Q1.
ABC is a right angle triangle with right angle at B. D is a point on AC such that ABD = 45°. If AC = 6 cm and AD = 2cm then AB is
(A) 6/√5 cm (B) 3√2 cm (C) 12/√5 cm (D) 2 cm (E) 6√5 cm
Q2.
Let F be an arbitrary point on the side AB of an acute angled triangle ABC. Let D be the point of intersection of BC with the straight line AD drawn parallel to FC through A. Let E be the point of intersection of AC with straight line BE drawn parallel to FC through B. If AD = 2 cm and BE = 3 cm then FC is:
(A) 1 cm (B) 6/5 cm (C) 7/6 cm (D) 9/5 cm (E) 11/6 cm
Q3.
For a regular octagon inscribed in a circle of radius 1, the product of the distances from a fixed vertex to the other seven vertices is:
(A) 4 (B) 8 (C) 12 (D) 16 (E) 20
please give the solutions.......
n/a
Q. 1
BD is angle bisector.
AB/BC = AD/DC
X / V(36 - x2) = 2 / 4
X = 6 / v5 cm
Q.2
FC = AD.BE/(AD+BE) =6/5 cm.
hey thanks alot buddy
but please tell me how this relation holds true...FC=AD.BE/(AD+BE)
also please tell me from where can i brush up these basics ?
thanks again
n/a
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