The side lengths of the base of a triangular prism are 5 meters, 8 meters, and 10 meters
1 answer:
The lateral surface area of the triangular prism is
Explanation:
The side lengths of the base of the triangular prism are 5 meters, 8 meters, and 10 meters.
It is given that the height of the prism is 16.5 meters.
To determine the lateral surface area of the prism, let us use the formula
where a, b,c are the side lengths of the base of the triangular prism and h is the height of the prism.
Here and
Substituting these values in the formula, we have,
Simplifying, we get,
Multiplying, we get,
Thus, the lateral surface area of the triangular prism is
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Solution:
x y ║ z w→→Given
Also, x z is a transversal, that intercepts x y and z w.
So, ∠ x z w=∠z x y→→Alternate interior angles as, x y ║ z w.
Also, v is point of intersection of x z and y w.
∠ x v y ≅ ∠ z v w→→[ Vertically opposite angles]
So,→→ Δ x y v ~ Δ z w v⇒⇒[Angle-Angle Similarity]
x f(X) g(x)
2.5 2 1.25
1.5 0 -2.75
2.75 2.5 2.56
2.8 2.6 2.84
2.77 2.54 2.67
2.73 2.46 2.45 <<<<<<< Answer.
I've included a graph to show you another way that you could get closer quicker.
The red line is y = 2x - 3
The blue line is y = x^2 - 5
2.5 2 1.25
1.5 0 -2.75
2.75 2.5 2.56
2.8 2.6 2.84
2.77 2.54 2.67
2.73 2.46 2.45 <<<<<<< Answer.
I've included a graph to show you another way that you could get closer quicker.
The red line is y = 2x - 3
The blue line is y = x^2 - 5
Refer to the figure shown below.
Because the maximum height of the parabola is 50 m, its equation is of the form
y = ax² + 50
This equation places the vertex at (0,50). The constant a should be negative for the vertex to be the maximum of y.
The base of the parabola is 10 m wide. Therefore the x-intercepts are (5,0) and (-5,0).
Set x=5 and y=0 to obtain
a(5²) + 50 = 0
25a = -50
a = -2
The equation of the parabola is
y = - 2x² + 50
At 2 m from the edge of the tunnel, x = 5 - 2 = 3 m.
Therefore the height of the tunnel (vertical clearance) at x = 3 m is
h = y(3)
= -2(3²) + 50
= - 18 + 50
= 32 m
Answer: 32 m
Because the maximum height of the parabola is 50 m, its equation is of the form
y = ax² + 50
This equation places the vertex at (0,50). The constant a should be negative for the vertex to be the maximum of y.
The base of the parabola is 10 m wide. Therefore the x-intercepts are (5,0) and (-5,0).
Set x=5 and y=0 to obtain
a(5²) + 50 = 0
25a = -50
a = -2
The equation of the parabola is
y = - 2x² + 50
At 2 m from the edge of the tunnel, x = 5 - 2 = 3 m.
Therefore the height of the tunnel (vertical clearance) at x = 3 m is
h = y(3)
= -2(3²) + 50
= - 18 + 50
= 32 m
Answer: 32 m