Question

Question 1
$\Delta$ABC with vertices A(- 2,0), B(2,0) and C(0,2) is similar to $\Delta$DEF with vertices D(-4, 0), E(4,0) and F(0,4). State true or false and justify your answer.

Answer 1

True
$\therefore \text{Distance between A(-2,0) and B(2,0),},AB=\sqrt{[2-(-2){]}^2+(0-0{)}^2}=4\left[\begin{array}{c}\because \text{distance between the points}(x_1,y_1)and(x_2,y_2),d\\ =\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}\end{array}\right]$
Similarly, distance between B(2,0) and C(0,2) BC
$=\sqrt{(0-2{)}^2+(2-0{)}^2}=\sqrt{4+4}=2\sqrt{2}In\Delta ABC,\text{distance between C(0,2) and A(-2,0),}CA=\sqrt{\left[\right(0-\left(2\right){)}^2+(2-0{)}^2]}=\sqrt{4+4}=2\sqrt{2}$
Distance between F(0,4) and D(-4,0),
FD$=\sqrt{(0+4{)}^2+(0-4{)}^2}=\sqrt{4^2+(-4{)}^2}=4\sqrt{2}$
Distance between F(0,4) and E(4,0),}
FE =$\sqrt{(4-0{)}^2+(0-4{)}^2}=\sqrt{4^4+4^2}=4\sqrt{2}$
And distance between E(4,0) and D(-4,0),
ED = $\sqrt{(4-(-4){)}^2+(0-0{)}^2}=\sqrt{(4+4{)}^2}=8$
Now, $\frac{AB}{DE}=\frac{4}{8}=\frac{1}{2}\frac{AC}{DF}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{1}{2}\frac{BC}{EF}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{1}{2}\therefore \frac{AB}{DE}=\frac{AC}{DF}=\frac{BC}{EF}$
Here, we see that sides of $\Delta ABC$ and $\Delta FDE$ are proportional.


$\text{Hence, both the triangles are similar[ by SSS rule].}$