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?

aboslutelydestroying Apr 7, 2024

#1**+1 **

Probably an easier way, but.....

Height of triangle EAD =10

(1/2) the base = 5

Slope of line through AE = 10/5 = 2

So....Let the line through AE have the equation y =2(x - 10) = 2x -20

Slope of line through CD = [ 10 -0 ] / [0 -20 ] = -1/2

So....Let the line through CD have the equation y = (-1/2)x + 10

Find the intersection of these lines = x coordinate of N

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

(5/2)x = 30

x = 12

The y coordinate of A is y = (-1/2)(10) + 10 = 5 = base of triangle formed by shaded region

(12 -10) = 2 = height of triangle formed by the shaded region

Shaded region = (1/2) (5) * 2 = 5

CPhill Apr 7, 2024