∆ABC is an equilateral triangle. Point P is on base BC such that PC = [frac{1}{3}] BC, if AB = 6 cm find AP.
Chapter 2 – Pythagoras Theorem- Text Book Solution
Problem Set 2 | Q 7 | Page 45
∆ABC is an equilateral triangle. Point P is on base BC such that PC = [frac{1}{3}] BC, if AB = 6 cm find AP.
solution
∆ABC is an equilateral triangle.
It is given that,
[PC = frac{1}{3}BC]
[ Rightarrow PC = frac{1}{3} times 6]
[ Rightarrow PC = 2 cm]
[ Rightarrow BP = 4 cm]
Since, ABC is an equilateral triangle, OA is the perpendicular bisector of BC.
∴ OC = 3 cm
⇒ OP = OC − PC
= 3 − 2
= 1 …(1)
Now, According to Pythagoras theorem,
In ∆AOB,
[{AB}^2 = {AO}^2 + {OB}^2 ]
[ Rightarrow left( 6 right)^2 = {AO}^2 + left( 3 right)^2 ]
[ Rightarrow 36 – 9 = {AO}^2 ]
[ Rightarrow {AO}^2 = 27]
[ Rightarrow AO = 3sqrt{3} cm . . . left( 2 right)]
In ∆AOP,
[{AP}^2 = {AO}^2 + {OP}^2 ]
[ Rightarrow {AP}^2 = left( 3sqrt{3} right)^2 + left( 1 right)^2 left( text{From} left( 1 right) text{and} left( 2 right) right)]
[ Rightarrow {AP}^2 = 27 + 1]
[ Rightarrow {AP}^2 = 28]
[ Rightarrow AP = 2sqrt{7} cm]
Hence, AP = 2[sqrt{7}] cm.
Explanation:-
In an equilateral triangle, all sides are equal, so if AB = 6 cm, then AC = BC = 6 cm.
Let’s denote BP as x. Then, using the fact that PC = (1/3)BC, we have:
PC = BC – BP (1/3)BC = 6 – x BC = 3(6 – x)
Since BC = 6 in an equilateral triangle, we can substitute this value and solve for x:
6 = 3(6 – x) 2 = 6 – x x = 4
Therefore, BP = 4. Using the Pythagorean theorem in right triangle APB, we have:
AP^2 = AB^2 – BP^2 AP^2 = 6^2 – 4^2 AP^2 = 20 AP = 2sqrt(5)
Therefore, the length of AP is 2sqrt(5) cm.
Chapter 2 – Pythagoras Theorem- Text Book Solution
Problem Set 2 | Q 7 | Page 45