Question

1. find the area and perimeter of the composite figure. leave answers in exact form or to the hundredth.

(image of composite figure with dimensions: 5 in, 6 in, 7 in, 4 in, 10 in, 15 in)
area: ______ perimeter: ______

Explanation:

Step1: Decompose the composite figure

The composite figure can be divided into a trapezoid (or a triangle and a rectangle) and a rectangle. Let's split it into a trapezoid (with bases \(7 + 5=12\) in and \(15\) in, height \(4\) in) and a rectangle (length \(5\) in, height \(10 - 4 = 6\) in? Wait, no, better to split into a trapezoid (left part) and a rectangle (right part). Wait, actually, the left part is a trapezoid with bases \(7 + 5=12\) in? Wait, no, let's re - examine. The bottom base is \(15\) in, the top horizontal segment of the trapezoid part: \(7\) in, and the rectangle on top has length \(5\) in, height \(6\) in (since the total height is \(10\) in, and the trapezoid height is \(4\) in, so \(10 - 4=6\) in). The trapezoid has bases \(15\) in and \(7 + 5 = 12\) in? Wait, no, the left - most part is a triangle? Wait, the slanted side: the horizontal difference between \(15\) in and \(7 + 5=12\) in is \(15-(7 + 5)=3\) in. Wait, maybe it's better to calculate the area as the sum of the area of the trapezoid (with bases \(7 + 5 = 12\) in and \(15\) in, height \(4\) in) and the area of the rectangle (length \(5\) in, height \(6\) in).

The formula for the area of a trapezoid is \(A_{trapezoid}=\frac{(a + b)h}{2}\), where \(a\) and \(b\) are the bases, and \(h\) is the height. The formula for the area of a rectangle is \(A_{rectangle}=l\times w\).

Step2: Calculate the area of the trapezoid

For the trapezoid: \(a = 7+5 = 12\) in, \(b = 15\) in, \(h = 4\) in.
\(A_{trapezoid}=\frac{(12 + 15)\times4}{2}=\frac{27\times4}{2}=27\times2 = 54\) square inches.

Step3: Calculate the area of the rectangle

For the rectangle: length \(l = 5\) in, width \(w = 6\) in (since \(10-4 = 6\) in).
\(A_{rectangle}=5\times6 = 30\) square inches.

Step4: Calculate the total area

Total area \(A=A_{trapezoid}+A_{rectangle}=54 + 30=84\) square inches? Wait, no, wait. Wait, the height of the rectangle: the vertical side of the rectangle is \(6\) in (because the total height is \(10\) in, and the trapezoid height is \(4\) in, so \(10 - 4 = 6\) in). But wait, the rectangle's length is \(5\) in, and its height is \(6\) in. The trapezoid: the two parallel sides (bases) are \(15\) in (bottom) and \(7 + 5=12\) in (top of the trapezoid part), and height \(4\) in. But wait, another way: the figure can be seen as a rectangle (length \(15\) in, height \(4\) in) plus a rectangle (length \(5\) in, height \(6\) in) plus a triangle? Wait, no, the slanted side: the horizontal difference between \(15\) and \(7 + 5\) is \(15-(7 + 5)=3\) in, and the height of the triangle (if it's a triangle) would be \(4\) in. Wait, maybe my initial decomposition is wrong. Let's try again.

Alternative decomposition: The figure is composed of a trapezoid (with bases \(15\) in and \(7\) in, height \(4\) in), a rectangle (length \(5\) in, height \(4\) in) and a rectangle (length \(5\) in, height \(6\) in). Wait, no. Let's use the correct method. The total area can be calculated as the area of the large rectangle - the area of the missing part? No, the figure is built by adding.

Wait, let's calculate the area of the trapezoid part: the trapezoid has bases \(7+5 = 12\) in and \(15\) in, height \(4\) in. Area of trapezoid: \(\frac{(12 + 15)\times4}{2}=54\) in². The rectangle on top: length \(5\) in, height \(6\) in (since \(10 - 4=6\) in). Area of rectangle: \(5\times6 = 30\) in². Total area: \(54+30 = 84\) in²? Wait, no, wait, the height of the rectangle: the vertical side of the rectangle is \(6\) in, and the length is \(5\) in. But also, the left - most part: the horizontal segment from \(7 + 5=12\) in to \(15\) in is \(3\) in, and the height is \(4\) in, so that's a triangle with base \(3\) in and height \(4\) in. Area of triangle: \(\frac{1}{2}\times3\times4 = 6\) in². Then the trapezoid part (without the triangle) would have bases \(7\) in and \(12\) in? Wait, I'm getting confused. Let's use the correct approach:

The composite figure can be divided into three parts: a triangle, a rectangle (middle), and a rectangle (top).

