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Arlecino [84]
3 years ago
8

In triangle ABC above, AD is perpendicular to CB. If the lengths of AD and CB were both increased by 50%, how would the area of

ABC change?
Mathematics
1 answer:
astra-53 [7]3 years ago
3 0

Answer:

The answer is below

Step-by-step explanation:

A triangle is a polygon with three sides and three angles. There are different types of triangles such as scalene triangle, equilateral triangle, isosceles triangle.

The area of a triangle is given as:

A = (1/2) * b * h

where b = base of triangle, h = height of triangle , A = area

Let BC = base = x, and AD = height = y, hence:

A = (1/2) * x * y = 0.5xy

If the lengths of AD and CB were both increased by 50%, hence:

new AD = y + 0.5y = 1.5y, new CB = x + 0.5x = 1.5x

The new area= (1/2) * 1.5y * 1.5x = 1.125xy

Increased area / area = 1.125xy / 0.5xy = 2.25

If the lengths are increased by 50%, the area would also increase and the new area would be 2.25 times

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your question is incomplete, here you can find the complete question below:

An angles vertex is located at (0,0) and the initial ray of the angle points in the 3 oclock direction. Let theta represent the varying radian measure of the angle.

a. if the terminal ray of the angle passes through the point(3.38,2.32), whta is the slope of the terminal ray of the angle?

b. if the terminal ray of the angle passes through the point(1.49,3.82), whta is the slope of the terminal ray of the angle?

c. if the terminal ray of the angle passes through the point(1.1,3.95), what is the slope of the terminal ray of the angle?

Answer:

a. 1.45

b. 2.56

c. 3.59

Step-by-step explanation:

As we know that formula to find the slope is change in y-coordinates divided by change in x-coordinates.

slope=\frac{chnage\ in\ y}{chnage\ in\ x}=\frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

It is given that initial point is (x1,y1)=(0,0)

Part(a)

initial point is (x1,y1)=(0,0)

terminal point is (x2,y2)=(3.38,2.32)

so the slope of the terminal ray is

slope=\frac{3.38-0}{2.32-0}=\frac{3.38}{2.32}=  1.45

Part(b)

initial point is (x1,y1)=(0,0)

terminal point is (x2,y2)=(1.49,3.82)

so the slope of the terminal ray is

slope=\frac{3.82-0}{1.49-0}=\frac{3.82}{1.49}=2.56

Part(c)

initial point is (x1,y1)=(0,0)

terminal point is (x2,y2)=(1.1,3.95)

so the slope of the terminal ray is

slpoe=\frac{3.95-0}{1.1-0}=\frac{3.95}{1.1}=  3.59

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3 years ago
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