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A solid is in the form of cone mounted on a hemisphere in such a way that the centre of the base of the cone just coincide with the centre of the base of the cone is 1 upon 2 hour where r is the radius of the hemisphere prove that the total surface area of the solid is pie upon 4 (11 r 2 l) r ?
Verified Answer
A solid is in the form of cone mounted on a hemisphere in such a way t...
Given: Radius of base of cone = 1/2 r
and slant height of cone = l
To prove: TSA of solid =
Now, the cone is mounted on hemisphere , which implies SA of hemisphere is half of sphere
SA of hemisphere=
3pir^2
....(ii)
3pir^2
....(ii)
Now, to get the new SA of cone and new SA of hemisphere, subtract the area of base of cone twice because once the cone is placed at top ,it will remove the SA from cone as well as hemisphere.
From (i) and (ii),
New SA of cone =
Hence proved.
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Most Upvoted Answer
A solid is in the form of cone mounted on a hemisphere in such a way t...
Proof:
Given:
Radius of the hemisphere = r
To prove:
Total surface area of the solid = π/4 * (11r^2 + 2rl)
Let's break down the solid into its individual components and calculate their surface areas separately.
1. Surface area of the cone:
The slant height of the cone can be found using the Pythagorean theorem in the right triangle formed by the radius of the hemisphere, the height of the cone, and the slant height. Since the center of the base of the cone coincides with the center of the base of the hemisphere, the height of the cone is equal to the radius of the hemisphere.
Let h be the height of the cone, then h = r.
The slant height, l, can be found using the Pythagorean theorem:
l^2 = r^2 + r^2
l^2 = 2r^2
l = √(2r^2)
l = r√2
The surface area of the cone can be calculated using the formula:
Surface area = πrl
Substituting the values of r and l:
Surface area of the cone = π * r * (r√2)
Surface area of the cone = πr^2√2
2. Surface area of the hemisphere:
The surface area of a hemisphere is given by the formula:
Surface area = 2πr^2
3. Surface area of the base of the hemisphere:
The surface area of the base of the hemisphere is simply the area of a circle with radius r:
Surface area = πr^2
Adding up the surface areas of the cone, hemisphere, and the base of the hemisphere, we get the total surface area of the solid:
Total surface area = Surface area of the cone + Surface area of the hemisphere + Surface area of the base of the hemisphere
Total surface area = πr^2√2 + 2πr^2 + πr^2
Total surface area = 4πr^2 + πr^2√2
Factoring out πr^2, we have:
Total surface area = πr^2(4 + √2)
Total surface area = π/4 * (11r^2 + 2rl)
Thus, we have proved that the total surface area of the solid is π/4 * (11r^2 + 2rl).
Given:
Radius of the hemisphere = r
To prove:
Total surface area of the solid = π/4 * (11r^2 + 2rl)
Let's break down the solid into its individual components and calculate their surface areas separately.
1. Surface area of the cone:
The slant height of the cone can be found using the Pythagorean theorem in the right triangle formed by the radius of the hemisphere, the height of the cone, and the slant height. Since the center of the base of the cone coincides with the center of the base of the hemisphere, the height of the cone is equal to the radius of the hemisphere.
Let h be the height of the cone, then h = r.
The slant height, l, can be found using the Pythagorean theorem:
l^2 = r^2 + r^2
l^2 = 2r^2
l = √(2r^2)
l = r√2
The surface area of the cone can be calculated using the formula:
Surface area = πrl
Substituting the values of r and l:
Surface area of the cone = π * r * (r√2)
Surface area of the cone = πr^2√2
2. Surface area of the hemisphere:
The surface area of a hemisphere is given by the formula:
Surface area = 2πr^2
3. Surface area of the base of the hemisphere:
The surface area of the base of the hemisphere is simply the area of a circle with radius r:
Surface area = πr^2
Adding up the surface areas of the cone, hemisphere, and the base of the hemisphere, we get the total surface area of the solid:
Total surface area = Surface area of the cone + Surface area of the hemisphere + Surface area of the base of the hemisphere
Total surface area = πr^2√2 + 2πr^2 + πr^2
Total surface area = 4πr^2 + πr^2√2
Factoring out πr^2, we have:
Total surface area = πr^2(4 + √2)
Total surface area = π/4 * (11r^2 + 2rl)
Thus, we have proved that the total surface area of the solid is π/4 * (11r^2 + 2rl).
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A solid is in the form of cone mounted on a hemisphere in such a way that the centre of the base of the cone just coincide with the centre of the base of the cone is 1 upon 2 hour where r is the radius of the hemisphere prove that the total surface area of the solid is pie upon 4 (11 r 2 l) r ?
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A solid is in the form of cone mounted on a hemisphere in such a way that the centre of the base of the cone just coincide with the centre of the base of the cone is 1 upon 2 hour where r is the radius of the hemisphere prove that the total surface area of the solid is pie upon 4 (11 r 2 l) r ? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about A solid is in the form of cone mounted on a hemisphere in such a way that the centre of the base of the cone just coincide with the centre of the base of the cone is 1 upon 2 hour where r is the radius of the hemisphere prove that the total surface area of the solid is pie upon 4 (11 r 2 l) r ? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solid is in the form of cone mounted on a hemisphere in such a way that the centre of the base of the cone just coincide with the centre of the base of the cone is 1 upon 2 hour where r is the radius of the hemisphere prove that the total surface area of the solid is pie upon 4 (11 r 2 l) r ?.
A solid is in the form of cone mounted on a hemisphere in such a way that the centre of the base of the cone just coincide with the centre of the base of the cone is 1 upon 2 hour where r is the radius of the hemisphere prove that the total surface area of the solid is pie upon 4 (11 r 2 l) r ? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about A solid is in the form of cone mounted on a hemisphere in such a way that the centre of the base of the cone just coincide with the centre of the base of the cone is 1 upon 2 hour where r is the radius of the hemisphere prove that the total surface area of the solid is pie upon 4 (11 r 2 l) r ? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solid is in the form of cone mounted on a hemisphere in such a way that the centre of the base of the cone just coincide with the centre of the base of the cone is 1 upon 2 hour where r is the radius of the hemisphere prove that the total surface area of the solid is pie upon 4 (11 r 2 l) r ?.
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