In Delta ABC,mangle B = 90^{circ} and overlin | vedclass

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

Question

(B) In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.Here,$\overline{BE}$ is the median to t

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.Here,$\overline{BE}$ is the median to the hypotenuse $\overline{AC}$ because $\angle B = 90^{\circ}$.Therefore,$BE = \frac{1}{2} AC$.Given $BE = 8.5$,we have $8.5 = \frac{1}{2} AC$,which implies $AC = 17$.In $\Delta ABC$,by the Pythagorean theorem,$AB^2 + BC^2 = AC^2$.Substituting the known values: $15^2 + BC^2 = 17^2$.$225 + BC^2 = 289$.$BC^2 = 289 - 225 = 64$.$BC = \sqrt{64} = 8$.

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BE}$ is a median. If $AB = 15$ and $BE = 8.5$,find $BC$.

Similar Questions

In $\Delta ABC$,$m \angle B = 90^{\circ}$,$\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$,and $AM > CM$. If $BM = 12$ and $AC = 25$,find $AB$ and $BC$.

In $\Delta ABC$,$\overline{AD}$ is a median. If $AB = 10$,$AC = 24$,and $AD = 13$,find $BC$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude. If $AB = 8$ and $BC = 6$,then $BM = \dots$

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is a median. If $AB = 15$ and $BC = 20$,find $BM$.

In a triangle $PQR$,$N$ is a point on $PR$ such that $QN \perp PR$. If $PN \cdot NR = QN^2$,prove that $\angle PQR = 90^{\circ}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module