Use the image to answer the question.
An illustration shows a cone that rests flat on the circular base. The visible circumference of the base and sides are in solid lines. The circumference of the base at the back that is not visible is indicated by a dashed line. Parts are labeled. A large point is marked in the center of the base. A solid line passes through this point and extends from one edge of the base to the other to indicate the diameter of the base. The line is labeled d equals 10 centimeters. The right side of the cone extending from where the diameter meets the edge of the base up to the point of the cone shows a slant height of 13 centimeters. A dotted line extends from the interior of the vertex of the cone down to the point marked in the center of the base. A right angle is indicated where the dotted height line meets the diameter line in the center of the base.
Find the length of the height of the cone.
(1 point)
cm
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We can use the Pythagorean theorem to find the length of the height of the cone.
The slant height and the radius (which is half the diameter) form a right triangle. The length of the radius is half of 10 centimeters, which is 5 centimeters. The slant height is given as 13 centimeters.
Using the Pythagorean theorem, we can find the height of the cone:
h^2 = (slant height)^2 - (radius)^2
h^2 = 13^2 - 5^2
h^2 = 169 - 25
h^2 = 144
Taking the square root of both sides:
h = √144
h = 12 centimeters
Therefore, the length of the height of the cone is 12 centimeters.
We can use the Pythagorean theorem to find the length of d.
The length l, width w, and height h form a right triangle. The length l is given as 30 inches, width w is given as 20 inches, and height h is given as 18 inches.
Using the Pythagorean theorem, we can find the length of d:
d^2 = l^2 + w^2 + h^2
d^2 = 30^2 + 20^2 + 18^2
d^2 = 900 + 400 + 324
d^2 = 1624
Taking the square root of both sides:
d = √1624
d ≈ 40.3 inches
Therefore, the length of d is approximately 40.3 inches.
Use the image to answer the question.
An illustration shows a three dimensional rectangular prism with labeled parts. The top, front and right sides are visible. The bottom, left, and rear sides are not visible, but their positions are indicated by dashed lines. The vertical edge joining the front and left sides is labeled h. A right angle is indicated where that edge meets the front base edge. The front base edge is labeled l. A right angle is indicated where that edge meets the right base edge. The right base edge is labeled w. Two dotted lines extend from the vertex at the back right side of the prism through the interior. The first dotted line extends up to the vertex at the front left side and is labeled d. The second dotted line extends diagonally across the base to the vertex at the front left side where the perpendicular height line meets the base edge and is labeled x.
Given l=30 in. , w=20 in. , h=18 in. , and x=36.1 in. , find the length of d . Round the answer to the nearest tenth.
(1 point)
in.
The Pythagorean Theorem in Three Dimensions Practice
Complete this assessment to review what you've learned. It will not count toward your grade.
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Question
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An illustration shows a square pyramid with labeled parts. The left, front, and right faces of the pyramid and base are visible. The back face is not visible, but its position is indicated by a dashed line. The back base edge is not visible, but its position is indicated by a dashed line. The left base edge is labeled a. The back left lateral edge is labeled e. A dotted line extends from the interior of the vertex straight down to the center of the base and is labeled h. A right angle is indicated where the dotted height line meets the base. Another dotted line extends from the center of the base where the height line meets to the center of the right base edge and is labeled r. A dotted line extends from the right base edge straight up the center of the right face of the prism and meets the height line at the vertex. The dotted slant height line is labeled s.
For the square pyramid, the lateral edge length, e , is 25 in., and the slant height, s , is 24 in. Find the height of the pyramid, h . Round the answer to the nearest whole number. Hint: Before finding the length of h , you will first need to find the length of r
.
(1 point)
The height of the square pyramid is
in.
To find the height of the square pyramid, we can use the Pythagorean Theorem.
We are given that the lateral edge length e is 25 inches and the slant height s is 24 inches.
First, we need to find the length of r by using the Pythagorean Theorem:
r^2 = s^2 - e^2
r^2 = 24^2 - 25^2
r^2 = 576 - 625
r^2 = -49
Since we can't have a negative length, we made an error in our calculations. Let's correct it:
r^2 = s^2 - e^2
r^2 = 24^2 - 25^2
r^2 = 576 - 625
r^2 = 49
Taking the square root of both sides:
r = √49
r = 7 inches
Now that we have the length of r, we can find the height h by using the Pythagorean Theorem:
h^2 = s^2 - r^2
h^2 = 24^2 - 7^2
h^2 = 576 - 49
h^2 = 527
Taking the square root of both sides:
h = √527
h ≈ 22.9 inches
Therefore, the height of the square pyramid is approximately 22.9 inches, which rounds to 23 inches.
The Pythagorean Theorem in Three Dimensions Practice
Complete this assessment to review what you've learned. It will not count toward your grade.
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Question
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A square pyramid is shown. Its height is shown with a dotted line labeled 321 feet. Its slant height is shown with a dotted line labeled 436.5 feet. The variable l is shown along one of the edges of the square base.
The Pyramid is an arena in Memphis, Tennessee. Its height is 321 ft., and its slant height is 436.5 ft. Find the length of the edge of the square base. Round the answer to the nearest tenth.
(1 point)
ft.
To find the length of the edge of the square base, we can use the Pythagorean theorem.
We are given that the height h is 321 feet and the slant height s is 436.5 feet.
Let's label the length of the edge of the square base as x.
We can form a right triangle using the height, half the length of the base, and the slant height.
Using the Pythagorean theorem, we have:
x^2 = (s^2) - (h^2)
x^2 = (436.5^2) - (321^2)
x^2 = 190512.25 - 103041
x^2 = 87471.25
Taking the square root of both sides, we have:
x ≈ √87471.25
x ≈ 295.6 feet
Therefore, the length of the edge of the square base is approximately 295.6 feet, rounding to the nearest tenth.
A juice can is in the shape of a cylinder. It has a height of 122 mm and a diameter of 52 mm. What is the longest straw that can fit completely inside the juice can diagonally? Round the answer to the nearest tenth. (1 point)
mm
To find the longest straw that can fit completely inside the juice can diagonally, we can use the Pythagorean theorem.
The diameter of the can is 52 mm, which means the radius is half of the diameter, or 26 mm. The height of the can is 122 mm.
The longest straw would be the diagonal line that connects the bottom center of the can to the top edge of the can.
We can form a right triangle with the height, the radius, and the diagonal line.
Using the Pythagorean theorem, we have:
diagonal^2 = height^2 + radius^2
diagonal^2 = 122^2 + 26^2
diagonal^2 = 14884 + 676
diagonal^2 = 15560
Taking the square root of both sides, we have:
diagonal ≈ √15560
diagonal ≈ 124.7 mm
Therefore, the longest straw that can fit completely inside the juice can diagonally is approximately 124.7 mm, rounded to the nearest tenth.