Let a=BC, b=CA, c=AB be the side lengths of a | vedclass

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

Question

(B) Let $AD$ be the median to side $BC$. By Apollonius Theorem,we have:$AB^2 + AC^2 = 2(AD^2 + BD^2)$Given $a = BC = 8$,so $BD = DC = 4$. Also $m = AD

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) Let $AD$ be the median to side $BC$. By Apollonius Theorem,we have:$AB^2 + AC^2 = 2(AD^2 + BD^2)$Given $a = BC = 8$,so $BD = DC = 4$. Also $m = AD = 6$.Let $AC = b$ and $AB = c$. Given $b - c = 2$,so $c = b - 2$.Substituting these values into the theorem:$(b - 2)^2 + b^2 = 2(6^2 + 4^2)$$b^2 - 4b + 4 + b^2 = 2(36 + 16)$$2b^2 - 4b + 4 = 2(52)$$2b^2 - 4b + 4 = 104$$2b^2 - 4b - 100 = 0$$b^2 - 2b - 50 = 0$Using the quadratic formula $b = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-50)}}{2(1)}$:$b = \frac{2 \pm \sqrt{4 + 200}}{2} = \frac{2 \pm \sqrt{204}}{2} = 1 \pm \sqrt{51}$Since $b$ must be positive,$b = 1 + \sqrt{51} \approx 1 + 7.14 = 8.14$.The nearest integer to $b$ is $8$.

Let $a=BC, b=CA, c=AB$ be the side lengths of a $\triangle ABC$ and $m$ be the length of the median through $A$. If $a=8, b-c=2, m=6$,then the nearest integer to $b$ is

Similar Questions

With usual notation,in a triangle $ABC$,if $\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13}$,then the value of $\cos B$ is equal to

In a $\triangle ABC$,the expression $(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$ is equal to:

In $\triangle ABC$,if $a=2(\sqrt{3}+1)$,$B=45^{\circ}$ and $C=60^{\circ}$,then the area (in sq.units) of that triangle is

The area of the triangle $ABC$ is $10\sqrt{3} \text{ cm}^2$,angle $B$ is $60^{\circ}$ and its perimeter is $20 \text{ cm}$. Then $\ell(AC) = $ (in $\text{ cm}$)

In a $\triangle ABC$,if $a=5, b=6, c=7$,then the length of the median drawn from $B$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module