Envision Math 2.0: Grade 8, Volume 2
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Exercise 1 Page 394

Practice makes perfect

We are told that Javier is standing near a tree and uses an electronic tape measure to measure the distances from his eyes to the top of the tree, from his eyes to the tree horizontally, and from his eyes to the base of the tree. These measurements create two triangles.

Javier claims that he can use the Pythagorean Theorem to find the height of the tree. We want to explain how he can do that. First, let's note that the height of the tree can be divided into two sections. Let's call the height of the lower section h_1 and the height of the upper section h_2.

Notice that the upper and the lower triangles are both right triangles. We know two out of three side lengths for both triangles.
We can use the Pythagorean Theorem on each of the triangles to find h_1 and h_2. The height of the tree is the sum of h_1 and h_2.
Now we are asked to find the height of the tree, rounding to the nearest tenth. Let's follow our plan from Part A. We will divide the height of the tree into two segments and use the Pythagorean Theorem to find the height of each one.
Let's start with h_1. By the Pythagorean Theorem, the sum of the squares of the lengths of the legs of the triangle is equal to the square of the length of the hypotenuse. Here, h_1 is the length of one of the legs. The other leg is 7 feet long and the is 9 feet long.
Let's write an equation. h_1^2 + 7^2 = 9^2 Let's solve this equation for h_1.
h_1^2 + 7^2 = 9^2
h_1^2 + 49 = 81
h_1^2 +49 - 49= 81 - 49
h_1 ^2 = 32
sqrt(h_1 ^2) = sqrt(32)
h_1 = sqrt(32)
h_1 = 5.656854...
h_1 ≈ 5.7
The value of h_1 is approximately 5.7 feet. Now let's find the value of h_2. Like the previous triangle, h_2 is the length of one of the legs of a right angle. The other leg is 7 feet long and the hypotenuse is 25 feet long.
We can write another equation using the Pythagorean Theorem. h_2^2 + 7^2 = 25^2 Let's solve this equation for h_2.
h_2^2 + 7^2 = 25^2
h_2^2 + 49 = 625
h_2^2 +49 - 49= 625 - 49
h_2 ^2 = 576
sqrt(h_2 ^2) = sqrt(576)
h_2 = sqrt(576)
h_2 = 24
The value of h_2 is 24 feet. Now we can find the height of the tree by adding h_1 and h_2.
h_1 + h_2
5.7 + 24
29.7
The height of the tree is about 29.7 feet.

Javier has now moved backward so that the horizontal distance from him to the tree is 3 feet greater. We want to determine whether the distance from his eyes to the top of the tree also increases by 3 feet. Notice that the vertical distance from Javier's eyes to the top of the tree does not change.

Let's use the Pythagorean Theorem to find the unknown length x. As we can see in the second triangle, x is the length of the hypotenuse of a right triangle with leg lengths 10 and 24. This lets us write an equation. 10^2 + 24^2 = x^2 Let's solve this equation for x!
10^2 + 24^2 = x^2
100 + 576 = x^2
676 = x^2
sqrt(676) = sqrt(x^2)
sqrt(676) = x
26 = x
x = 26
The distance from Javier's eyes to the top of the tree increased to 26 feet after he moved backward. This means it only increased by 1 foot, not 3 feet.
Now we want to find out if Javier could change his horizontal distance from the tree so that the distance from his eyes to the top of the tree is only 20 feet. As we found in Part B, the height from Javier's eye level to the top of the tree is 24 feet.
The horizontal line from Javier's eyes to the tree, the height of the tree, and the line from Javier's eyes to the top of the tree always form a right triangle, no matter how Javier changes his distance from the tree. The line from Javier's eyes to the top of the tree is the hypotenuse of this triangle.
In a right triangle, the hypotenuse is always longer than either leg. Here, the length of one of the legs is the vertical distance between Javier's eyes and the top of the tree, or 24 feet. No matter how Javier moves horizontally, this distance will never change. Then, since 20 can never be greater than 24, he cannot move so that the distance from his eyes to the top of the tree is only 20 feet.