To name the parallelogram having the greatest area for a given perimeter.
Answer to Problem 30PSC
The parallelogram having the greatest area for a given perimeter is square.
Explanation of Solution
Given information:
A parallelogram having greatest area and certain perimeter.
The parallelogram having the greatest area for a given perimeter is square because the length and width are the same.
Area of square is given by product of length and width.
In square, all sides are equal.
The magnitude of area will be larger because both values of length and width are same.
In case of other parallelogram, the length and width are different.This will result into area of certain value which is not as great as square.
Chapter 11 Solutions
Geometry For Enjoyment And Challenge
Additional Math Textbook Solutions
A Problem Solving Approach To Mathematics For Elementary School Teachers (13th Edition)
Elementary Statistics
Elementary Statistics (13th Edition)
Algebra and Trigonometry (6th Edition)
Calculus for Business, Economics, Life Sciences, and Social Sciences (14th Edition)
Elementary Statistics: Picturing the World (7th Edition)
- In the graph below triangle I'J'K' is the image of triangle UK after a dilation. 104Y 9 CO 8 7 6 5 I 4 3 2 J -10 -9 -8 -7 -6 -5 -4 -3 -21 1 2 3 4 5 6 7 8 9 10 2 K -3 -4 K' 5 -6 What is the center of dilation? (0.0) (-5. 2) (-8. 11 (9.-3) 6- 10arrow_forwardSelect all that apply. 104 8 6 4 2 U U' -10 -8 -6 4 -2 2 4 6 10 -2 V' W' -4 -6 -8 -10 W V Select 2 correct answerts! The side lengths are equal in measure. The scale factor is 1/5. The figure has been enlarged in size. The center of dilation is (0.0) 8 10 Xarrow_forwardIn the graph below triangle I'J'K' is the image of triangle UK after a dilation. 104Y 9 CO 8 7 6 5 I 4 3 2 J -10 -9 -8 -7 -6 -5 -4 -3 -21 1 2 3 4 5 6 7 8 9 10 2 K -3 -4 K' 5 -6 What is the center of dilation? (0.0) (-5. 2) (-8. 11 (9.-3) 6- 10arrow_forward
- Qll consider the problem -abu+bou+cu=f., u=0 ondor I prove atu, ul conts. @ if Blu,v) = (b. 14, U) + ((4,0) prove that B244) = ((c- — ob)4;4) ③if c±vbo prove that acuius v. elliptic.arrow_forwardQ3: Define the linear functional J: H₁(2) R by ¡(v) = a(v, v) - L(v) Л Let u be the unique weak solution to a(u,v) = L(v) in H(2) and suppose that a(...) is a symmetric bilinear form on H(2) prove that 1- u is minimizer. 2- u is unique. 3- The minimizer J(u) can be rewritten under 1(u) = u Au-ub, algebraic form 1 2 Where A, b are repictively the stiffence matrix and the load vector Q4: A) Answer 1- show that the solution to -Au = f in A, u = 0 on a satisfies the stability Vullfll and show that ||V(u u)||||||2 - ||vu||2 2- Prove that Where lu-ul Chuz - !ull = a(u, u) = Vu. Vu dx + fu. uds B) Consider the bilinea forta Л a(u, v) = (Au, Av) (Vu, Vv + (Vu, v) + (u,v) Show that a(u, v) continues and V- elliptic on H(2)arrow_forward7) In the diagram below of quadrilateral ABCD, E and F are points on AB and CD respectively, BE=DF, and AE = CF. Which conclusion can be proven? A 1) ED = FB 2) AB CD 3) ZA = ZC 4) ZAED/CFB E B D 0arrow_forward
- 1) In parallelogram EFGH, diagonals EG and FH intersect at point I such that EI = 2x - 2 and EG = 3x + 11. Which of the following is the length of GH? a) 15 b) 28 c) 32 d) 56arrow_forward5) Which of the following are properties of all squares: 1. Congruent diagonals 2. Perpendicular diagonals 3. Diagonals that bisect vertex angles a) 1 and 2 only b) 1 and 3 only c) 2 and 3 only d) 1, 2, and 3arrow_forward6) In an isosceles trapezoid HIJK it is known that IJ || KH. Which of the following must also be true? a) IJ = KH b) HIJK c) HIJK d) IJ KHarrow_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