independent practice
1. triangle abc
a(0, 9)
b(4, 9)
c(0, 1)
5. square efgh
g(3, -1)
h(1, -3)
f(5, -3)
e(3, -5)
6. triangle abc
a(0, 4)
c(-4, 0)
b(2, 0)
exit tickets:
find the perimeter and area of the figure if each unit on the graph measures 1 centimeter. round answers to the nearest tenth, if necessary.
a(-2, 7)
b(4, 4)
d(-4, 3)
c(2, 0)

Question

Accepted Answer

1. **For triangle $ABC$ in the first - part ($A(0,9)$, $B(4,9)$, $C(0,1)$)**:
   - **Area ($A$)**:
     - # Explanation:
       - ## Step1: Identify base and height
         - The base of the triangle is the horizontal distance between $A(0,9)$ and $B(4,9)$. Using the distance formula for points with the same $y$ - coordinate $d=\vert x_2 - x_1\vert$, the base $b=\vert4 - 0\vert = 4$ cm. The height of the triangle is the vertical distance between $A(0,9)$ and $C(0,1)$. Using the distance formula for points with the same $x$ - coordinate $d=\vert y_2 - y_1\vert$, the height $h=\vert9 - 1\vert=8$ cm.
       - ## Step2: Apply the area formula for a triangle
         - The area formula for a triangle is $A=\frac{1}{2}bh$. Substituting $b = 4$ and $h = 8$, we get $A=\frac{1}{2}\times4\times8=16$ $cm^2$.
   - **Perimeter ($P$)**:
     - ## Step1: Find the lengths of the sides
       - The length of side $AC$: Using the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$ for $A(0,9)$ and $C(0,1)$, $AC=\sqrt{(0 - 0)^2+(9 - 1)^2}=\sqrt{0 + 64}=8$ cm.
       - The length of side $AB$: For $A(0,9)$ and $B(4,9)$, $AB=\sqrt{(4 - 0)^2+(9 - 9)^2}=\sqrt{16+0}=4$ cm.
       - The length of side $BC$: For $B(4,9)$ and $C(0,1)$, $BC=\sqrt{(0 - 4)^2+(1 - 9)^2}=\sqrt{(-4)^2+(-8)^2}=\sqrt{16 + 64}=\sqrt{80}\approx8.9$ cm.
     - ## Step2: Calculate the perimeter
       - $P=AC + AB+BC=8 + 4+\sqrt{80}\approx8 + 4+8.9 = 20.9$ cm.
     - # Answer:
       - $A:16$ $cm^2$
       - $P:20.9$ $cm$
2. **For square $EFGH$ ($E(3,-5)$, $F(5,-3)$, $G(3,-1)$, $H(1,-3)$)**:
   - **Area ($A$)**:
     - # Explanation:
       - ## Step1: Find the side - length
         - First, find the distance between two adjacent vertices, say $E(3,-5)$ and $F(5,-3)$. Using the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$, $d=\sqrt{(5 - 3)^2+(-3+5)^2}=\sqrt{4 + 4}=\sqrt{8}=2\sqrt{2}$ cm.
       - ## Step2: Calculate the area
         - The area of a square is $A = s^2$, where $s$ is the side - length. Substituting $s = 2\sqrt{2}$, we get $A=(2\sqrt{2})^2=8$ $cm^2$.
   - **Perimeter ($P$)**:
     - ## Step1: Use the side - length to find the perimeter
       - The perimeter of a square is $P = 4s$. Since $s = 2\sqrt{2}$, $P=4\times2\sqrt{2}=8\sqrt{2}\approx11.3$ cm.
     - # Answer:
       - $A:8$ $cm^2$
       - $P:11.3$ $cm$
3. **For triangle $ABC$ ($A(0,4)$, $B(2,0)$, $C(-4,0)$)**:
   - **Area ($A$)**:
     - # Explanation:
       - ## Step1: Identify base and height
         - The base of the triangle is the horizontal distance between $B(2,0)$ and $C(-4,0)$. Using the distance formula for points with the same $y$ - coordinate $d=\vert x_2 - x_1\vert$, the base $b=\vert2+4\vert = 6$ cm. The height of the triangle is the vertical distance between $A(0,4)$ and the line $y = 0$, so $h = 4$ cm.
       - ## Step2: Apply the area formula for a triangle
         - Using $A=\frac{1}{2}bh$, substituting $b = 6$ and $h = 4$, we get $A=\frac{1}{2}\times6\times4 = 12$ $cm^2$.
   - **Perimeter ($P$)**:
     - ## Step1: Find the lengths of the sides
       - The length of side $AB$: For $A(0,4)$ and $B(2,0)$, $AB=\sqrt{(2 - 0)^2+(0 - 4)^2}=\sqrt{4 + 16}=\sqrt{20}\approx4.5$ cm.
       - The length of side $AC$: For $A(0,4)$ and $C(-4,0)$, $AC=\sqrt{(-4 - 0)^2+(0 - 4)^2}=\sqrt{16 + 16}=\sqrt{32}\approx5.7$ cm.
       - The length of side $BC$: For $B(2,0)$ and $C(-4,0)$, $BC=\sqrt{(-4 - 2)^2+(0 - 0)^2}=\sqrt{36}=6$ cm.
     - ## Step2: Calculate the perimeter
       - $P=AB + AC+BC\approx4.5+5.7 + 6=16.2$ cm.
     - # Answer:
       - $A:12$ $cm^2$
       - $P:16.2$ $cm$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Answer: