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Question

Show that ABC, where A(2,0),B(2,0),C(0,2) and PQR, where P(4,0),Q(4,0),R(0,4) are similar.

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Solution


In ABC,
AB=(x2x1)2+(y2y1)2
=(2(2))2+(00)2
=(2+2)2
=(4)2
=4units
AB=4units
Similarly, BC=(02)2+(20)2
=(2)2+(2)2
=4+4
=8
=22units
Similarly, AC=(20)2+(02)2
=4+4
=22units
In PQR$,
PQ=(x2x1)2+(y2y1)2
=(4(4))2+(00)2
=(4+4)2
=(8)2
=8units
Similarly, QR=(04)2+(40)2
=(4)2+(4)2
=16+16
=32
=42units
Similarly, PR=(4)2+(4)2
=16+16
=32
=42units
Now as pe BPT,
APQ=ACPR=BCQR
48=2242=2242
12=12=12
ABCPQR ---- Hence proved

997186_1062253_ans_bf5529755996497f9d2013c25caec127.png

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