Find that length!

Geometry Level 4

There exists a triangle $ABC$ with $AB=5$ and $AC=7$ . Let $D$ be a point on $BC$ such that $CD=2BD$ . Given that $\angle A$ is obtuse, the area of $\triangle ABC = \frac{21\sqrt{11}}{4}$ , and the length of $AD$ can be expressed in the form $\sqrt{n}$ where $n$ is an integer, find the value of $n$ .

The answer is 15.

3 solutions

Niranjan Khanderia
May 24, 2015

$\dfrac{21\sqrt{11}}{4} = \dfrac{\sqrt{11*21*21}}{4} = \dfrac{\sqrt{11*3*7*21}}{4}\\Hero's~ formula~ largest~ factor~ is=parameter=21\\=AB+BC+CA=5+BC+7 .~~~~~~~~~~ BC=9.\\Applying ~Cos~ Rule, ~to~\Delta s~~ABC ~and~ABD,\\7^2=9^2+5^2-9*5*(2CosB)~~and~~AD^2=3^2+5^2-3*5*(2CosB)\\Equating ~~(2CosB)~from ~both~we~have\\\dfrac{49-81-25} 3 =AD^2-9-25. ~~Gives~~AD^2=~~~~~~\large \color{#D61F06}{15}$ ,

Ayush Choubey
Dec 5, 2014

By Stewart's theorem -

$AD^{2}$ = 49BD+50BD/3BD - 2 $BD^{2}$ = 33 - 2 $BD^{2}$

Find $BD^{2}$ by heron's formula = 9 and some rational root

$AD^{2}$ = 33 - $BD^{2}$ = 33-2*9 = `15

$AD^{2}$ = n = 15

Blah Blah
Aug 3, 2014

this problem has two answers. both lengths of $\sqrt{15}$ and $\sqrt{\frac{55}{3}}$ work.

how did you find 15^1/2

hritik agarwal - 6 years, 9 months ago

Find that length!

The answer is 15.

3 solutions

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