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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+(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√2In Δ ABC,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].  Suggest Corrections  0      Similar questions  Related Videos   Visualisation of Pythagoras' Theorem
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