"One leg of a right triangle is 50.A line parallel to the other leg and at a distance of 20 from that leg cuts off a right trapezoid of area 320. What is the area of the original right triangle?"
To find the area of the original right triangle, we need to first find the length of the other leg.
Let's assume the length of the other leg is 'x'. We know that one leg is 50, so the legs of the right triangle are 50 and x.
The line parallel to the second leg and at a distance of 20 from it cuts off a right trapezoid of area 320.
Since the line is parallel to the second leg, the trapezoid's height is 20. Let's call the lengths of the parallel sides 'a' and 'b'.
The formula for calculating the area of a trapezoid is:
Area = (1/2)(a + b)(h)
where 'h' is the height of the trapezoid.
We know the area of the trapezoid is 320, so we can write the equation:
320 = (1/2)(a + b)(20)
Now, let's simplify the equation. Since we have a right trapezoid, we know that the two parallel sides are equal in length. So, we can replace 'a' with 'b'.
320 = (1/2)(x + x)(20)
320 = (1/2)(2x)(20)
320 = 20x
Now, let's solve for 'x':
20x = 320
x = 320/20
x = 16
Therefore, the length of the other leg of the right triangle is 16.
To find the area of the right triangle, we can use the formula:
Area = (1/2)(base)(height)
In this case, the base is 50, and the height is 16.
Area = (1/2)(50)(16)
Area = 400
Therefore, the area of the original right triangle is 400.