Question
In a triangle ABC, a point D lies on AB and points E and
F lie on BC such that DF is parallel to AC and DE is parallel to AF. If BE = 4 cm. EF = 6cm. then find the length (in cm) of BC.Solution
Δ BDE ~ Δ BAF -------- (i) Δ BDF ~ Δ ABC ------- (ii) From (i), => x/(x+y) = 4/10 ------- (iii) From (ii), => x/(x+y) = 10/(10+t) ------- (iv) Now using (iii) in (iv) we get, => 4/10 = 10/(10+t) => 20 + 2t = 50 => t = 15 Therefore, BC = 4 + 6 + 15 = 25 cm
13 22 40 67 103 ?
...92,  51,  21,  13.5,  x,  4.25
find the value of (10x + x -5)?
60  61  126  387  ?   7845
2, A, 9, B, 6, C, 13, D, ?
40 30 20 ? 7.5 4.375
9 15 90 95 475 479
6 a b c d e
Find the value of c.
...45    47    97    296    ?     5966
11Â Â Â 12.1Â Â Â Â 14.3Â Â Â Â 17.6Â Â Â Â Â 22Â Â Â Â Â ?
77 ? 190 257 331 412
...979.001 + 4.0087 × 82.79 = ?
...