Advertisements
Advertisements
Question
Show that ΔABC, where A(–2, 0), B(2, 0), C(0, 2) and ΔPQR where P(–4, 0), Q(4, 0), R(0, 2) are similar triangles
Advertisements
Solution
In ΔABC, the coordinates of the vertices are A(–2, 0), B(2, 0), C(0, 2).
`AB = sqrt((2+2)^2 + (0 - 0)^2) = 4`
`BC = sqrt((0 - 2)^2 + (2 - 0)^2) =sqrt8 = 2sqrt2`
`CA = sqrt((0 + 2)^2 + (2 - 0)^2) = sqrt8 = 2sqrt2)`
In ΔPQR, the coordinates of the vertices are P(–4, 0), Q(4, 0), R(0, 4).
`PQ = sqrt((4+4)^2 + (0-0)^2) = 8`
`QR = sqrt((0 - 4)^2 + (4 - 0)^2) = 4sqrt2`
`PR= sqrt((0 + 4)^2 + (4 - 0)^2) = 4sqrt2`
Now, for ΔABC and ΔPQR to be similar, the corresponding sides should be proportional
So, `(AB)/(PQ) = (BC)/(QR) = (CA)/(PR)`
`=> 4/8 = (2sqrt2)/(4sqrt2) = (2sqrt2)/(4sqrt2) = 1/2`
Thus, ΔABC is similar to ΔPQR
shaalaa.com
Is there an error in this question or solution?
