Make a triangle ABC with angle A at 45 degrees Make the height AD of this triangle where D falls on the BC side where BD=6 and CD=4 Find AD

To find AD, we can use the formula for the area of a triangle:

Area = (base * height) / 2

In this case, the base of the triangle is BC, and the height is AD. We are given that BD = 6 and CD = 4. Therefore, BC = BD + CD = 6 + 4 = 10.

Now, we can substitute the values into the formula and solve for AD:

Area = (BC * AD) / 2 Area = (10 * AD) / 2 Area = 5 * AD

Since we are given that angle A is 45 degrees, we can also find the area of the triangle using the formula for the area of a triangle:

Area = (1/2) * AB * AC * sin(A)

Since angle A is 45 degrees, we know that sin(A) is 1/sqrt(2).

Therefore, (1/2) * AB * AC * (1/sqrt(2)) = 5 * AD

Simplifying, AB * AC = 10 * AD * sqrt(2)

We can solve for AD by finding the value of AB * AC. We can use the Pythagorean theorem to find the length of AB and AC.

In triangle ABC, AB^2 + AC^2 = BC^2

Let's assume AB = x, then AC = 10 - x

(x)^2 + (10 - x)^2 = 10^2

Simplifying this equation:

x^2 + (100 - 20x + x^2) = 100

2x^2 - 20x = 0

2x(x - 10) = 0

x = 0 or x = 10

Since x cannot be 0 (as it represents a side length), we have x = 10.

Therefore, AB = 10 and AC = 10 - x = 0.

Now we can substitute the values into the equation AB * AC = 10 * AD * sqrt(2):

10 * 0 = 10 * AD * sqrt(2)

0 = 10 * AD * sqrt(2)

0 = AD * sqrt(2)

AD = 0

Therefore, the height AD of the triangle is 0