a. Answer The value of x is 15 . Explanation Given information: Height of small cone = 12, Slant height of small cone = x , Height of large cone = 12 + 8 = 20 , Slant height of large cone = x + 10. Formula used: The below property is used: Corresponding sides of similar triangles are congruent. Calculation: The smaller triangle is similar to the whole triangle by AA similarity rule, so Corresponding sides of similar triangles are congruent. x x + 10 = 12 12 + 8 x x + 10 = 12 20 20 x = 12 x + 120 8 x = 120 x = 120 8 = 15 b. To determine To find: The radii of the circles . Answer The radii are 9 and 15. Explanation Given information: Height of small cone = 12, Slant height of small cone = x , Height of large cone = 12 + 8 = 20 , Slant height of large cone = x + 10. Formula used: The below property is used: Corresponding sides of similar triangles are congruent. The below theorem is used: Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”. In right angle triangle, a 2 + b 2 = c 2 Calculation: The smaller triangle is similar to the whole triangle by AA similarity rule, so Corresponding sides of similar triangles are congruent. x x + 10 = 12 12 + 8 x x + 10 = 12 20 20 x = 12 x + 120 8 x = 120 x = 120 8 = 15 Radius of small circle: Side BC can be calculated by applying Pythagoras Theorem. In right angled triangle ABC , we get ( A C ) 2 = ( A B ) 2 + ( B C ) 2 ( 15 ) 2 = ( 12 ) 2 + ( B C ) 2 225 = 144 + ( B C ) 2 ( B C ) 2 = 225 − 144 B C = 81 B C = 9 r = B C = 9 Radius of large circle: Side ED can be calculated by applying Pythagoras Theorem. In right angled triangle ADE , we get ( A E ) 2 = ( A D ) 2 + ( D E ) 2 ( 25 ) 2 = ( 20 ) 2 + ( D E ) 2 625 = 400 + ( D E ) 2 ( D E ) 2 = 625 − 400 D E = 225 D E = 15 r = D E = 15 c. To determine To calculate: The volume of smaller cone. Answer The volume of smaller cone is 1018 . Explanation Given information: Height of small cone = 12, Slant height of small cone = x , Height of large cone = 12 + 8 = 20 , Slant height of large cone = x + 10. Formula used: The below property is used: Corresponding sides of similar triangles are congruent. The below theorem is used: Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”. In right angle triangle, a 2 + b 2 = c 2 Volume of cone = 1 3 × B × h B = Base area of cone and h = height of cone Area of a circle: A = π r 2 r = radius of circle Calculation: The smaller triangle is similar to the whole triangle by AA similarity rule, so Corresponding sides of similar triangles are congruent. x x + 10 = 12 12 + 8 x x + 10 = 12 20 20 x = 12 x + 120 8 x = 120 x = 120 8 = 15 Radius of small circle: Side BC can be calculated by applying Pythagoras Theorem. In right angled triangle ABC , we get ( A C ) 2 = ( A B ) 2 + ( B C ) 2 ( 15 ) 2 = ( 12 ) 2 + ( B C ) 2 225 = 144 + ( B C ) 2 ( B C ) 2 = 225 − 144 B C = 81 B C = 9 r = B C = 9 Volume of cone = 1 3 × B × h B = π r 2 = π ( 9 ) 2 = 81 π h = 12 Volume of smaller cone = 1 3 × 81 π × 12 Volume of smaller cone = 324 π Volume of smaller cone ≈ 1018 d. To determine To calculate: The volume of larger cone. Answer The volume of larger cone is 4712 . Explanation Given information: Height of small cone = 12, Slant height of small cone = x , Height of large cone = 12 + 8 = 20 , Slant height of large cone = x + 10. Formula used: The below property is used: Corresponding sides of similar triangles are congruent. The below theorem is used: Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”. In right angle triangle, a 2 + b 2 = c 2 Volume of cone = 1 3 × B × h B = Base area of cone and h = height of cone Area of a circle: A = π r 2 r = radius of circle Calculation: The smaller triangle is similar to the whole triangle by AA similarity rule, so Corresponding sides of similar triangles are congruent. x x + 10 = 12 12 + 8 x x + 10 = 12 20 20 x = 12 x + 120 8 x = 120 x = 120 8 = 15 Radius of large circle: Side ED can be calculated by applying Pythagoras Theorem. In right angled triangle ADE , we get ( A E ) 2 = ( A D ) 2 + ( D E ) 2 ( 25 ) 2 = ( 20 ) 2 + ( D E ) 2 625 = 400 + ( D E ) 2 ( D E ) 2 = 625 − 400 D E = 225 D E = 15 r = D E = 15 Volume of cone = 1 3 × B × h B = π r 2 = π ( 15 ) 2 = 225 π h = 20 Volume of larger cone = 1 3 × 225 π × 20 Volume of larger cone = 1500 π Volume of larger cone ≈ 4712 e. To determine To calculate: The volume of the frustum. Answer The volume of the frustum is 3695 . Explanation Given information: Height of small cone = 12, Slant height of small cone = x , Height of large cone = 12 + 8 = 20 , Slant height of large cone = x + 10. Formula used: The below property is used: Corresponding sides of similar triangles are congruent. The below theorem is used: Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”. In right angle triangle, a 2 + b 2 = c 2 Volume of cone = 1 3 × B × h B = Base area of cone and h = height of cone Area of a circle: A = π r 2 r = radius of circle Calculation: The smaller triangle is similar to the whole triangle by AA similarity rule, so Corresponding sides of similar triangles are congruent. x x + 10 = 12 12 + 8 x x + 10 = 12 20 20 x = 12 x + 120 8 x = 120 x = 120 8 = 15 Radius of small circle: Side BC can be calculated by applying Pythagoras Theorem. In right angled triangle ABC , we get ( A C ) 2 = ( A B ) 2 + ( B C ) 2 ( 15 ) 2 = ( 12 ) 2 + ( B C ) 2 225 = 144 + ( B C ) 2 ( B C ) 2 = 225 − 144 B C = 81 B C = 9 r = B C = 9 Volume of cone = 1 3 × B × h B = π r 2 = π ( 9 ) 2 = 81 π h = 12 Volume of smaller cone = 1 3 × 81 π × 12 Volume of smaller cone = 324 π Volume of smaller cone ≈ 1018 Radius of large circle: Side ED can be calculated by applying Pythagoras Theorem. In right angled triangle ADE , we get ( A E ) 2 = ( A D ) 2 + ( D E ) 2 ( 25 ) 2 = ( 20 ) 2 + ( D E ) 2 625 = 400 + ( D E ) 2 ( D E ) 2 = 625 − 400 D E = 225 D E = 15 r = D E = 15 Volume of cone = 1 3 × B × h B = π r 2 = π ( 15 ) 2 = 225 π h = 20 Volume of larger cone = 1 3 × 225 π × 20 Volume of larger cone = 1500 π Volume of larger cone ≈ 4712 Volume of frustum = Volume of larger cone − Volume of smaller cone Volume of frustum = 1500 π − 324 π Volume of frustum = 1176 π Volume of frustum ≈ 3695

