In trapezium K L M N, K L and M N are parallel sides. A line is drawn, from the point A on K N, parallel to M N meeting L M at B . K N: L M is equal to

Question

In trapezium K L M N, K L and M N are parallel sides. A line is drawn, from the point A on K N, parallel to M N meeting L M at B . K N: L M is equal to (A) K L: N M (B) (K L+K A):(N M+B M) (C) (K A-A N):(L B-B M) (D) K L2: M N2

Tardigrade Admin · Accepted Answer

Apply BPT and use componondo-dividendo after drawing the complete figure.

Q. In trapezium $K L MN, K L$ and $MN$ are parallel sides. $A$ line is drawn, from the point $A$ on $K N$ , parallel to $MN$ meeting $L M$ at $B . K N : L M$ is equal to

$K L : NM$

$(K L + K A) : (NM + BM)$

$(K A - A N) : (L B - BM)$

$K L^{2} : M N^{2}$

Apply BPT and use componondo-dividendo after drawing the complete figure.

Q. In trapezium KLMN,KL and MN are parallel sides. A line is drawn, from the point A on KN, parallel to MN meeting LM at B.KN:LM is equal to

KL:NM

(KL+KA):(NM+BM)

(KA−AN):(LB−BM)

KL2:MN2

Apply BPT and use componondo-dividendo after drawing the complete figure.

Q. In trapezium $K L MN, K L$ and $MN$ are parallel sides. $A$ line is drawn, from the point $A$ on $K N$ , parallel to $MN$ meeting $L M$ at $B . K N : L M$ is equal to

$K L : NM$

$(K L + K A) : (NM + BM)$

$(K A - A N) : (L B - BM)$

$K L^{2} : M N^{2}$