
Concept explainers
a.
To find the height of the larger cylinder.
a.

Answer to Problem 26PPS
Explanation of Solution
Given:
Two similar cylinders.
The height of the cylinders are in the ratio
Lateral surface area of the larger cylinder is
Diameter of the smaller cylinder is
Concept used:
Two
Formula used:
Lateral surface area of a cylinder
Calculation:
Since the two cylinders are similar, and the heights are in the ratio
So, the diameter of the larger cylinder is
Now, using this radius and the lateral surface area to find the height of the larger cylinder.
Lateral surface area of the larger cylinder
Conclusion:
Therefore, the height of the larger cylinder is
b.
To sketch and label the two cylinders.
b.

Explanation of Solution
Given:
Two similar cylinders.
The height of the cylinders are in the ratio
Lateral surface area of the larger cylinder is
Diameter of the smaller cylinder is
Diameter of large cylinder is
Height of large cylinder is
Height of small cylinder is
Concept used:
Two solids are similar if they are the same type of solid and their corresponding radii and heights are proportional.
Calculation:
c.
To find the ratio of volumes of both cylinders.
c.

Answer to Problem 26PPS
Explanation of Solution
Given:
Two similar cylinders.
The height of the cylinders are in the ratio
Lateral surface area of the larger cylinder is
Diameter of the smaller cylinder is
Formula used:
Volume of a cylinder
Calculation:
First find the height of the smaller cylinder.
Since the two cylinders are similar, and the heights are in the ratio
Simplify the ratio of the volume of the larger cylinder to the volume of the smaller cylinder.
Conclusion:
Therefore, the volume of the larger cylinder is
Chapter 12 Solutions
Geometry, Student Edition
Additional Math Textbook Solutions
A Problem Solving Approach To Mathematics For Elementary School Teachers (13th Edition)
University Calculus: Early Transcendentals (4th Edition)
Algebra and Trigonometry (6th Edition)
Elementary Statistics (13th Edition)
- find the value of each variablearrow_forwardName: Date: Bell: Unit 11: Volume & Surface Area Homework 2: Area of Sectors Directions: Find the area of each shaded sector. Round to the hundredths place. 1. GH 11 in 2. KL 20 ft H F 64 G L 119 M K 3. BA 6.5 cm 4. YZ 14.2 m B 23 X 87° Y Z 5. KL = 27.1 mm J 32 L X:360-32.1 K A-3 360 7. BD 18 cm E 136 B X=32.8 127.0 (271) A: 069.13 Amm² 19=2102.13 A-136 360.16912 A:300cm² A=96.13 6. PQ = 2.8 in P R 311° 8. WZ 5.3 km V = Z 108 W D 9. HK = 25 ft G H KO 26 X 10. SR 26 m = S 73 T R Gina Wilson (All Things Algebarrow_forward538 Chapter 13 12. Given: Points E(-4, 1), F(2, 3), G(4, 9), and H(-2, 7) a. Show that EFGH is a rhombus. b. Use slopes to verify that the diagonals are perpendicular. 13. Given: Points R(-4, 5), S(-1, 9), T(7, 3) and U(4, -1) a. Show that RSTU is a rectangle. b. Use the distance formula to verify that the diagonals are congruent. 14. Given: Points N(-1, -5), O(0, 0), P(3, 2), and 2(8, 1) a. Show that NOPQ is an isosceles trapezoid. b. Show that the diagonals are congruent. Decide what special type of quadrilateral HIJK is. Then prove that your answer is correct. 15. H(0, 0) 16. H(0, 1) 17. H(7, 5) 18. H(-3, -3) I(5, 0) I(2,-3) 1(8, 3) I(-5, -6) J(7, 9) K(1, 9) J(-2, -1) K(-4, 3) J(0, -1) K(-1, 1) J(4, -5) K(6,-2) 19. Point N(3, - 4) lies on the circle x² + y² = 25. What is the slope of the (Hint: Recall Theorem 9-1.) - line that is tangent to the circle at N? 20. Point P(6, 7) lies on the circle (x + 2)² + (y − 1)² = 100. What is the slope of the line that is tangent to the circle at…arrow_forward
- Can you cut the 12 glass triangles from a sheet of glass that is 4 feet by 8 feet? If so, how can it be done?arrow_forwardCan you cut 12 glass triangles from a sheet of glass that is 4 feet by 8 feet? If so, draw a diagram of how it can be done.arrow_forwardIn triangle with sides of lengths a, b and c the angle a lays opposite to a. Prove the following inequality sin a 2√bc C α b a Warrow_forward
- Find the values of x, y, and z. Round to the nearest tenth, if necessary. 8, 23arrow_forward11 In the Pharlemina's Favorite quilt pattern below, vega-pxe-frame describe a motion that will take part (a) green to part (b) blue. Part (a) Part (b)arrow_forward5. 156 m/WXY = 59° 63 E 7. B E 101 C mFE = 6. 68° 8. C 17arrow_forward
- 1/6/25, 3:55 PM Question: 14 Similar right triangles EFG and HIJ are shown. re of 120 √65 adjacent E hypotenuse adjaca H hypotenuse Item Bank | DnA Er:nollesup .es/prist Sisupe ed 12um jerit out i al F 4 G I oppe J 18009 90 ODPO ysma brs & eaus ps sd jon yem What is the value of tan J? ed on yem O broppo 4 ○ A. √65 Qx oppoEF Adj art saused taupe ed for yem 4 ○ B. √65 29 asipnisht riod 916 zelprisht rad √65 4 O ○ C. 4 √65 O D. VIS 9 OD elimiz 916 aelonsider saused supsarrow_forwardFind all anglesarrow_forwardFind U V . 10 U V T 64° Write your answer as an integer or as a decimal rounded to the nearest tenth. U V = Entregararrow_forward
- Elementary Geometry For College Students, 7eGeometryISBN:9781337614085Author:Alexander, Daniel C.; Koeberlein, Geralyn M.Publisher:Cengage,Elementary Geometry for College StudentsGeometryISBN:9781285195698Author:Daniel C. Alexander, Geralyn M. KoeberleinPublisher:Cengage Learning