1. Triangle: base \(=15-(7 + 5)=3\) in, height \(=4\) in. Area \(=\frac{1}{2}\times3\times4 = 6\) in².
2. Middle rectangle: length \(=7 + 5=12\) in, height \(=4\) in. Area \(=12\times4 = 48\) in².
3. Top rectangle: length \(=5\) in, height \(=6\) in (since \(10 - 4 = 6\) in). Area \(=5\times6 = 30\) in².

Total area \(=6 + 48+30=84\) in².

Now for the perimeter:

We need to sum up all the outer sides.

- The bottom side: \(15\) in.
- The right - hand vertical side: \(10\) in.
- The top - right horizontal side: \(5\) in.
- The top - left vertical side: \(6\) in.
- The top - left horizontal side: \(7\) in.
- The slanted side: we can calculate its length using the Pythagorean theorem. The horizontal component is \(15-(7 + 5)=3\) in, the vertical component is \(4\) in. So the length of the slanted side is \(\sqrt{3^{2}+4^{2}}=\sqrt{9 + 16}=\sqrt{25}=5\) in.
- The left - hand vertical side: \(4\) in? Wait, no. Wait, let's list all the sides:

Starting from the bottom right corner:

1. Bottom side: \(15\) in (right to left).
2. Left - slanted side: length \(5\) in (as calculated above, since \(3 - 4-5\) triangle).
3. Left - vertical side: \(4\) in (upwards).
4. Top - left horizontal side: \(7\) in (rightwards).
5. Top - left vertical side: \(6\) in (upwards).
6. Top - right horizontal side: \(5\) in (rightwards).
7. Right - vertical side: \(10\) in (downwards).

Now sum them up: \(15+5 + 4+7+6+5+10\)

\(15+5=20\); \(20 + 4=24\); \(24+7 = 31\); \(31+6=37\); \(37+5 = 42\); \(42+10=52\) in. Wait, but let's check again.

Wait, the vertical side between the top rectangle and the middle part: the height of the top rectangle is \(6\) in, and the middle part (trapezoid) has height \(4\) in, total height \(10\) in. The horizontal sides:

- Bottom: \(15\) in.
- Top of the trapezoid part: \(7\) in (left) and \(5\) in (right), so total top of trapezoid: \(7 + 5=12\) in.
- Top of the rectangle: \(5\) in.

The vertical sides:

- Right side: \(10\) in.
- Left - most vertical side: \(4\) in (height of trapezoid).
- The vertical side between the trapezoid and the top rectangle: \(6\) in (height of top rectangle).

The slanted side: the difference in the horizontal lengths: \(15-(7 + 5)=3\) in, vertical length \(4\) in, so slanted side length \(5\) in (3 - 4-5 triangle).

Now perimeter:

Bottom: \(15\) in.

Slanted side: \(5\) in.

Left vertical: \(4\) in.

Top - left horizontal: \(7\) in.

Top - middle vertical: \(6\) in.

Top - right horizontal: \(5\) in.

Right vertical: \(10\) in.

Now sum: \(15 + 5+4 + 7+6+5+10=52\) in.

Wait, but let's recalculate the area correctly. The area of the trapezoid (with bases \(15\) in and \(12\) in ( \(7 + 5\) ), height \(4\) in) is \(\frac{(15 + 12)\times4}{2}=\frac{27\times4}{2}=54\) in². The area of the rectangle (length \(5\) in, height \(6\) in) is \(5\times6 = 30\) in². Total area \(=54 + 30=84\) in².

Wait, but let's check with another method. The figure can be considered as a rectangle of length \(15\) in and height \(10\) in minus the area of the missing rectangle? No, the missing part would be a rectangle of length \(15-(7 + 5)=3\) in and height \(6\) in? No, that doesn't make sense.

Wait, maybe my area calculation is wrong. Let's use the triangle - rectangle - rectangle method. The triangle has base \(3\) in, height \(4\) in, area \(6\) in². The middle rectangle (length \(12\) in, height \(4\) in) has area \(12\times4 = 48\) in². The top rectangle (length \(5\) in, height \(6\) in) has area \(30\) in². Total area \(=6 + 48+30 = 84\) in². That matches the previous result.

Answer:

Area: \(\boldsymbol{84}\) square inches, Perimeter: \(\boldsymbol{52}\) inches