In ∆ABC seg AP is a median. If BC = 18, AB2 + AC2 = 260 Find AP.
Chapter 2 – Pythagoras Theorem- Text Book Solution
Problem Set 2 | Q 6 | Page 44
In ∆ABC seg AP is a median. If BC = 18, AB2 + AC2 = 260 Find AP
In ∆ABC, point P is the midpoint of side BC.
\[ \Rightarrow 260 = 2 {AP}^2 + 2\left( 9^2 \right)\]
\[ \Rightarrow 260 = 2 {AP}^2 + 2\left( 81 \right)\]
\[ \Rightarrow 260 = 2 {AP}^2 + 162\]
\[ \Rightarrow 2 {AP}^2 = 260 - 162\]
\[ \Rightarrow 2 {AP}^2 = 98\]
\[ \Rightarrow {AP}^2 = 49\]
\[ \Rightarrow AP = 7\]
Hence, AP = 7.
Explanation:-
In a triangle, a median from a vertex divides the opposite side into two equal segments. Therefore, in triangle ABC, AP is a median from vertex A, which means BP = PC.
Let’s denote BP and PC as x. Then we have:
BC = BP + PC 18 = x + x 18 = 2x x = 9
Therefore, BP = PC = 9.
Now let’s use the median formula, which states that the length of a median from vertex A in triangle ABC is given by:
AP^2 = (AB^2 + AC^2)/2 – BC^2/4
We are given that AB^2 + AC^2 = 260 and BC = 18, so we can substitute these values and simplify:
AP^2 = (260/2) – (18^2/4) AP^2 = 130 – 81 AP^2 = 49 AP = 7
Therefore, the length of the median AP is 7 units.
Chapter 2 – Pythagoras Theorem- Text Book Solution
Problem Set 2 | Q 6 | Page 44