Question

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

8.6 units
8.6 units

9.7 units
9.7 units

13.7 units

Answer 1

To find the perimeter of the right isosceles triangle, we start by identifying the vertices of the triangle. The right-angle vertex is at (4, 2), and one of the other vertices is at (8, 2).

Since the triangle is isosceles and right-angled, the two legs of the triangle will be equal. The distance from (4, 2) to (8, 2) is one leg of the triangle.

1. **Calculating the length of one leg:**
\[
\text{Distance} = x_2 - x_1 = 8 - 4 = 4 \text{ units}
\]
So one leg is 4 units long.

2. **Since this is a right isosceles triangle, the other leg will also be 4 units long.**

3. **Finding the length of the hypotenuse:**
The hypotenuse can be calculated using the Pythagorean theorem:
\[
c = \sqrt{a^2 + b^2} = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}
\]

4. **Calculating the perimeter:**
The perimeter \( P \) of the triangle is:
\[
P = \text{leg}_1 + \text{leg}_2 + \text{hypotenuse} = 4 + 4 + 4\sqrt{2}
\]

5. **Calculating \( 4\sqrt{2} \):**
The approximate value of \( \sqrt{2} \) is about 1.414. Thus:
\[
4\sqrt{2} \approx 4 \times 1.414 \approx 5.656
\]

6. **Now summing the lengths:**
\[
P \approx 4 + 4 + 5.656 \approx 13.656
\]

7. **Rounding to the nearest tenths place:**
\[
P \approx 13.7
\]

Thus, the perimeter of the triangle is approximately **13.7 units**.