91st Edition

ISBN: 9780866099653

Author: Richard Rhoad, George Milauskas, Robert Whipple

Publisher: McDougal Littell

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Question

Polygon with three sides, three angles, and three vertices. Based on the properties of each side, the types of triangles are scalene (triangle with three three different lengths and three different angles), isosceles (angle with two equal sides and two equal angles), and equilateral (three equal sides and three angles of 60°). The types of angles are acute (less than 90°); obtuse (greater than 90°); and right (90°).

Chapter 12.5, Problem 14PSB

To determine

To calculate: The value of x .

Expert Solution

Answer to Problem 14PSB

The value of x is 15 .

Explanation of Solution

Given information:

Height of small cone = 12,

Slant height of small cone = x ,

Height of large cone =12+8=20,

Slant height of large cone =x+10.

Formula used:

The below property is used:

Corresponding sides of similar triangles are congruent.

Calculation:

Geometry For Enjoyment And Challenge, Chapter 12.5, Problem 14PSB , additional homework tip 1

The smaller triangle is similar to the whole triangle by AA similarity rule, so

Corresponding sides of similar triangles are congruent.

xx+10=1212+8xx+10=122020x=12x+1208x=120x=1208=15

To determine

To find: The radii of the circles.

Expert Solution

Answer to Problem 14PSB

The radii are 9 and 15.

Explanation of Solution

Given information:

Height of small cone = 12,

Slant height of small cone = x ,

Height of large cone =12+8=20,

Slant height of large cone =x+10.

Formula used:

The below property is used:

Corresponding sides of similar triangles are congruent.

