ABCD is a square The triangle PDC is constructed such that P

ABCD is a square. The triangle PDC is constructed such that P is able to me moved along AB. No matter where P is moved on AB, the area of the triangle PDC is 1/2 of the area of ABCD. I need to prove that no matter where P in on AB the area of triangle PDC will always be 1/2 of ABCD. I would begin by constructing the altitude.

Solution

Draw an altitude from the tip i.e. Point P to the base CD

Area of Square ABCD = (side)^2 = (AB)^2 = (CD)^2

Area of Triangle PDC = 1/2 * base * height

=> 1/2 * CD * BC

(since the height of altitude will also be equal to the side of square)

Since the figure is a squar, hence all sides must be equal,therefore we can write

AB = BC = CD = AD

substituting BC = CD, in the area of triangle formula we get

Area of triangle PDC = 1/2 * CD * CD = 1/2 * (CD)^2 = 1/2 * area of square (ABCD)