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Find the dimensions of a right triangle if it’s area is 40m^2 where the height is 2 meters less than the base.

base = _ meters
height= _ meters

User Obmerk Kronen
by
3.3k points

1 Answer

23 votes
23 votes

Answer:

The dimensions of the right triangle are 9 meters by 7 meters.
or
The base of the triangle is 9 meters, and the height is 7 meters (since the height is 2 meters less than the base).

Explanation:

To find the dimensions of the right triangle, we can use the formula for the area of a triangle, which is:

A = 1/2 * b * h

Where b is the base and h is the height of the triangle. In this case, we know that the area of the triangle is 40 square meters, and the height is 2 meters less than the base, so we can write the equation as:

40 = 1/2 * b * (b - 2)

To solve for b, we can rearrange the equation to get b by itself:

40 = 1/2 * b^2 - b

Then, we can move all the terms involving b to the left-hand side of the equation and all the constants to the right-hand side:

1/2 * b^2 - b - 40 = 0

Next, we can use the quadratic formula to solve for b:

b = (-(-1) +/- sqrt((-1)^2 - 4 * (1/2) * -40)) / (2 * (1/2))

Which simplifies to:

b = (1 +/- sqrt(1 + 80)) / 1

Since b must be a positive number, we take the positive solution:

b = (1 + sqrt(81)) / 1

Therefore, the base of the triangle is 9 meters, and the height is 7 meters (since the height is 2 meters less than the base). Thus, the dimensions of the right triangle are 9 meters by 7 meters.

User Manoj Talreja
by
3.0k points
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