What is the surface area

Mathematics

1 answer:

anygoal [31]2 years ago

3 0

Bear in mind that the area of a "parallelogram" is base * height.

notice the picture, the front and back parallelograms, have an altitude/height of 3 cm, and a base of 8 cm, so the area of one of those two is 3*8.

as far as the left and right parallelograms, they have a base of 7 cm, but the altitude is also 3 cm, so the area of one of those is 3 *7.

bear in mind the altitude/height is the distance from the base to the top, and in this case is just 3 cm.

anyway, you have 4 parallelograms, get the area of each, add them up, and that's the surface area of the figure.

2(3*8) + 2(3*7).

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Olin [163]

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$\frac{\sin \:4A}{\cos \:2A} \times \frac{1 - \cos \:2A}{1 - \cos\:4A} = \tan \:A \\ \\ LHS = \frac{\sin \:4A}{\cos \:2A} \times \frac{1 - \cos \:2A}{1 - \cos\:4A} \\ \\ = \frac{2\sin \:2A.\cos \:2A}{\cos \:2A} \times \frac{1 - (2 { \cos}^{2}A - 1) }{1 - (2 { \cos}^{2}2A - 1) } \\ \\ = 2\sin \:2A \times \frac{1 - 2 { \cos}^{2}A + 1}{1 - 2 { \cos}^{2}2A + 1 } \\ \\ = 2\sin \:2A \times \frac{2- 2 { \cos}^{2}A }{2 - 2 { \cos}^{2}2A } \\ \\ = 2\sin \:2A \times \frac{2(1 - { \cos}^{2}A) }{2 (1- { \cos}^{2}2A) } \\ \\ = 2\sin \:2A \times \frac{1 - { \cos}^{2}A}{1- { \cos}^{2}2A } \\ \\ = 2\sin \:2A \times \frac{ { \sin}^{2}A}{{ \sin}^{2}2A } \\ \\ = 2 \times \frac{ { \sin}^{2}A}{{ \sin}2A } \\ \\ = 2 \times \frac{ { \sin}^{2}A}{{ 2\sin}A. \cos \: A } \\ \\ = \frac{ { \sin}A}{ \cos \: A } \\ \\ = tan \: A \\ \\ = RHS \\$