Question

1. perimeter =

area =
student manual
math ready. unit 3. lesson 5

1. perimeter =

area =

Explanation:

Step1: Calculate the perimeter of the first figure

The figure can be thought of as a rectangle - like shape. The lengths of the sides are added. The horizontal lengths are \(5 + 6=11\)m and the vertical lengths are \(10\)m and \(4\)m. The perimeter \(P\) is the sum of all outer - side lengths.
\[P=(5 + 6)+10+(4 + 10)+5=36\space m\]

Step2: Calculate the area of the first figure

The figure can be divided into two rectangles. One rectangle has dimensions \(6\times10\) and the other has dimensions \(5\times4\). The area \(A\) is the sum of the areas of these two rectangles.
\[A=(6\times10)+(5\times4)=60 + 20=80\space m^{2}\]

Step3: Calculate the perimeter of the second figure

The perimeter of the second figure is the sum of all outer - side lengths. We know the lengths of the sides: \(4\) in, \(14\) in, \(4\) in, and we need to find the lengths of the non - horizontal and non - vertical sides. Using the Pythagorean theorem for the slanted sides. The height of the slanted part is \(7 - 4=3\) in and the base of the slanted part is \(\frac{14 - 8}{2}=3\) in. The length of each slanted side \(l=\sqrt{3^{2}+3^{2}}=\sqrt{9 + 9}=\sqrt{18}=3\sqrt{2}\) in.
\[P = 4+14+4 + 2\times3\sqrt{2}=22 + 6\sqrt{2}\approx22+6\times1.414=22 + 8.484 = 30.484\approx30.48\space in\]

Step4: Calculate the area of the second figure

The figure can be divided into a rectangle and a trapezoid. The rectangle has dimensions \(4\times14\) and the trapezoid has bases \(8\) in and \(14\) in and height \(7 - 4 = 3\) in.
The area of the rectangle \(A_{1}=4\times14 = 56\) in², and the area of the trapezoid \(A_{2}=\frac{(8 + 14)\times3}{2}=\frac{22\times3}{2}=33\) in².
\[A=A_{1}+A_{2}=56+33 = 89\space in^{2}\]

Answer:

For the first figure:
Perimeter \( = 36\space m\)
Area \( = 80\space m^{2}\)
For the second figure:
Perimeter \(\approx30.48\space in\)
Area \( = 89\space in^{2}\)