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2-51. plot triangle abc with vertices a(0,0), b(3,4), and c(3,0) on gra…

Question

2-51. plot triangle abc with vertices a(0,0), b(3,4), and c(3,0) on graph paper. using the origin as the point of dilation, enlarge it by a factor of 2 (imagine using two rubber bands). label this new triangle abc.
a. what are the side lengths of the original triangle, δabc?
hint (a):
find ab by using the pythagorean theorem. for this triangle, (ac)^2+(cb)^2=(ab)^2.
b. what are the side lengths of the enlarged triangle, δabc?
hint (b):
c. calculate the area and the perimeter of δabc.
use the etool below to plot triangle abc and then enlarge from the origin.
click the link at right for the full version of the etool: int2 2-51 hw etool

Explanation:

Step1: Find side - lengths of $\triangle ABC$

For points $A(0,0)$, $B(3,4)$, $C(3,0)$:
The length of $AC$: Using the distance formula for two - points $(x_1,y_1)$ and $(x_2,y_2)$ which is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. For $A(0,0)$ and $C(3,0)$, $AC=\sqrt{(3 - 0)^2+(0 - 0)^2}=3$.
The length of $CB$: For $C(3,0)$ and $B(3,4)$, $CB=\sqrt{(3 - 3)^2+(4 - 0)^2}=4$.
Using the Pythagorean theorem $AB=\sqrt{AC^{2}+CB^{2}}=\sqrt{3^{2}+4^{2}}=\sqrt{9 + 16}=\sqrt{25}=5$.

Step2: Find side - lengths of $\triangle A'B'C'$

When a figure is dilated by a factor of $k$ from the origin, the coordinates of a point $(x,y)$ become $(kx,ky)$. Here $k = 2$.
The new coordinates are $A'(0,0)$, $B'(6,8)$, $C'(6,0)$.
The length of $A'C'$: $A'C'=\sqrt{(6 - 0)^2+(0 - 0)^2}=6$.
The length of $C'B'$: $C'B'=\sqrt{(6 - 6)^2+(8 - 0)^2}=8$.
The length of $A'B'$: $A'B'=\sqrt{(6 - 0)^2+(8 - 0)^2}=\sqrt{36+64}=\sqrt{100}=10$.

Step3: Calculate the area of $\triangle A'B'C'$

The area of a triangle is given by $A=\frac{1}{2}\times base\times height$. For $\triangle A'B'C'$ with base $A'C' = 6$ and height $C'B' = 8$, $A=\frac{1}{2}\times6\times8 = 24$.

Step4: Calculate the perimeter of $\triangle A'B'C'$

The perimeter $P$ of a triangle is the sum of its side - lengths. $P=A'C'+C'B'+A'B'=6 + 8+10=24$.

Answer:

a. $AB = 5$, $AC = 3$, $CB = 4$.
b. $A'B' = 10$, $A'C' = 6$, $C'B' = 8$.
c. Area = 24, Perimeter = 24.