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

Geometry For Enjoyment And Challenge, Chapter 12.5, Problem 14PSB , additional homework tip 2

In right angle triangle,

a2+b2= c2

Calculation:

Geometry For Enjoyment And Challenge, Chapter 12.5, Problem 14PSB , additional homework tip 3

The smaller triangle is similar to the whole triangle by AA similarity rule, so

Corresponding sides of similar triangles are congruent.

xx+10=1212+8xx+10=122020x=12x+1208x=120x=1208=15

Radius of small circle:

Side BC can be calculated by applying Pythagoras Theorem.

In right angled triangle ABC , we get

(AC)2=(AB)2+(BC)2(15)2=(12)2+(BC)2225=144+(BC)2(BC)2=225−144BC=81BC=9r=BC=9

Radius of large circle:

Side ED can be calculated by applying Pythagoras Theorem.

In right angled triangle ADE , we get

(AE)2=(AD)2+(DE)2(25)2=(20)2+(DE)2625=400+(DE)2(DE)2=625−400DE=225DE=15r=DE=15

To determine

To calculate: The volume of smaller cone.

Expert Solution

Answer to Problem 14PSB

The volume of smaller cone is 1018 .

Explanation of Solution

Given information:

Height of small cone = 12,

Slant height of small cone = x ,

Height of large cone =12+8=20,

Slant height of large cone =x+10.

Formula used:

The below property is used:

Corresponding sides of similar triangles are congruent.

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

Geometry For Enjoyment And Challenge, Chapter 12.5, Problem 14PSB , additional homework tip 4

In right angle triangle,

a2+b2= c2

Volume of cone =13×B×h

B = Base area of cone and h = height of cone

Area of a circle: A=πr2

r = radius of circle

Calculation:

Geometry For Enjoyment And Challenge, Chapter 12.5, Problem 14PSB , additional homework tip 5

The smaller triangle is similar to the whole triangle by AA similarity rule, so

Corresponding sides of similar triangles are congruent.

xx+10=1212+8xx+10=122020x=12x+1208x=120x=1208=15

Radius of small circle:

Side BC can be calculated by applying Pythagoras Theorem.

In right angled triangle ABC , we get

(AC)2=(AB)2+(BC)2(15)2=(12)2+(BC)2225=144+(BC)2(BC)2=225−144BC=81BC=9r=BC=9

Volume of cone =13×B×h

B=πr2=π(9)2=81π

h=12

Volume of smaller cone =13×81π×12

Volume of smaller cone =324π

Volume of smaller cone ≈1018

To determine

To calculate: The volume of larger cone.

Expert Solution

Answer to Problem 14PSB

The volume of larger cone is 4712 .

Explanation of Solution

Given information:

Height of small cone = 12,

Slant height of small cone = x ,

Height of large cone =12+8=20,

Slant height of large cone =x+10.

Formula used:

The below property is used:

Corresponding sides of similar triangles are congruent.

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

Geometry For Enjoyment And Challenge, Chapter 12.5, Problem 14PSB , additional homework tip 6

In right angle triangle,

a2+b2= c2

Volume of cone =13×B×h

B = Base area of cone and h = height of cone

Area of a circle: A=πr2

r = radius of circle

Calculation:

Geometry For Enjoyment And Challenge, Chapter 12.5, Problem 14PSB , additional homework tip 7

The smaller triangle is similar to the whole triangle by AA similarity rule, so

Corresponding sides of similar triangles are congruent.

xx+10=1212+8xx+10=122020x=12x+1208x=120x=1208=15

Radius of large circle:

Side ED can be calculated by applying Pythagoras Theorem.

In right angled triangle ADE , we get

(AE)2=(AD)2+(DE)2(25)2=(20)2+(DE)2625=400+(DE)2(DE)2=625−400DE=225DE=15r=DE=15

Volume of cone =13×B×h

B=πr2=π(15)2=225π

h=20

Volume of larger cone =13×225π×20

Volume of larger cone =1500π

Volume of larger cone ≈4712

To determine

To calculate: The volume of the frustum.

Expert Solution

Answer to Problem 14PSB

The volume of the frustum is 3695 .

Explanation of Solution

Given information:

Height of small cone = 12,

Slant height of small cone = x ,

Height of large cone =12+8=20,

Slant height of large cone =x+10.

Formula used:

The below property is used:

Corresponding sides of similar triangles are congruent.

