In Delta ABC,overline{AB} cong overline{AC} a | vedclass

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

Question

(A) In $\Delta ABC$,$\overline{AD}$ is a median to the base $\overline{BC}$.Since $\Delta ABC$ is an isosceles triangle with $\overline{AB} \cong \ove

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(A) In $\Delta ABC$,$\overline{AD}$ is a median to the base $\overline{BC}$.Since $\Delta ABC$ is an isosceles triangle with $\overline{AB} \cong \overline{AC}$,the median $\overline{AD}$ to the base is also the altitude to the base.Therefore,$\overline{AD} \perp \overline{BC}$ and $\angle ADB = 90^{\circ}$.Let $AB = AC = x$. The perimeter of $\Delta ABC$ is $AB + AC + BC = 36$.$x + x + BC = 36 \implies BC = 36 - 2x$.Since $D$ is the midpoint of $\overline{BC}$,$BD = \frac{BC}{2} = \frac{36 - 2x}{2} = 18 - x$.In right-angled $\Delta ABD$,by the Pythagorean theorem:$AB^2 = AD^2 + BD^2$$x^2 = 12^2 + (18 - x)^2$$x^2 = 144 + 324 - 36x + x^2$$36x = 468$$x = 13$.Thus,$BD = 18 - 13 = 5$,so $BC = 2 	imes 5 = 10$.The area of $\Delta ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes BC 	imes AD$.Area $= \frac{1}{2} 	imes 10 	imes 12 = 60$.

In $\Delta ABC$,$\overline{AB} \cong \overline{AC}$ and $\overline{AD}$ is a median. If $AD = 12$ and the perimeter of $\Delta ABC$ is $36$,find the area of $\Delta ABC$.

Similar Questions

$\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow RPQ$. If $m \angle A + m \angle C = m \angle B$,then in $\Delta PQR$,$\ldots \ldots \ldots$ is a right angle.

In $\Delta ABC$,$m \angle B = 90^{\circ}$,$\overline{BM}$ is an altitude to the hypotenuse $AC$,and $AM < CM$. If $BM = 6$ and $AC = 13$,find $AB$.

In $\Delta ABC$,$\overline{AB} \cong \overline{AC}$ and $\overline{AD}$ is a median. If $BC = 12$ and $AD = 8$,find $AB$.

In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$. If $\frac{AB}{5} = \frac{AC}{3}$ and $BD = 4.5$,find $BC$.

In $\Delta GBS$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{GS}$. Prove that $\frac{GB^{2}}{BS^{2}} = \frac{GM}{SM}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module