Question

$\Delta ABC$ is an equilateral triangle. Point $P$ is on base $BC$ such that $PC=\frac{1}{3}$BC ,if $AB=6$ cm find $AP$.

Answer 1

In the given figure$,$

$\because$ All the sides of the Equilateral Triangles are equal$,$

$\therefore AB=BC=AC=6$ units.

Also$,$

$PC=1/3BC$

$=1/3\times 6$

$=2$ units.

We know Perpendicular in the Equilateral triangles bisects the opposite sides$,$

$\therefore QB=QC$

$=1/2\times BC$

$=1/2\times 6$

$=3$ units.

In $\Delta AQB,$

Applying Pythagoras Theorem$,$

$A{B}^2=B{Q}^2+A{Q}^2$

$\Rightarrow (6{)}^2=(3{)}^2+A{Q}^2$

$\Rightarrow A{Q}^2=36-9$

$\Rightarrow A{Q}^2=27$

$\Rightarrow AQ=3\surd 3$ units.

In $\Delta AQP,$

$PQ=QC-PC$

$=3-2$

$=1$ units.

Applying Pythagoras Theorem$,$

$A{P}^2=P{Q}^2+A{Q}^2$

$\Rightarrow A{P}^2=(1{)}^2+(3\surd 3{)}^2$

$\Rightarrow A{P}^2=1+27$

$\Rightarrow AP=\surd 28$

$\Rightarrow AP=2\surd 7$units.

Hence$,$ the length of the AP is $2\surd 7$ units.