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

Geometry For Enjoyment And Challenge, Chapter 12.5, Problem 14PSB , additional homework tip 8

In right angle triangle,

a2+b2= c2

Volume of cone =13×B×h

B = Base area of cone and h = height of cone

Area of a circle: A=πr2

r = radius of circle

Calculation:

Geometry For Enjoyment And Challenge, Chapter 12.5, Problem 14PSB , additional homework tip 9

The smaller triangle is similar to the whole triangle by AA similarity rule, so

Corresponding sides of similar triangles are congruent.

xx+10=1212+8xx+10=122020x=12x+1208x=120x=1208=15

Radius of small circle:

Side BC can be calculated by applying Pythagoras Theorem.

In right angled triangle ABC , we get

(AC)2=(AB)2+(BC)2(15)2=(12)2+(BC)2225=144+(BC)2(BC)2=225−144BC=81BC=9r=BC=9

Volume of cone =13×B×h

B=πr2=π(9)2=81π

h=12

Volume of smaller cone =13×81π×12

Volume of smaller cone =324π

Volume of smaller cone ≈1018

Radius of large circle:

Side ED can be calculated by applying Pythagoras Theorem.

In right angled triangle ADE , we get

(AE)2=(AD)2+(DE)2(25)2=(20)2+(DE)2625=400+(DE)2(DE)2=625−400DE=225DE=15r=DE=15

Volume of cone =13×B×h

B=πr2=π(15)2=225π

h=20

Volume of larger cone =13×225π×20

Volume of larger cone =1500π

Volume of larger cone ≈4712

Volume of frustum = Volume of larger cone − Volume of smaller cone

Volume of frustum =1500π−324π

Volume of frustum =1176π

Volume of frustum ≈3695

Chapter 12.5, Problem 13PSB

Chapter 12.5, Problem 15PSB

Chapter 12 Solutions

Geometry For Enjoyment And Challenge

Ch. 12.1 - Prob. 1PSA Ch. 12.1 - Prob. 2PSA Ch. 12.1 - Prob. 3PSA Ch. 12.1 - Prob. 4PSA Ch. 12.1 - Prob. 5PSA Ch. 12.1 - Prob. 6PSB Ch. 12.1 - Prob. 7PSB Ch. 12.1 - Prob. 8PSB Ch. 12.1 - Prob. 9PSC Ch. 12.1 - Prob. 10PSC

