Question

In a ABC seg AP in median. If BC = 18, AB²+ AC² =260 find AP.​

Answer 1

$\large\text{\underline{Let's begin:-}}$

The Apollonius theorem is a theorem about triangles and median.

In $\triangle ABC$, if $\overline{AP}$ is the median, the following equation is satisfied.

$\hookrightarrow\overline{AB}^{2}+\overline{AC}^{2}=2(\overline{AP}^{2}+\overline{BP}^{2})$

$\large\text{\underline{Solution:-}}$

We can find that $\overline{BP}=\dfrac{1}{2}\overline{BC}=9$.

$\hookrightarrow\overline{AB}^{2}+\overline{AC}^{2}=2(\overline{AP}^{2}+\overline{BP}^{2})$

$\hookrightarrow 260=2(9^{2}+\overline{AP}^{2})$

$\hookrightarrow \overline{AP}^{2}+162=260$

$\hookrightarrow\overline{AP}^{2}=98$

We only count positive values as lengths. So, rejecting the negative solution,

$\hookrightarrow\overline{AP}=\sqrt{2\cdot7^{2}}=7\sqrt{2}$

$\large\text{\underline{Conclusion:-}}$

The answer is $\overline{AP}=7\sqrt{2}$.

Answer 2

Step-by-step explanation:

Question :

In a ABC seg AP in median. If BC = 18, AB²+ AC² = 260 find AP.

❥ For better understanding please refer to the attachment

Solution :

According to the question,

${\bold{\sf{⟼ \: \: \: \: \: PC \: = \: \frac{1}{2} \: BC \: \: \: \: \: \: \: \: \: \: [Midpoint \: theorem]}}}$

${\bold{\sf{⟼ \: \: \: \: \: PC \: = \: \frac{1}{2} \: × \: 18 }}}$

${\bold{\sf{⟼ \: \: \: \: \: PC \: = \: 9 }}}$

We know that in ∆ABC, segment AP is the median

${\bold{\sf{Now, \: AB² \: + \: AC² \: = \: 2A² \: + \: 2 PC² \: \: \: \: \: \: \: \: \: \: [Apollonius \: theorem]}}}$

${\bold{\sf{⟼ \: \: \: \: 260 \: = \: 2AP² \: + \: 2 (9)²}}}$

${\bold{\sf{⟼ \: \: \: \: 130 \: = \: AP² \: + \: 81 \: \: \: \: \: \: \: \: \: [Dividing \: both \: sides \: by \: 2] }}}$

${\bold{\sf{⟼ \: \: \: \: \: AP² \: = \: 130 \: \: – \: \: 81 }}}$

${\bold{\sf{⟼ \: \: \: \: \: AP² \: = \: 49 }}}$

${\bold{\sf{⟼ \: \: \: \: \: AP \: = √49 \: \: \: \: \: \: \: \: \: [Taking \: square \: root \: of \: both \: sides] }}}$

$⟼ \: \: \: \: \: {\bold{\underline{\boxed{\sf{\color{cyan}{AP \: = \: 7}}}}}}$