In the figure bellow the trapezium ABCD is isosceles. AB is parallel to CD and the diagonals AC and BD are crossing in the point P. The triangle ABP has an area of 4 cm2 and the triangle DCP has an area of 9 cm2. What is the area of the triangle BCP?
The two triangles, triangle(APB) and triangle(CPD) are similar by AA.
The corresponding lengths of sides of similar figures are in the ratio of the square roots of the areas.
So: AP / PC = sqrt(4) / sqrt(9) ---> AP / PC = 2 / 3.
Let's call the length of AP = 2x, which makes the length PC = 3x; also, BP = 2x and PD = 3x.
A formula for the area of a triangle is: A = ½·a·b·sin(C).
Using this formula on triangle(APB): a = AP = 2x b = BP = 2x A = 4 cm2
---> A = ½·a·b·sin(angle( APB) ) ---> 4 = ½·(2x)·(2x)·sin(angle( APB )
---> sin( angle(APB) ) = 2 / x2
Similarly, sin( angle(CPD) = 2 / x2
Area of triangle(APD) = ½·(2x)·(3x)·sin(angle( APB ) = (3x2)·sin(angle( APB )
(since sin( angle(CPD) = 2 / x2 ) ---> Area of triangle(APD) = (3x2)·(2 / x2) = 6 cm2
Therefore, the total area will be 4 cm2 + 9 cm2 + 6 cm2 + 6 cm2 = 25 cm2
Let M and N be the midpoints of AB and CD.
The ratio in which each diagonal is divided is equal to the ratio of the lengths of the parallel sides.
AB/CD = AP/PC = BP/PD = 2/3 The ratio of MP to NP is 2/3 also
If CD = 8 then AB = 8 * (2/3) = 5.33333333333
NP = 9 / 4 = 2.25
MP = 2.25 * (2/3) = 1.5
The height of the trapezium is h = MP + NP = 3.75
Area of the trapezium is A = h/2 ( AB + CD ) = 25 cm2
The area of the triangle BCP Abcp = ( 25 - 13 ) / 2