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

Guest Oct 5, 2019

#1**+2 **

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.**

Guest Oct 5, 2019

#2**+2 **

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

hectictar Oct 5, 2019