A right isosceles triangle is drawn on a coordinate plane with the right-angle vertex at (4,2)

Question

and another vertex at (8,2)
. Using the coordinates given, what is the perimeter of the triangle? Round the answer to the nearest tenths place.(1 point)
Responses

16.7 units
16.7 units

9.7 units
9.7 units

13.7 units
13.7 units

8.6 units

Answer 1

Given the right isosceles triangle with the right-angle vertex at (4, 2) and another vertex at (8, 2), we can find the perimeter by identifying the lengths of its sides.

1. Calculate the length of the base:
   The base goes from (4, 2) to (8, 2).
   Length = $|8-4|=4$ units.

2. Determine the height:
   Since the triangle is right isosceles, the other vertex must be directly above or below the right-angle vertex. The length of the other two sides (height) will also be equal to the base.
   Thus, the height = 4 units.

3. Identifying the vertices of the triangle:

   • Right angle vertex: $(4,2)$
   • Vertex along the base: $(8,2)$
   • Vertex at the tip, which will have the same x-coordinate as the right-angle vertex (4), and the y-coordinate will be 4 units up from 2 (because it's an isosceles triangle), making it $(4,6)$.

4. Calculating the lengths of the two equal sides: The two equal sides stretch from $(4,2)$ to $(4,6)$ and from $(8,2)$ to $(4,6)$.

   Length of (4, 2) to (4, 6):
   Length is $|6-2|=4$ units.

   Length of (8, 2) to (4, 6):
   Using the distance formula:
   $$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}=\sqrt{(4-8{)}^2+(6-2{)}^2}=\sqrt{(-4{)}^2+(4{)}^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}\approx 5.656\text{ units}$$

5. Determine the total perimeter:
   Now we can compute the perimeter as follows:
   $$\text{Perimeter}=\text{Base}+2\times \text{Length of Equal Sides}=4+4+4\sqrt{2}$$ Approximating $4\sqrt{2}\approx 4\times 1.414\approx 5.656$: $$\text{Perimeter}\approx 4+4+5.656\approx 13.656\text{ units}$$

Rounding this to the nearest tenths place gives us approximately $13.7$ units.

Therefore, the answer is 13.7 units.