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
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.
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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
