Question

lt2.3 i can find the area of compound figures.

1. show how you can divide the compound figures into polygons that you know how to find the area of. show the use of area formulas to find the area of the shaded portion of the figure

Explanation:

Step1: Find the area of the triangle

The triangle has sides 16 yd, 19 yd, and we can assume it's a triangle (maybe isoceles or scalene, but to find area, we can use the formula for the area of a triangle with inradius. Wait, alternatively, maybe it's a triangle with base 16 yd and let's check the perimeter? Wait, the inradius is 5 yd. The formula for the area of a triangle with inradius \( r \) is \( A = r \times s \), where \( s \) is the semi - perimeter. First, find the semi - perimeter. Let's assume the triangle has sides \( a = 16 \), \( b = 19 \), and let's find the third side? Wait, maybe it's an isoceles triangle? Wait, no, the inradius is 5. Wait, maybe the triangle is a triangle with base 16 and the other two sides 19? Wait, no, let's re - examine. Wait, the formula for the area of a triangle with inradius \( r \) is \( A=\frac{1}{2}\times perimeter\times r \). Wait, perimeter \( P=16 + 19+19 = 54 \) (assuming two sides are 19). Then semi - perimeter \( s=\frac{54}{2}=27 \). Then area of triangle \( A_{triangle}=r\times s=5\times27 = 135 \) square yards. Alternatively, if we use the formula \( A=\frac{1}{2}\times base\times height \), but we can also use the inradius formula.

Step2: Find the area of the circle

The radius of the circle \( r = 5 \) yd. The formula for the area of a circle is \( A_{circle}=\pi r^{2}\). So \( A_{circle}=\pi\times5^{2}=25\pi\approx25\times3.14 = 78.5 \) square yards.

Step3: Find the area of the shaded region

The shaded region is the area of the triangle minus the area of the circle. So \( A_{shaded}=A_{triangle}-A_{circle}\). From step 1, if \( A_{triangle}=135 \) (using the inradius formula: \( A = r\times s \), \( s=\frac{16 + 19+19}{2}=27 \), \( A = 5\times27 = 135 \)), and \( A_{circle}=25\pi\approx78.5 \). Then \( A_{shaded}=135 - 78.5=56.5 \) square yards (if we use \( \pi\approx3.14 \)). Wait, let's check the triangle area again. Alternatively, if the triangle has base 16 and height \( h \), and the inradius is 5. The formula \( A=\frac{1}{2}\times base\times height=r\times s \). Let's find the semi - perimeter \( s=\frac{16 + 19+19}{2}=27 \), so \( A = 5\times27 = 135 \), and \( \frac{1}{2}\times16\times h=135 \), then \( 8h = 135 \), \( h=\frac{135}{8}=16.875 \), which is a valid height. Then the area of the circle is \( \pi r^{2}=\pi\times5^{2}=25\pi\approx78.5 \). Then the shaded area is \( 135 - 25\pi\approx135 - 78.5 = 56.5 \) square yards.

Answer:

The area of the shaded region is \( 135 - 25\pi\approx56.5 \) square yards (or \( 135 - 78.5 = 56.5 \) square yards if \( \pi\approx3.14 \)).