Question

for exercises 2–5, find the values of x and y.
2.
3.
4.
5.

1. devin says that since the diagonals of the kite intersect at right angles, the small right triangles are similar to both the left half and the right half of the kite. is he correct? explain.
2. isabel and helena have built a frame and covered it with cloth. the frame is in the shape of a right triangle, △abc, with side lengths 6 ft, 8 ft, and 10 ft. they use a vertical pole (overline{ae}) to raise corner a 3 ft, as shown. what is the distance ed from the base of the pole to the edge of the frame? round to the nearest foot.

Explanation:

Problem 2

Step1: Find hypotenuse of big triangle

First, calculate the hypotenuse of the large right triangle to use for altitude calculation.
$$\text{Hypotenuse} = \sqrt{6^2 + 14^2} = \sqrt{36 + 196} = \sqrt{232} = 2\sqrt{58}$$

Step2: Calculate x via area equality

Use area of right triangle (legs vs hypotenuse+altitude) to find $x$.
$$\frac{1}{2} \times 6 \times 14 = \frac{1}{2} \times 2\sqrt{58} \times x \implies 42 = \sqrt{58}x \implies x = \frac{42}{\sqrt{58}} = \frac{42\sqrt{58}}{58} = \frac{21\sqrt{58}}{29} \approx 5.5$$

Step3: Calculate y via geometric mean

Use geometric mean for altitude to hypotenuse: $y = \frac{6 \times 14}{\text{Hypotenuse}}$.
$$y = \frac{84}{2\sqrt{58}} = \frac{42}{\sqrt{58}} = \frac{21\sqrt{58}}{29} \approx 5.5$$

Step1: Find hypotenuse of big triangle

Calculate the hypotenuse of the large right triangle.
$$\text{Hypotenuse} = \sqrt{1^2 + y^2}$$

Step2: Calculate x via area equality

Use area equality to find $x$ first, then solve for $y$.
$$\frac{1}{2} \times 1 \times y = \frac{1}{2} \times (1+9) \times x \implies y = 10x$$

Step3: Use Pythagorean theorem on big triangle

Substitute $y=10x$ into Pythagorean theorem.
$$1^2 + y^2 = 10^2 \implies 1 + 100x^2 = 100 \implies 100x^2=99 \implies x = \frac{3\sqrt{11}}{10} \approx 0.995$$

Step4: Solve for y

Use $y=10x$ to find $y$.
$$y = 10 \times \frac{3\sqrt{11}}{10} = 3\sqrt{11} \approx 9.95$$

Step1: Find hypotenuse of big triangle

The large triangle is right-angled, hypotenuse = 12, one leg =4.
$$x = \sqrt{12^2 - 4^2} = \sqrt{144 - 16} = \sqrt{128} = 8\sqrt{2} \approx 11.31$$

Step2: Calculate altitude (unlabeled, but x solved)

The question asks for $x$, so we stop here.

Answer:

$x = \frac{21\sqrt{58}}{29} \approx 5.5$, $y = \frac{21\sqrt{58}}{29} \approx 5.5$

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Problem 3