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.
Open in App
Solution
True ∴Distance between A(-2,0) and B(2,0),,AB=√[2−(−2)]2+(0−0)2=4[∵distance between the points (x1,y1)and(x2,y2),d=√(x2−x1)2+(y2−y1)2]
Similarly, distance between B(2,0) and C(0,2) BC =√(0−2)2+(2−0)2=√4+4=2√2InΔABC,distance between C(0,2) and A(-2,0),CA=√[(0−(2))2+(2−0)2]=√4+4=2√2
Distance between F(0,4) and D(-4,0),
FD=√(0+4)2+(0−4)2=√42+(−4)2=4√2
Distance between F(0,4) and E(4,0),}
FE =√(4−0)2+(0−4)2=√44+42=4√2
And distance between E(4,0) and D(-4,0),
ED = √(4−(−4))2+(0−0)2=√(4+4)2=8
Now, ABDE=48=12ACDF=2√24√2=12BCEF=2√24√2=12∴ABDE=ACDF=BCEF
Here, we see that sides of ΔABC and ΔFDE are proportional.
Hence, both the triangles are similar[ by SSS rule].