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

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Solution

True
Distance between A(-2,0) and B(2,0),,AB=[2(2)]2+(00)2=4[distance between the points (x1,y1) and (x2,y2),d=(x2x1)2+(y2y1)2]
Similarly, distance between B(2,0) and C(0,2) BC
=(02)2+(20)2=4+4=22In Δ ABC,distance between C(0,2) and A(-2,0),CA=[(0(2))2+(20)2]=4+4=22
Distance between F(0,4) and D(-4,0),
FD=(0+4)2+(04)2=42+(4)2=42
Distance between F(0,4) and E(4,0),}
FE =(40)2+(04)2=44+42=42
And distance between E(4,0) and D(-4,0),
ED = (4(4))2+(00)2=(4+4)2=8
Now, ABDE=48=12ACDF=2242=12BCEF=2242=12ABDE=ACDF=BCEF
Here, we see that sides of ΔABC and ΔFDE are proportional.


Hence, both the triangles are similar[ by SSS rule].

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