What is the approximate value of x in this figure?
19 POINTS
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Answer:
4.69 inches (2 decimal places)
Step-by-step explanation:
Refer to image attached to this answer:
Using Pythagoras Theorem:
[tex]AB^2+AC^2=BC^2[/tex]
[tex]2^2+3^2=BC^2[/tex]
[tex]4+9=BC^2[/tex]
[tex]BC^2=13[/tex]
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Using Pythagoras Theorem,
[tex]BC^2+CD^2=BD^2\ (x)[/tex]
[tex]13+3^2=BD^2[/tex]
[tex]13+9=BD^2[/tex]
[tex]BD=\sqrt{22} = 4.69\ (2d.p.)[/tex]
First, we have to find the last remaining side of the leftmost triangle.
We can use the Pythagorean theorem.
The Pythagorean theorem: a^2 + b^2 = c^2
a = one leg
b = the second leg
c = the hypotenuse
In this problem,
a = 3
b = 2
c = ?
Let's plug our numbers into the Pythagorean theorem.
3^2 + 2^2 = c^2
Simplify the left side
9 + 4 = c^2
Add together like terms
13 = c^2
Take the square root of both sides
[tex]\sqrt{13}[/tex] = c
Now we can use the Pythagorean theorem on the rightmost triangle to find x.
a = 3
b = [tex]\sqrt{13}[/tex]
c = x
let's plug our values into the Pythagorean theorem.
3^2 + ([tex]\sqrt{13}[/tex])^2 = (x)^2
Simplify the left side
9 + 13 = x^2
Add together like terms
22 = x^2
Take the square root of both sides
[tex]\sqrt{22}[/tex] = x
But the problem is asking for the approximate value, the approximate value of [tex]\sqrt{22}[/tex] = 4.69 in