∆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.
∆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,
\[{AB}^2 = {AO}^2 + {OB}^2 \]
\[ \Rightarrow \left( 6 \right)^2 = {AO}^2 + \left( 3 \right)^2 \]
\[ \Rightarrow 36 - 9 = {AO}^2 \]
\[ \Rightarrow {AO}^2 = 27\]
\[ \Rightarrow AO = 3\sqrt{3} cm . . . \left( 2 \right)\]
In ∆AOP,
\[{AP}^2 = {AO}^2 + {OP}^2 \]
\[ \Rightarrow {AP}^2 = \left( 3\sqrt{3} \right)^2 + \left( 1 \right)^2 \left( \text{From} \left( 1 \right) \text{and} \left( 2 \right) \right)\]
\[ \Rightarrow {AP}^2 = 27 + 1\]
\[ \Rightarrow {AP}^2 = 28\]
\[ \Rightarrow AP = 2\sqrt{7} cm\]
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