In trapezium ABCD has AD parallel to BC-AC and BD intersect P- If dfrac -ADP-BCP-dfrac -1-2- dfrac -ADP-ABCD- -Here the notion -P-1-P-n- denotes the area of the polygon - -P-1-P-n-

Question

Toppr Admin · Accepted Answer

Correct option is A- -2-sqrt -3-

In trapezium $$ABCD$$ has $$AD$$ parallel to $$BC,AC$$ and $$BD$$ intersect at $$P$$. If $$\dfrac {[ADP]}{[BCP]}=\dfrac {1}{2}$$, find $$\dfrac {[ADP]}{[ABCD]}$$. (Here the notion $$[P_{1}...P_{n}]$$ denotes the area of the polygon ) $$[P_{1}...P_{n}]$$

Correct option is A. $$2-\sqrt {3}$$