Question

triangle ABC is an equilateral triangle. Point P is on base BC such that PC=1/3BC. If AB=12cm, find AP

Answer 1

Answer:

△ABC is an equilateral triangle. Point P is on base BC. [ Given ]

$pc = \frac{1}{3} bc.ab = 6cm$

[ Given ]

$Here, PC= \frac{1}{3} \times BC= \frac{1}{3} \times 6$

∴ PC=2cm

$RC= \frac{1}{2} \times BC= \frac{1}{2} \times 6$

∴ RC=3cm

RP=RC−PC=3−2

∴ RP=1cm

$⇒  Altitude AR= \frac{ \sqrt{3} }{2} \times a$

[ a=length of side ]

$⇒  AR= \frac{ \sqrt{3} }{2} \times 6 = 3 \sqrt{3}$

Now, in △ARP,

(AP)² = (AR)² + (RB)²

$AP= \sqrt{(3 { \sqrt{3)} }^{2} + {(1)}^{2} }$

$AP= \sqrt{27 + 1}$

$∴  AP=2 \sqrt{7} cm$