Question

In a parallelogram ABCD, a point K is taken on diagonal BD, when AK is extended it intersects CD and BC at M and L, respectively. If AK = $6$ units, MK = $4$ units, then LM equals

• A. $9$
• B. $6$
• C. $5$
• D. $\sqrt{24}$

Answer 1

The correct option is C $5$

The base of an isosceles triangle is half the length of each of its congruent sides.

Here the length $AM=A{K}_1+K_{1}M$

$=6+4$

$=10$ units.

We need to find length of $LM$ which is the bases

i.e, $LM=\frac{1}{2}AM$

$=\frac{1}{2}\times 10$

$=5$ units.

Hence, the answer is $5.$