Determine each segment length in right triangle
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Triangle ABC with right angle marked at vertex B. Side AC, opposite vertex B, is labeled 14. Dashed segment is drawn from vertex B to point D on side AC. Angle BDA is marked right angle. Angles A and C both marked 45 degrees. Segment AD is labeled 7.
In a right triangle, the side opposite the right angle is called the hypotenuse, and the other two sides are called the legs.
Let's label the sides of the triangle:
Leg AB = x
Leg BC = y
Hypotenuse AC = 14
We are given that angle A and C are both 45 degrees and that segment AD is labeled 7.
Since angle A is 45 degrees, triangle ABD is also a right triangle with a right angle at vertex D. Therefore, segment BD is also a leg of a right triangle and segment AD is its hypotenuse.
Using the Pythagorean theorem for triangle ABD, we can write:
x^2 + y^2 = 7^2
Since angles A and C are both 45 degrees, triangle BCD is also a right triangle. Therefore, we can use the Pythagorean theorem for triangle BCD:
y^2 + (14 - x)^2 = 14^2
Now we have a system of two equations:
1) x^2 + y^2 = 7^2
2) y^2 + (14 - x)^2 = 14^2
We can solve this system of equations to find the values of x and y.
From equation 1), we can rearrange it to get:
x^2 = 7^2 - y^2
Substituting this into equation 2), we get:
y^2 + (14 - (7^2 - y^2))^2 = 14^2
Simplifying this equation, we get:
y^2 + (14 - 49 + y^2)^2 = 196
y^2 + (y^2 - 35)^2 = 196
y^2 + y^4 - 70y^2 + 1225 = 196
y^4 - 69y^2 + 1029 = 0
This is a quadratic equation in terms of y^2. We can solve it by factoring:
(y^2 - 21)(y^2 - 48) = 0
This gives us two possible values for y^2:
1) y^2 - 21 = 0, which gives y^2 = 21
2) y^2 - 48 = 0, which gives y^2 = 48
Taking the square root of both sides, we get two possible values for y:
1) y = √21
2) y = √48
Let's substitute these values back into equation 1) to find the corresponding values of x.
1) For y = √21:
x^2 + (√21)^2 = 7^2
x^2 + 21 = 49
x^2 = 28
x ≈ ±√28
2) For y = √48:
x^2 + (√48)^2 = 7^2
x^2 + 48 = 49
x^2 = 1
x = ±1
Since all sides of the triangle are positive lengths, we discard the negative values of x. Therefore, we have:
For y = √21: x = √28
For y = √48: x = 1
So the lengths of the segments are:
Segment AB = x = √28
Segment BC = y = √21
Segment BD = 14 - x = 14 - √28