Question

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In isosceles △ABC, AB = BC and CH is an altitude. Find the perimeter of △ABC, if CH = 84 cm and m∠HBC = m∠BAC +m∠BCH?

Answer 1

Answer:

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Have fun and feel free to ask me something new.

Or we can prove some properities without calculating by details

Step-by-step explanation:

I think I could help you

Have fun and feel free to ask me something new.

Or we can prove some properities without calculating by details

Answer 2

Answer:  The perimeter of triangle ABC is (112√3 + 168) cm.

Step-by-step explanation:  Given that in isosceles triangle △ABC, AB = BC and CH is an altitude. Also,

CH = 84 cm   and  

$m\angle HBC = m\angle BAC+m\angle BCH~~~~~~~~~~~~~~~~~~~~~~~(i)$

We are to find the perimeter of triangle ABC

Since AB = BC, so the angles opposite to them are congruent and have equal measures.

That is, $m\angle BAC=m\angle ACB~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

Now, since angle HBC is an exterior angle of triangle ABC and triangles BAC and ACB are remote interior angles.

So, we have

$m\angle HBC=m\angle BAC+m\angle ACB\\\\\Rightarrow m\angle BAC+m\angle BCH=m\angle BAC+m\angle ACB\\\\\Rightarrow m\angle ACB=m\angle BCH.$      

Therefore, from equation (i) implies that

$m\angle HBC = m\angle BAC+m\angle BCH\\\\\Rightarrow m\angle HBC=m\angle BCH+m\angle BCH~~~~~~~[\textup{applying equations (ii) and (iii)}]\\\\\Rightarrow m\angle HBC=2m\angle BCH.$

Now, from angle sum property in triangle BCH, we have

$m\angle HBC+m\angle BCH+m\angle BHC=180^\circ\\\\\Rightarrow 3m\angle BCH+90^\circ=180^\circ\\\\\Rightarrow 3m\angle BCH=90^\circ\\\\\Rightarrow m\angle BCH=30^\circ.$

So, we get

$m\angle BAC=m\angle ACB=m\angle BCH=30^\circ,\\\\m\angle HBC=2\times30^\circ=60^\circ.$

In right-angled triangle ACH, we have

$\tan 30^\circ=\dfrac{CH}{AH}\\\\\\\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{84}{AH}\\\\\\\Rightarrow AH=84\sqrt3.$

In right-angled triangle BCH, we have

$\tan 60^\circ=\dfrac{CH}{BH}\\\\\\\Rightarrow \sqrt3=\dfrac{84}{BH}\\\\\\\Rightarrow BH=\dfrac{84}{\sqrt3}.$

And,

$\sin 60^\circ=\dfrac{CH}{BC}\\\\\\\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{84}{BC}\\\\\\\Rightarrow BC==\dfrac{168}{\sqrt3}=56\sqrt3.$

Therefore,

$AB=AH-BH=84\sqrt3-\dfrac{84}{\sqrt3}=\sqrt3(84-28)=56\sqrt3.$

Now, in triangle ACH,

$\sin 30^\circ=\dfrac{CH}{AC}\\\\\\\Rightarrow \dfrac{1}{2}=\dfrac{84}{AC}\\\\\Rightarrow AC=168.$

Thus, the perimeter of triangle ABC is given by

$P=AB+BC+CA=56\sqrt3+56\sqrt3+168=112\sqrt3+168.$

The perimeter of triangle ABC is (112√3 + 168) cm.