Answer:
[tex] \displaystyle \rm A_{ \text{c - pentagon}} =41[/tex]
[tex] \rm \displaystyle P _{ \rm c - pentagon} = 20+ 4 \sqrt{2} [/tex]
Step-by-step explanation:
we have a square and a triangle
we want to figure out the area and the perimeter of the pentagon
to figure out the area of the pentagon we can use the given formula:
[tex] \displaystyle \rm A_{ \text{c - pentagon}} = A_{ \text{squre}} - A_{ \text{triangle}}[/tex]
let's figure out [tex]A_{\rm square}[/tex]:
since the given shape is a square Every angle of its 90° Thus the triangle is a right angle triangle
therefore the height is 4
[tex] \displaystyle A _{ \rm triangle} = \frac{1}{2} \times b \times h[/tex]
substitute h and b
[tex] \displaystyle A _{ \rm triangle} = \frac{1}{2} \times 4 \times 4[/tex]
reduce fraction:
[tex] \displaystyle A _{ \rm triangle} = 2 \times 4[/tex]
simplify multiplication:
[tex] \displaystyle A _{ \rm triangle} = 8[/tex]
likewise square
[tex] \displaystyle A _{ \rm squre} = {s}^{2} [/tex]
substitute s:
[tex] \displaystyle A _{ \rm squre} = {7}^{2} [/tex]
simplify square
[tex] \displaystyle A _{ \rm squre} = 49[/tex]
hence,
[tex] \displaystyle \rm A_{ \text{c - pentagon}} =49- 8[/tex]
[tex] \displaystyle \rm A_{ \text{c - pentagon}} =41[/tex]
to figure out perimeter
let's figure out hypotenuse first
[tex] \displaystyle h = \sqrt{ {4}^{2} + {4}^{2} } [/tex]
[tex] \displaystyle h = \sqrt{ 16 + 16} [/tex]
[tex] \displaystyle h = 4 \sqrt{2} [/tex]
therefore,
[tex] \rm \displaystyle P _{ \rm c - pentagon} = 3 + 3 + 4 \sqrt{2} + 7 + 7[/tex]
[tex] \rm \displaystyle P _{ \rm c - pentagon} = 6 + 4 \sqrt{2} + 14[/tex]
[tex] \rm \displaystyle P _{ \rm c - pentagon} = 20+ 4 \sqrt{2} [/tex]
