Maharashtra State Board course SSC (English Medium) Class 10th Board Exam
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∆Abc is an Equilateral Triangle. Point P is on Base Bc Such that Pc = 1 3 Bc, If Ab = 6 Cm Find Ap. - Geometry

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Question

∆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,

\[{AB}^2 = {AO}^2 + {OB}^2 \]
\[ \Rightarrow \left( 6 \right)^2 = {AO}^2 + \left( 3 \right)^2 \]
\[ \Rightarrow 36 - 9 = {AO}^2 \]
\[ \Rightarrow {AO}^2 = 27\]
\[ \Rightarrow AO = 3\sqrt{3} cm . . . \left( 2 \right)\]

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.

  Is there an error in this question or solution?
Solution ∆Abc is an Equilateral Triangle. Point P is on Base Bc Such that Pc = 1 3 Bc, If Ab = 6 Cm Find Ap. Concept: Similarity in Right Angled Triangles.
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