A 10 ft slide is attached to a deck that is 5 ft high. Find the nearest distance from the bottom of the deck to the bottom of the slide to the nearest tenth.
1. Draw a picture. What triangle is formed by this picture? (I think right. I don't have graph paper with me right now.)
2. Is the unknown length a leg or hypotenuse of the triangle?
3. Mason is on the southwest corner of a 90° intersection. One street in the intersection is 23 feet wide. If Mason crosses diagonally to the northeast corner, he will walk 34 feet. Find the width of the other street. If necessary, round your answer to the nearest tenth.
Thank you in advance! I was absent when this was taught, plus, I can't wrap my head around it. :))
a^2 + b^2 = c^2
a^2 + 5^2 = 10^2
a^2 + 25 = 100
- 25 -25
______________
a^2 = 75 find square root
a = about 8.6
or
a = square root of 75.
1. We have a rt. triangle:
X = ? = Hor. leg.
Y = 5 Ft. = Ver. leg.
r = 10 Ft. = hyp.
x^2 + y^2 = r^2.
x^2 + 5^2 = 10^2,
1. To visualize the situation, imagine a right triangle formed by the deck, the slide, and the ground. The deck forms the vertical side of the triangle, while the slide forms the hypotenuse, and the ground forms the horizontal side.
2. The unknown length is the distance from the bottom of the deck to the bottom of the slide. This length is the vertical side of the triangle, also known as the leg.
3. To solve for the width of the other street, we can apply the Pythagorean theorem. According to the theorem, the sum of the squares of the lengths of the two legs of a right triangle is equal to the square of the length of the hypotenuse.
In this case, the leg representing the width of one street is 23 feet, the hypotenuse is 34 feet, and we need to find the other leg. Let's use the formula:
a^2 + b^2 = c^2
where a is the width of one street, b is the width of the other street, and c is the distance Mason walked diagonally.
Substituting the given values:
23^2 + b^2 = 34^2
529 + b^2 = 1156
b^2 = 1156 - 529
b^2 = 627
Taking the square root of both sides:
b ≈ sqrt(627)
b ≈ 25 feet (rounded to the nearest tenth)
Therefore, the width of the other street is approximately 25 feet.
1. You can draw a simple triangle to represent the scenario. The deck forms the vertical side of the triangle, the slide forms the hypotenuse, and the unknown distance forms the horizontal side of the triangle. Make sure the triangle is a right triangle with the deck height as the vertical side and the length of the slide as the hypotenuse.
2. The unknown length, which is the distance from the bottom of the deck to the bottom of the slide, is a leg of the triangle.
3. In this question, we are given the diagonal distance (34 ft) and one leg of the right triangle (23 ft). We need to find the other leg, which represents the width of the other street.
To solve this, we can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (c) is equal to the sum of the squares of the lengths of the other two sides (a and b).
In this case, the diagonal distance is the hypotenuse (c), the width of one street is one side (a), and the width of the other street (unknown side) is the other side (b).
So, we have the equation:
c^2 = a^2 + b^2
Substituting the given values:
34^2 = 23^2 + b^2
Simplifying:
1156 = 529 + b^2
Subtracting 529 from both sides:
627 = b^2
Taking the square root of both sides:
b ≈ √627
b ≈ 25 ft (rounded to the nearest tenth)
Therefore, the width of the other street is approximately 25 feet.