Question

Suppose given a trapezoid ABCD with bases BC=5 and AD=8, and leg CD=3. The lines AB and CD meet at a point P.
Find DP

Answer 1

The value of $DP$ is $8\ units$

Step-by-step explanation:

Given,

Trapezium $ABCD$ with base $BC=5,AD=8$ and leg $CD=3$

The lines $AB$ and $CD$ meet at a point $P$

From figure,

In $\triangle BPC\ and\ \triangle APD$,

• $\angle DAP=\angle CBP$ (Corresponding angles)
• $\angle APD=\angle BPC$ (Same angle for both triangle)

• $\angle BCP=\angle ADP$ (Corresponding angles)

∴ $\triangle BPC$≅$\triangle APD$

So, We can write,

$\frac{BP}{AP}=\frac{CP}{DP}=\frac{BC}{AD}$

⇒$\frac{CP}{CP+3}=\frac{5}{8}$

⇒$8CP=5CP+15$

⇒$3CP=15$

∴$CP=5$

∴ $DP=CP+CD=5+3=8\ units$

So, The value of $DP$ is $8\ units$