In Delta XYZ,overline{XM} is a median. If XY | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) According to Apollonius's Theorem for $\Delta XYZ$ with median $\overline{XM}$ to side $\overline{YZ}$:$XY^2 + XZ^2 = 2(XM^2 + YM^2)$Given $XY = 2

Vedclass · Accepted Answer

$29$

Vedclass · Answer

(D) According to Apollonius's Theorem for $\Delta XYZ$ with median $\overline{XM}$ to side $\overline{YZ}$:$XY^2 + XZ^2 = 2(XM^2 + YM^2)$Given $XY = 20$,$XZ = 21$,and $XM = 14.5$.Substituting the values:$20^2 + 21^2 = 2(14.5^2 + YM^2)$$400 + 441 = 2(210.25 + YM^2)$$841 = 2(210.25 + YM^2)$$420.5 = 210.25 + YM^2$$YM^2 = 420.5 - 210.25 = 210.25$$YM = \sqrt{210.25} = 14.5$Since $\overline{XM}$ is a median,$M$ is the midpoint of $\overline{YZ}$,so $YZ = 2 	imes YM$.$YZ = 2 	imes 14.5 = 29$.

In $\Delta XYZ$,$\overline{XM}$ is a median. If $XY = 20$,$XZ = 21$ and $XM = 14.5$,find $YZ$.

Similar Questions

In $\Delta ABC$ and $\Delta PQR$,$\angle A \cong \angle Q$ and $\angle B \cong \angle R$. Which correspondence between them is a similarity?

In $\Delta ABC$,$m \angle B = 90^{\circ}$. If $AB = 20$ and $AC = 29$,find the perimeter of $\Delta ABC$.

In $\Delta ABC$,the bisector of $\angle B$ intersects $\overline{AC}$ at $D$. If $\frac{AD}{DC} = \frac{3}{4}$ and $AB = 7.5$,then $BC = \ldots$

In $\Delta ABC$,$M$ lies on $AB$ and $N$ lies on $AC$ such that $\overline{MN} \parallel \overline{BC}$. If $\frac{AM}{MB} = \frac{3}{4}$ and $AC = 21$,find $AN$.

In $\Delta ABC$,$m\angle B = 90^{\circ}$,$N \in \overline{AB}$ and $M \in \overline{BC}$. Prove that $AM^{2} + CN^{2} = AC^{2} + MN^{2}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module