Answer:
9. [tex] A = 4608~in.^2 [/tex]
10. [tex] V = \sqrt{5}~ft^3 [/tex]
11. (a) [tex] 2\sqrt{6}~ft [/tex]
11. (b) [tex] perimeter = (6 + 2\sqrt{6})~ft [/tex]
11. (c) [tex] area = 3~ft^2 [/tex]
Step-by-step explanation:
9. Area of square:
[tex] A = s^2 [/tex]
[tex] A = (48\sqrt{2}~in.)^2 [/tex]
[tex] A = 48^2 \times (\sqrt{2})^2~in.^2 [/tex]
[tex] A = 2304 \times 2~in.^2 [/tex]
[tex] A = 4608~in.^2 [/tex]
10. Volume of prism:
V = LWH
[tex] V = LWH [/tex]
[tex] V = \sqrt{5} \times (2 + \sqrt{3}) \times (2 - \sqrt{3})~ft^3 [/tex]
The last two factors are a sum and a difference, so I will multiply them first. The product of a sum and a difference is the difference of two squares.
[tex] V = \sqrt{5} \times (4 - 3)~ft^3 [/tex]
[tex] V = \sqrt{5} \times 1~ft^3 [/tex]
[tex] V = \sqrt{5}~ft^3 [/tex]
11.
(a) Use the Pythagorean theorem.
[tex] a^2 + b^2 = c^2 [/tex]
[tex] c^2 = a^2 + b^2 [/tex]
[tex] c^2 = (3 + \sqrt{3})^2 + (3 - \sqrt{3})^2 [/tex]
[tex] c^2 = 9 + 6\sqrt{3} + 3 + 9 - 6\sqrt{3} + 3 [/tex]
[tex] c^2 = 9 + 3 + 9 + 3 + 6\sqrt{3} - 6\sqrt{3} [/tex]
[tex] c^2 = 24 [/tex]
[tex] c = \sqrt{24} [/tex]
[tex] c = \sqrt{4 \times 6} [/tex]
[tex] c = 2\sqrt{6}~ft [/tex]
(b) perimeter = sum of lengths of three sides
[tex] perimeter = 3 + \sqrt{3} + 3 - \sqrt{3} + 2\sqrt{6} [/tex]
[tex] perimeter = (6 + 2\sqrt{6})~ft [/tex]
(c) area = base * height/2
The base and height are the legs.
[tex] area = \dfrac{bh}{2} [/tex]
[tex] area = \dfrac{(3 + \sqrt{3})(3 - \sqrt{3})}{2} [/tex]
[tex] area = \dfrac{9 - 3}{2} [/tex]
[tex] area = \dfrac{6}{2} [/tex]
[tex] area = 3~ft^2 [/tex]