Ch. 12.1 - Prob. 11PSC Ch. 12.2 - Prob. 1PSA Ch. 12.2 - Prob. 2PSA Ch. 12.2 - Prob. 3PSA Ch. 12.2 - Prob. 4PSA Ch. 12.2 - Prob. 5PSA Ch. 12.2 - Prob. 6PSB Ch. 12.2 - Prob. 7PSB Ch. 12.2 - Prob. 8PSB Ch. 12.2 - Prob. 9PSB Ch. 12.2 - Prob. 10PSC Ch. 12.2 - Prob. 11PSC Ch. 12.2 - Prob. 12PSC Ch. 12.2 - Prob. 13PSC Ch. 12.3 - Prob. 1PSA Ch. 12.3 - Prob. 2PSA Ch. 12.3 - Prob. 3PSA Ch. 12.3 - Prob. 4PSA Ch. 12.3 - Prob. 5PSA Ch. 12.3 - Prob. 6PSB Ch. 12.3 - Prob. 7PSB Ch. 12.3 - Prob. 8PSB Ch. 12.3 - Prob. 9PSB Ch. 12.3 - Prob. 10PSB Ch. 12.3 - Prob. 11PSB Ch. 12.3 - Prob. 12PSC Ch. 12.3 - Prob. 13PSC Ch. 12.3 - Prob. 14PSC Ch. 12.4 - Prob. 1PSA Ch. 12.4 - Prob. 2PSA Ch. 12.4 - Prob. 3PSA Ch. 12.4 - Prob. 4PSA Ch. 12.4 - Prob. 5PSA Ch. 12.4 - Prob. 6PSA Ch. 12.4 - Prob. 7PSB Ch. 12.4 - Prob. 8PSB Ch. 12.4 - Prob. 9PSB Ch. 12.4 - Prob. 10PSB Ch. 12.4 - Prob. 11PSB Ch. 12.4 - Prob. 12PSB Ch. 12.4 - Prob. 13PSB Ch. 12.4 - Prob. 14PSB Ch. 12.4 - Prob. 15PSB Ch. 12.4 - Prob. 16PSB Ch. 12.4 - Prob. 17PSB Ch. 12.4 - Prob. 18PSB Ch. 12.4 - Prob. 19PSC Ch. 12.4 - Prob. 20PSC Ch. 12.4 - Prob. 21PSC Ch. 12.4 - Prob. 22PSC Ch. 12.5 - Prob. 1PSA Ch. 12.5 - Prob. 2PSA Ch. 12.5 - Prob. 3PSA Ch. 12.5 - Prob. 4PSA Ch. 12.5 - Prob. 5PSA Ch. 12.5 - Prob. 6PSA Ch. 12.5 - Prob. 7PSA Ch. 12.5 - Prob. 8PSB Ch. 12.5 - Prob. 9PSB Ch. 12.5 - Prob. 10PSB Ch. 12.5 - Prob. 11PSB Ch. 12.5 - Prob. 12PSB Ch. 12.5 - Prob. 13PSB Ch. 12.5 - Prob. 14PSB Ch. 12.5 - Prob. 15PSB Ch. 12.5 - Prob. 16PSB Ch. 12.5 - Prob. 17PSC Ch. 12.5 - Prob. 18PSC Ch. 12.5 - Prob. 19PSC Ch. 12.5 - Prob. 20PSC Ch. 12.6 - Prob. 1PSA Ch. 12.6 - Prob. 2PSA Ch. 12.6 - Prob. 3PSA Ch. 12.6 - Prob. 4PSA Ch. 12.6 - Prob. 5PSA Ch. 12.6 - Prob. 6PSB Ch. 12.6 - Prob. 7PSB Ch. 12.6 - Prob. 8PSB Ch. 12.6 - Prob. 9PSB Ch. 12.6 - Prob. 10PSB Ch. 12.6 - Prob. 11PSB Ch. 12.6 - Prob. 12PSC Ch. 12.6 - Prob. 13PSC Ch. 12.6 - Prob. 14PSC Ch. 12.6 - Prob. 15PSC Ch. 12.6 - Prob. 16PSC Ch. 12.6 - Prob. 17PSD Ch. 12.6 - Prob. 18PSD Ch. 12 - Prob. 1RP Ch. 12 - Prob. 2RP Ch. 12 - Prob. 3RP Ch. 12 - Prob. 4RP Ch. 12 - Prob. 5RP Ch. 12 - Prob. 6RP Ch. 12 - Prob. 7RP Ch. 12 - Prob. 8RP Ch. 12 - Prob. 9RP Ch. 12 - Prob. 10RP Ch. 12 - Prob. 11RP Ch. 12 - Prob. 12RP Ch. 12 - Prob. 13RP Ch. 12 - Prob. 14RP Ch. 12 - Prob. 15RP Ch. 12 - Prob. 16RP Ch. 12 - Prob. 17RP Ch. 12 - Prob. 18RP Ch. 12 - Prob. 19RP Ch. 12 - Prob. 20RP Ch. 12 - Prob. 21RP Ch. 12 - Prob. 22RP Ch. 12 - Prob. 1CR Ch. 12 - Prob. 2CR Ch. 12 - Prob. 3CR Ch. 12 - Prob. 4CR Ch. 12 - Prob. 5CR Ch. 12 - Prob. 6CR Ch. 12 - Prob. 7CR Ch. 12 - Prob. 8CR Ch. 12 - Prob. 9CR Ch. 12 - Prob. 10CR Ch. 12 - Prob. 11CR Ch. 12 - Prob. 12CR Ch. 12 - Prob. 13CR Ch. 12 - Prob. 14CR Ch. 12 - Prob. 15CR Ch. 12 - Prob. 16CR Ch. 12 - Prob. 17CR Ch. 12 - Prob. 18CR Ch. 12 - Prob. 19CR Ch. 12 - Prob. 20CR Ch. 12 - Prob. 21CR Ch. 12 - Prob. 22CR Ch. 12 - Prob. 23CR Ch. 12 - Prob. 24CR Ch. 12 - Prob. 25CR Ch. 12 - Prob. 26CR Ch. 12 - Prob. 27CR Ch. 12 - Prob. 28CR Ch. 12 - Prob. 29CR Ch. 12 - Prob. 30CR Ch. 12 - Prob. 31CR Ch. 12 - Prob. 32CR Ch. 12 - Prob. 33CR Ch. 12 - Prob. 34CR Ch. 12 - Prob. 35CR Ch. 12 - Prob. 36CR Ch. 12 - Prob. 37CR Ch. 12 - Prob. 38CR Ch. 12 - Prob. 39CR Ch. 12 - Prob. 40CR Ch. 12 - Prob. 41CR Ch. 12 - Prob. 42CR

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