Question

find the area of the figure. (sides meet at right angles.)
3 cm
5 cm
7 cm
4 cm
7 cm
cm²

Explanation:

Step1: Divide the figure into two rectangles

We can split the L - shaped figure into a vertical rectangle and a horizontal rectangle. The vertical rectangle has dimensions \(3\space cm\times5\space cm\), and the horizontal rectangle has dimensions \(7\space cm\times(7 - 5)\space cm=7\space cm\times2\space cm\) (or we can also split it as a \(3\space cm\times7\space cm\) rectangle and a \(4\space cm\times(7 - 3)\space cm = 4\space cm\times4\space cm\) rectangle, but we will use the first way here).

Step2: Calculate the area of the vertical rectangle

The area of a rectangle is given by the formula \(A = l\times w\), where \(l\) is the length and \(w\) is the width. For the vertical rectangle with \(l = 5\space cm\) and \(w=3\space cm\), the area \(A_1=3\times5 = 15\space cm^{2}\).

Step3: Calculate the area of the horizontal rectangle

The height of the horizontal rectangle is \(7 - 5=2\space cm\) and the length is \(7\space cm\). Using the area formula for a rectangle \(A = l\times w\), the area \(A_2=7\times2 = 14\space cm^{2}\).

Step4: Sum the areas of the two rectangles

To find the total area of the figure, we add the areas of the two rectangles. So the total area \(A=A_1 + A_2=15 + 14=29\space cm^{2}\). (Alternative way: Another way to split is to consider the big rectangle of \(7\times7\) and subtract the missing rectangle. The missing rectangle has dimensions \(4\times(7 - 3)=4\times4 = 16\space cm^{2}\). The area of the big rectangle is \(7\times7 = 49\space cm^{2}\), then the area of the figure is \(49-16 = 33\space cm^{2}\)? Wait, no, I made a mistake in the first splitting. Let's re - do it.

Wait, the correct way: Let's look at the figure again. The vertical part: the height is \(5\space cm\), width \(3\space cm\). The horizontal part: the length is \(7\space cm\), and the height is \(7 - 5 = 2\space cm\), but the width of the horizontal part should be \(7\space cm\)? No, no. Wait, the bottom rectangle: length is \(7\space cm\), and the height is \(7 - 5=2\space cm\), and the left part of the bottom rectangle? Wait, no, the correct splitting is: the figure can be divided into a rectangle of \(3\space cm\times7\space cm\) and a rectangle of \(4\space cm\times(7 - 3)\space cm\). Let's calculate that.

First rectangle: \(3\times7 = 21\space cm^{2}\). Second rectangle: \(4\times(7 - 3)=4\times4 = 16\space cm^{2}\). Then total area \(21 + 16=37\space cm^{2}\)? No, I'm confused. Wait, let's use the coordinates. Let's assume the bottom - left corner is at \((0,0)\). The top - right corner of the vertical rectangle is at \((3,7)\), and the bottom - right corner of the horizontal rectangle is at \((7,2)\). Wait, the figure has a vertical segment of \(5\space cm\) (from \(y = 2\) to \(y = 7\)) with width \(3\space cm\) (from \(x = 4\) to \(x = 7\))? No, the given lengths: the bottom side is \(7\space cm\), the left - bottom horizontal segment is \(4\space cm\), the vertical segment on the left - middle is \(5\space cm\), the top - horizontal segment is \(3\space cm\), and the right - vertical segment is \(7\space cm\).

So the correct way is to split the figure into two rectangles:

Rectangle 1: Width \(3\space cm\), height \(7\space cm\) (vertical rectangle on the right). Area \(A_1=3\times7 = 21\space cm^{2}\).

