Please solve q 99?

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Please solve q 99?

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Answer 1 · 2018-06-17T19:49:06

$D P + E P + P F = 5 \sqrt{3}$ $DP+EP+PF=5sqrt3$ regardless where P is in the triangle.
This is called Viviani's theorem: https://en.wikipedia.org/wiki/Viviani%27s_theorem

Explanation:

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ABC is the triangle. It's height $h = 5 \sqrt{3}$ $h=5sqrt3$ (use Pythagoras on for instance ADC)

Remember that the area of a triangle is $\frac{b a s e \cdot h e i g h t}{2}$ $(base*height)/2$ .

Now all three triangles APC, BPC and APB have a base = 10 since all sides are equal and 10. Their total area is equal to the area of ABC

therefore $\frac{s i d e \cdot C M}{2} = \frac{s i d e \cdot h}{2}$ $(side*CM)/2=(side*h)/2$
= $\frac{s i d e \cdot D P}{2} + \frac{s i d e \cdot E P}{2} \cdot \frac{s i d e \cdot F P}{2}$ $(side*DP)/2+(side*EP)/2*(side*FP)/2$

Therefore $D P + E P + P F = M C = 5 \sqrt{3}$ $DP+EP+PF=MC=5sqrt3$ regardless where P is in the triangle.

Answer 2 · 2018-06-17T20:23:00

Another one from Rahul's Book!

I'd say that given we're given choices, the result doesn't depend on our choice of $P$ $P$ .

Let's pick $P = G,$ $P=G,$ the centroid/orthocenter. We get an altitude $5^{2} + h^{2} = 10^{2}$ $5^2+h^2=10^2$ or $h = 5 \sqrt{3}$ $h=5 sqrt{3}$ . The medians meet $\frac{2}{3}$ $2/3$ from each vertex, and the remaining piece of $\frac{1}{3} h$ $1/3h$ is perpendicular to the sides. We have three of these to add up, they add to $h$ $h$ , choice (2).

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Please solve q 99?

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