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three trees in a grouping of a landscape are positioned relative to eac…

Question

three trees in a grouping of a landscape are positioned relative to each other to form an isosceles right triangle as shown. what is the approximate distance in feet between the centers of the trees located at a and b? if your answer is (\frac{14sqrt{2}}{2}), please type 7sqrt2 - no space. the distance between a and b is (square) ft. (exact value - radical form) or (square) ft. (decimal) the approximate distance between a and b is (square) ft.

Explanation:

Step1: Identify triangle type and sides

$\triangle ABC$ is isosceles right triangle, hypotenuse $AC=14.6$, let $AB=BC=x$.

Step2: Apply Pythagorean theorem

$$x^2 + x^2 = 14.6^2$$
$$2x^2 = 213.16$$
$$x^2 = 106.58$$
$$x = \sqrt{106.58} = \frac{73\sqrt{2}}{10}$$

Step3: Calculate approximate value

$$x \approx \frac{73\times1.414}{10} \approx 10.32$$

Answer:

The exact distance between A and B is $\frac{73\sqrt{2}}{10}$ ft.
The approximate distance between A and B is $10.32$ ft.