Question

01/28 - special right triangle mixed practice l
three trees in a drawing of a landscape are positioned relative to each other to form an isosceles right triangle as shown. what is the shortest distance in feet between the centers of the trees located at a and b?
(nearest hundredth)

Explanation:

Step1: Identify triangle properties

$\triangle ABC$ is isosceles right triangle, $\angle B=90^\circ$, $AC=18.7$ ft, $AB=BC$.

Step2: Apply Pythagorean theorem

For right triangle: $AB^2 + BC^2 = AC^2$. Since $AB=BC$, substitute:
$$2AB^2 = (18.7)^2$$

Step3: Solve for $AB^2$

$$AB^2 = \frac{18.7^2}{2} = \frac{349.69}{2} = 174.845$$

Step4: Calculate $AB$

$$AB = \sqrt{174.845} \approx 13.22$$

Answer:

13.22 ft