Question

What is the approximate value of x in this figure? 19 POINTS

Answer 1

Answer:

4.69 inches (2 decimal places)

Step-by-step explanation:

Refer to image attached to this answer:

Using Pythagoras Theorem:

$AB^2+AC^2=BC^2$

$2^2+3^2=BC^2$

$4+9=BC^2$

$BC^2=13$

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Using Pythagoras Theorem,

$BC^2+CD^2=BD^2\ (x)$

$13+3^2=BD^2$

$13+9=BD^2$

$BD=\sqrt{22} = 4.69\ (2d.p.)$

Answer 2

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

$\sqrt{13}$ = c

Now we can use the Pythagorean theorem on the rightmost triangle to find x.

a = 3

b = $\sqrt{13}$

c = x

let's plug our values into the Pythagorean theorem.

3^2 + ($\sqrt{13}$)^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

$\sqrt{22}$ = x

But the problem is asking for the approximate value, the approximate value of $\sqrt{22}$ = 4.69 in