Rectangle 2: Width \(4\space cm\), height \((7 - 5)\space cm=2\space cm\) (horizontal rectangle on the left - bottom). Area \(A_2=4\times2 = 8\space cm^{2}\). Wait, no, \(7 - 5 = 2\)? No, the vertical length of the left - bottom rectangle: the total height is \(7\space cm\), the vertical part on the right is \(5\space cm\), so the height of the left - bottom rectangle is \(7 - 5=2\space cm\), and the width is \(4\space cm\). Then total area \(21+8 = 29\space cm^{2}\)? No, that's not right. Wait, let's use the formula for the area of a composite figure with right angles. The figure can be thought of as a rectangle of length \(7\) and width \(7\) minus a rectangle of length \(4\) and width \(4\) (since \(7 - 3 = 4\) and \(7 - 5 = 2\)? No, \(7-3 = 4\) and \(7 - 5 = 2\), no. Wait, the missing part is a rectangle with length \(4\) (horizontal) and height \(4\) (vertical)? No, let's calculate the correct dimensions.

Looking at the figure: the horizontal segments: the bottom is \(7\space cm\), the top horizontal is \(3\space cm\), so the horizontal extension on the left is \(7 - 3=4\space cm\). The vertical segments: the right vertical is \(7\space cm\), the middle vertical is \(5\space cm\), so the vertical extension at the bottom is \(7 - 5 = 2\space cm\).

So the figure can be divided into two rectangles:

1. Top - right rectangle: length \(3\space cm\), height \(5\space cm\). Area \(=3\times5 = 15\space cm^{2}\).

2. Bottom rectangle: length \(7\space cm\), height \(2\space cm\). Area \(=7\times2=14\space cm^{2}\).

Total area \(=15 + 14 = 29\space cm^{2}\). Wait, but if we check with another method: the area of the big rectangle (if we consider the outer rectangle) would be \(7\times7 = 49\space cm^{2}\). The missing rectangle has length \(4\space cm\) (horizontal) and height \(4\space cm\) (vertical, since \(7 - 3 = 4\) and \(7 - 5 = 2\)? No, \(7 - 5 = 2\), so the missing rectangle is \(4\times(7 - 5)=4\times2 = 8\space cm^{2}\)? No, I'm really confused. Wait, let's look at the lengths again.

The vertical side on the right is \(7\space cm\), the vertical side in the middle is \(5\space cm\), so the difference is \(7 - 5 = 2\space cm\) (the height of the bottom rectangle). The horizontal side at the bottom is \(7\space cm\), the horizontal side at the top is \(3\space cm\), so the difference is \(7 - 3 = 4\space cm\) (the width of the left - bottom rectangle).

So the two rectangles are:

- Rectangle 1: \(3\space cm\times5\space cm\) (area \(15\))

- Rectangle 2: \(4\space cm\times2\space cm\) (area \(8\))

Wait, no, that can't be. Wait, the bottom rectangle should have length \(7\space cm\) and height \(2\space cm\), because the bottom side is \(7\space cm\) and the height from the bottom to the middle vertical line is \(2\space cm\). And the right rectangle has length \(3\space cm\) and height \(5\space cm\). Then \(7\times2=14\) and \(3\times5 = 15\), sum is \(29\). Alternatively, the area can be calculated as the area of the rectangle with length \(7\) and height \(5\) plus the area of the rectangle with length \(4\) and height \(2\). \(7\times5=35\), \(4\times2 = 8\), \(35 + 8=43\)? No, I'm making a mess.

Wait, let's use the correct approach for composite figures with right angles:

We can find the area by adding the areas of two rectangles:

First rectangle: width \(3\), height \(7\) (vertical rectangle on the right). Area \(=3\times7 = 21\).

Second rectangle: width \(4\), height \((7 - 5)=2\) (horizontal rectangle on the left - bottom). Area \(=4\times2 = 8\).

Total area \(=21 + 8=29\). Yes, that's correct. Because the vertical rectangle on the right has height \(7\) (from bottom to top) and width \(3\). The left - bottom rectangle has width \(4\) (since \(7 - 3 = 4\)) and height \(2\) (since \(7 - 5 = 2\)). So \(3\times7+4\times2=21 + 8 = 29\space cm^{2}\).

Answer:

\(29\space cm^{2}\)