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Consider square ABCD and a point P on its diagonal AC such that AP:PC=1:3, If AP=10, what is the area of triangele BPC

 Oct 5, 2019
 #1
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Since the diagonal AC of the square ABCD splits the square into 2 EQUAL isosceles right triangles, each triangle having:90, 45, 45-degree angles and a hypotenuse(the diagonal) of:10 + 30 =40, then the area  of each of the 2 triangles =1/2 sin(45) x 40 x sqrt(40^2 - S^2=sqrt(800) = 400 - square units - the area of one of the two triangles.


The area is proportional to the ratio of the diagonal(1:3), then the area of triangle BPC =400 / 4 x 3 =300 square units.

 Oct 5, 2019
 #2
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See:  https://www.desmos.com/calculator/sln5zuurfx

 

We know that

 

AP / PC  =  1 / 3

                              We are also given that  AP = 10   so we can substitute  10  in for  AP

10 / PC  =  1 / 3

                              If it isn't already clear that  PC  must be  30 , we can invert both sides

PC / 10  =  3 / 1

                              and then multiply both sides of the equation by  10

PC  =  3 / 1 * 10

 

PC  =  30

 

And now we can find the length of  AC

 

AC   =   AP + PC   =   10 + 30   =   40

 

The diagonals of a square are the same length and bisect each other so...

 

h   =   AC / 2   =   40 / 2   =   20       where  h  is the height of triangle BPC

 

Area of triangle BPC   =   ( 1/2 )( PC )( h )   =   ( 1/2 )( 30 )( 20 )   sq units   =   300   sq units

 Oct 5, 2019

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