corresponding sides of both the triangles not in the same ratio. And also corresponding angles are not given. Therefore, $$\mathrm{\Delta }MPL$$ is not similar $$\mathrm{\Delta }EDF$$.
4. In $$\mathrm{\Delta }MNL$$ and $$\mathrm{\Delta }QPR$$
The corresponding angles are
$$\mathrm{\angle }M={70}^{\circ }$$
$$\mathrm{\angle }Q={70}^{\circ }$$
Now, the corresponding sides are
$${\displaystyle \frac{MN}{QP}}={\displaystyle \frac{2.5}{5}}={\displaystyle \frac{1}{2}}$$
$${\displaystyle \frac{ML}{QR}}={\displaystyle \frac{5}{10}}={\displaystyle \frac{1}{2}}$$
so, $${\displaystyle \frac{MN}{PQ}}={\displaystyle \frac{ML}{QR}}={\displaystyle \frac{1}{2}}$$
By, $$SAS$$ Similarity, $$\mathrm{\Delta }MNL$$ and $$\mathrm{\Delta }QPR$$ are similar
This is, $$\mathrm{\Delta }MNL\sim \mathrm{\Delta }QPR$$
5. In $$\mathrm{\Delta }ACB$$ and $$\mathrm{\Delta }FDE$$
The corresponding angles are
$$\mathrm{\angle }A={80}^{\circ }$$
$$\mathrm{\angle }F={80}^{\circ }$$
The corresponding sides are
$${\displaystyle \frac{AB}{FD}}={\displaystyle \frac{2.5}{5}}={\displaystyle \frac{1}{2}}$$
$${\displaystyle \frac{AC}{FE}}={\displaystyle \frac{AC}{6}}$$
So the sides are not corresponding because of $$AC$$ is not given
We cannot say the two triangles are under any of $$SAS,SSS,AAA$$ similarity.
So, $$\mathrm{\Delta }ACB$$ and $$\mathrm{\Delta }FDE$$ are not similar.
6. In$$\mathrm{\Delta }DEF$$ and $$\mathrm{\Delta }PQR$$
Here, the two triangles have only angles, for that we compare to similar triangles for corresponding angles are same or not. For that, There is one angle we need to find either $$\mathrm{\angle }F$$ or $$\mathrm{\angle }P$$.
Now, we find $$\mathrm{\angle }P$$ for $$\mathrm{\Delta }PQR$$
The angles are $$\begin{array}{c}\mathrm{\angle }Q={80}^{\circ }\\ \mathrm{\angle }R={30}^{\circ }\end{array}$$
We know that, sum of the angles of the triangle is $${180}^{\circ }$$
$$\mathrm{\angle }P+\mathrm{\angle }Q+\mathrm{\angle }R={180}^{\circ }$$
$$\mathrm{\angle }P+{80}^{\circ }+{30}^{\circ }={180}^{\circ }$$
$$\mathrm{\angle }P+{110}^{\circ }={180}^{\circ }$$
$$\mathrm{\angle }P={180}^{\circ }-{110}^{\circ }$$
$$\mathrm{\angle }P={70}^{\circ }$$
In $$\mathrm{\Delta }DEF$$
$$\begin{array}{c}\mathrm{\angle }D=\mathrm{\angle }P={70}^{\circ }\\ \mathrm{\angle }E=\mathrm{\angle }Q={80}^{\circ }\end{array}$$
By, $$AA$$similarity $$\mathrm{\Delta }DEF$$ and $$\mathrm{\Delta }PQR$$ are similar
Which is, $$\mathrm{\Delta }DEF\sim \mathrm{\Delta }PQR$$
Hence, we can say $$(i),(ii),(iv),(vi)$$ are similar triangles
The similarity criterion using is, $$SAS,SSS,AAA,AA$$
Symbolic form of triangles:
 $$\mathrm{\Delta }ABC\sim \mathrm{\Delta }PQR$$,$$\mathrm{\Delta }ABC\sim \mathrm{\Delta }QRP$$,$$\mathrm{\Delta }MNL\sim \mathrm{\Delta }QPR$$,$$\mathrm{\Delta }DEF\sim \mathrm{\Delta }PQR$$

Note: If two triangles, corresponding angles are equal, then their corresponding sides are same ratio and hence the two triangles are similar ($$AAA$$ similarity)
Also, if two angles of triangle are respectively equal to the two angles of another triangle, then the two triangles are similar ($$AA$$ similarity)
If two triangles, sides of one triangle are proportional to the sides of the others triangle, then their corresponding angles are equal and hence the two triangles are similar ($$SSS$$similarity)
If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar ($$SAS$$ similarity)
Here,$$SAS,SSS,AAA,AA$$ we are using this are called properties or similarities of similar triangles.