A square and isosceles triangle of equal height are side-by-side, as shown, with both bases on the x-axis. The lower right vertex of the square and the lower left vertex of the triangle are at \((10, 0)\). The side of the square and the base of the triangle on the x-axis each equal \(10\) units. A segment is drawn from the top left vertex of the square to the farthest vertex of the triangle, as shown. What is the area of the shaded region?

Logic Mar 2, 2019

#1**+2 **

The slope of the line drawn from the top left vertex of the square to the bottom right vertex of the triangle = -1/2

The equation of this line is y = (-1/2)x + 10 (1)

The point at the top vertex of the triangle is (15, 10)

The slope of the segmnent joining (10,0) and (15,10) = 10/5 = 2

So....the equation of the line joining these two points is y = 2(x - 10) ⇒ y = 2x - 20 (2)

The x coordinate of the intersection of these lines can be found by setting (1) = (2)

(-1/2)x + 10 = 2x - 20

30 = [5/2] x

x = 12 and y = 2(12) - 20 = 4

So....the height of the triangular shaded area = 4 and its base = 10

So...its area = (1/2)(4) (10) = 20 units^2

CPhill Mar 2, 2019