A point inside the triangle

Geometry Level pending

Triangle A B C ABC has side lengths of A C = 360 , C B = 330 AC=360,CB=330 and A B = 250 AB=250 . If the perpendicular distance of point P P to side A C AC is 50 50 and to side A B AB is 70 70 , what is the perpendicular distance of point P P to side C B CB ?

2200 329 − 17750 2200\sqrt{329}-17750 40 329 3 − 3550 33 \dfrac{40\sqrt{329}}{3}-\dfrac{3550}{33} 2200 329 − 17750 156 \dfrac{2200\sqrt{329}-17750}{156} 1592360000 120 \dfrac{\sqrt{1592360000}}{120}

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1 solution

Consider the diagram above.

A r e a A B C = A r e a A P C + A r e a A P B + A r e a B P C Area_{ABC}=Area_{APC}+Area_{APB}+Area_{BPC}

A r e a A B C Area_{ABC} can be found using the Heron’s Formula. We have

s = 360 + 250 + 330 2 = 470 s=\dfrac{360+250+330}{2}=470

Therefore, A r e a A B C Area_{ABC} is:

A r e a A B C = s ( s − a ) ( s − b ) ( s − c ) = 470 ( 470 − 360 ) ( 470 − 250 ) ( 470 − 330 ) = 1592360000 = 2200 329 Area_{ABC}=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{470(470-360)(470-250)(470-330)}=\sqrt{1592360000}=2200\sqrt{329}

Then we have

2200 329 = 1 2 ( 360 ) ( 50 ) + 1 2 ( 250 ) ( 70 ) + 1 2 ( 330 ) ( x ) 2200\sqrt{329}=\dfrac{1}{2}(360)(50)+\dfrac{1}{2}(250)(70)+\dfrac{1}{2}(330)(x)

2200 329 = 17750 + 165 x 2200\sqrt{329}=17750+165x

165 x = 2200 329 − 17750 165x=2200\sqrt{329}-17750

x = 2200 329 − 17750 165 = x=\dfrac{2200\sqrt{329}-17750}{165}= 40 329 3 − 3550 33 \boxed{\dfrac{40\sqrt{329}}{3}-\dfrac{3550}{33}}

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