Question

plot abc with vertices a(0,0), b(3, 5), and c(3,0). using the origin as the point of dilation, enlarge it by a factor of 2. label this new triangle. a. what are the side lengths of the original triangle, abc? b. what are the side lengths of the enlarged triangle, abc? how do they compare to the side lengths of abc? c. calculate the area and the perimeter of abc and abc. how do they compare?

Explanation:

Step1: Calculate side - lengths of original triangle

Use the distance formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. For side $AB$ with $A(0,0)$ and $B(3,5)$: $AB=\sqrt{(3 - 0)^2+(5 - 0)^2}=\sqrt{9 + 25}=\sqrt{34}$. For side $AC$ with $A(0,0)$ and $C(3,0)$: $AC=\sqrt{(3 - 0)^2+(0 - 0)^2}=3$. For side $BC$ with $B(3,5)$ and $C(3,0)$: $BC=\sqrt{(3 - 3)^2+(0 - 5)^2}=5$.

Step2: Calculate side - lengths of dilated triangle

When dilating by a factor of $k = 2$ with the origin as the center of dilation, the coordinates of the new points are $A'(0\times2,0\times2)=(0,0)$, $B'(3\times2,5\times2)=(6,10)$, $C'(3\times2,0\times2)=(6,0)$. Then for side $A'B'$: $A'B'=\sqrt{(6 - 0)^2+(10 - 0)^2}=\sqrt{36+100}=\sqrt{136}=2\sqrt{34}$. For side $A'C'$: $A'C'=\sqrt{(6 - 0)^2+(0 - 0)^2}=6$. For side $B'C'$: $B'C'=\sqrt{(6 - 6)^2+(0 - 10)^2}=10$. The side - lengths of the dilated triangle are twice the side - lengths of the original triangle.

Step3: Calculate perimeter of original triangle

$P_{ABC}=AB + AC+BC=\sqrt{34}+3 + 5=\sqrt{34}+8$.

Step4: Calculate perimeter of dilated triangle

$P_{A'B'C'}=A'B'+A'C'+B'C'=2\sqrt{34}+6 + 10=2\sqrt{34}+16$. The perimeter of the dilated triangle is twice the perimeter of the original triangle.

Step5: Calculate area of original triangle

Using the formula for the area of a triangle $A=\frac{1}{2}\times base\times height$. With base $AC = 3$ and height $BC = 5$, $A_{ABC}=\frac{1}{2}\times3\times5 = 7.5$.

Step6: Calculate area of dilated triangle

With base $A'C'=6$ and height $B'C' = 10$, $A_{A'B'C'}=\frac{1}{2}\times6\times10=30$. The area of the dilated triangle is $4$ times the area of the original triangle (since $A_{A'B'C'}=k^2A_{ABC}$ and $k = 2$, $2^2=4$).

Answer:

a. The side - lengths of $\triangle ABC$ are $AB=\sqrt{34}$, $AC = 3$, $BC = 5$.
b. The side - lengths of $\triangle A'B'C'$ are $A'B'=2\sqrt{34}$, $A'C'=6$, $B'C'=10$. The side - lengths of the dilated triangle $\triangle A'B'C'$ are twice the side - lengths of the original triangle $\triangle ABC$.
c. The perimeter of $\triangle ABC$ is $\sqrt{34}+8$, and the perimeter of $\triangle A'B'C'$ is $2\sqrt{34}+16$. The perimeter of the dilated triangle is twice the perimeter of the original triangle. The area of $\triangle ABC$ is $7.5$, and the area of $\triangle A'B'C'$ is $30$. The area of the dilated triangle is $4$ times the area of the original triangle.