In Delta ABC,m angle B = 90^{circ} and overli | vedclass

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

Question

(A) $1$. In $\Delta ABC$,since $\angle B = 90^{\circ}$,by Pythagoras theorem,$AC^{2} = AB^{2} + BC^{2}$ (Equation $1$).$2$. Since $\overline{AD}$ is a

Vedclass · Accepted Answer

Vedclass · Answer

(A) $1$. In $\Delta ABC$,since $\angle B = 90^{\circ}$,by Pythagoras theorem,$AC^{2} = AB^{2} + BC^{2}$ (Equation $1$).
$2$. Since $\overline{AD}$ is a median to side $\overline{BC}$,$D$ is the midpoint of $\overline{BC}$. Thus,$BD = DC = \frac{1}{2} BC$,which implies $BC = 2BD$.
$3$. In $\Delta ABD$,since $\angle B = 90^{\circ}$,by Pythagoras theorem,$AD^{2} = AB^{2} + BD^{2}$.
$4$. Rearranging for $BD^{2}$,we get $BD^{2} = AD^{2} - AB^{2}$.
$5$. Substitute $BC = 2BD$ into Equation $1$: $AC^{2} = AB^{2} + (2BD)^{2} = AB^{2} + 4BD^{2}$.
$6$. Substitute $BD^{2} = AD^{2} - AB^{2}$ into the expression: $AC^{2} = AB^{2} + 4(AD^{2} - AB^{2})$.
$7$. Simplify: $AC^{2} = AB^{2} + 4AD^{2} - 4AB^{2} = 4AD^{2} - 3AB^{2}$.
$8$. Hence,$AC^{2} = 4AD^{2} - 3AB^{2}$ is proved.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{AD}$ is a median. Prove that $AC^{2} = 4AD^{2} - 3AB^{2}$.

Similar Questions

In $\Delta MNP$,$\overline{MX}$ is a median. If $MN^2 + MP^2 = 50$ and $MX = 3$,then $NP = \ldots$

The perimeter of equilateral $\Delta ABC$ is $30$. Find its area. (in $\sqrt{3}$)

In $\Delta ABC$,$m \angle B = 90^{\circ}$. If $AB = 6$ and $AC = 10$,find the area of $\Delta ABC$.

In $\Delta ABC$,$m \angle A = 90^\circ$. If $b = 12$ and $c = 35$,then $a = \ldots$

In $\Delta PQR$,$P-X-Q$,$P-Y-R$ and $\overline{XY} \parallel \overline{QR}$. If $PY = 4$,$YR = 7$ and $XY = 11$,find $QR$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module