Question

done

Answer 1

Ok. Let's start.

Answer 2

#17

Answer 3

How can we find "x"?

Answer 4

Use Pyhagora's Theorem: \( 5^2=3^2+x^2\). Now what is \(x\)?

Answer 5

\( x=\sqrt{5^2-3^2}\). Now what's \(x\)?

Answer 6

\( x=\sqrt{25-9}=\sqrt{16}\)

Answer 7

Sure.

We will first determine \(x\) then determine the area of the triangle, then the area of the rectangle. We'll sum the two areas to get the total area.

Here is how we determined \(x\):

Answer 8

$$
x^2 +3^25^2\\
x^2 =5^2-3^2\\
x=\sqrt{5^2-3^2}\\
x=\sqrt{16}\\
x=4\\
$$
Now the area of the triangle:

Answer 9

Area of triangle is \(\dfrac 1 2bh\), where \(b\) is base and \(h\) is height.
$$\Large
A_{triangle}=\dfrac 1 2 \times x \times3\\
\Large A_{triangle}=\dfrac 1 2 4\times3\\
\Large A_{triangle}=2\times3=6\\
$$

This is the area of the triangle.

Next, the area of the rectangle:

Answer 10

Area of a rectangle is \(l\times w\), where \(l\) is length and \(w\) is width.
$$
\Large A_{rectangle}=l\times w\\
\Large A_{rectangle}=l\times x\\
\Large A_{rectangle}=6\times 4\\
\Large A_{rectangle}=24\\
$$

Now we have the areas of the rectangle and the triangle. We now just need to sum them:

$$
\Large Total~Area=A_{triangle}+A_{rectangle}
$$

Answer 11

@MayMay_69

Answer 12

@MayMay_69