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Answer:
a. The expression for the area of the gym is 15·x
b. The expression for the area of the cafeteria is 15·x + 45
c. The expression for the total area of the gym and the cafeteria is 30·x + 45
d. The expression for the total perimeter of the gym and the cafeteria is 36 + 4·x
Step-by-step explanation:
The parameters of the given figures are;
The width of the figure, w = 15
The breadth of the gym, = x
The breadth of the cafeteria, = x + 3
The shape of the area of the gym = A rectangle
The shape of the area of the cafeteria = A rectangle
a. The expression for the area, 'A₁', of the gym is given by the area of a rectangle which is Width × Breadth
The area of the gym, A₁ = The width of the figure × The breadth of the gym
Therefore, we have; The area of the gym, A₁ = w ×
A₁ = 15 × x
The area of the gym, A₁ = 15·x
b. The area of the cafeteria, A₂ = The width of the figure × The breadth of the cafeteria
The area of the cafeteria, A₂ = w ×
∴ A₂ = 15 × (x + 3) = 15·x + 45
The expression for the area of the cafeteria, A₂ = 15·x + 45
c. The expression for the total area of the gym and the cafeteria, 'A', is given as follows;
A = A₁ + A₂
∴ A = 15·x + 15·x + 45 = 30·x + 45
The expression for the total area of the gym and the cafeteria = 30·x + 45
d. The expression for the total perimeter of the gym and the cafeteria, 'P', is given as follows;
P = The perimeter of the rectangular area of the gym plus the area of the cafeteria
∴ P = 2 × 15 + 2 × (x + x + 3)
P = 30 + 4·x + 6
P = 36 + 4·x
The expression for the total perimeter of the gym and the cafeteria P = 36 + 4·x.
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You have the hypotenuse, so now you need to figure out where the angle is in relation to x and y (one at a time though).
45 degrees is opposite from x, so you would use sin (SOH) because you have the opposite and hypotenuse. This looks like:
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