Question

there is a point P inside an equliateral triangle ABC of side d whose distance from the vertices is 3,4,5.Rotate he triangle and P through 60 degree about C. Let A go to A' and P to P'. (i)The area of triangle PAP'

Answer 1

Answer:

Let △ ABC be the equilateral triangle, and the point P in it is such that the perpendicular distances from the sides BC, AB and AC are PD=3 cm, PE=4 cm and PF=5 cm respectively.

Let the length of each side of the equilateral triangle be L cm.

Now we have,

ΔABC=ΔPBC+ΔPAB+ΔPAC

=12PD.AC+12PE.AB+12PF.AC

=L2(PD+PE+PF)

=L2(3+4+5)=6L cm2.

Now, since ΔABC is an equilateral triangle with side L , its area is given by the formula 3√4L2 .

So 3√4L2=6L .

⟹L=83–√

Hence the area of the given equilateral triangle is 48 3–√ cm2 .

Hope it helps

Peace!!

Answer 2

Answer:

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