Question

A triangle with area 45 square inches has a height that is two less than four times the width. Find the width and height of the triangle.

Answer 1

Answer:

Step-by-step explanation:

45 = (b(4b - 2))/2

90 = 4b2 - 2b

4b2 - 2b - 90 = 0

(2b + 9)(2b - 10) = 0

 

2b + 9 = 0  

2b = -9

b = -9/2

If we were looking for zeroes, this would be an answer, but since we're looking for measurement, we don't use negative numbers.

90 = 4b2 - 2b

4b2 - 2b - 90 = 0

(2b + 9)(2b - 10) = 0

 

2b + 9 = 0  

2b = -9

b = -9/2

Answer 2

Answer:

The width is 5 inches and the height is 18 inches.

Step-by-step explanation:

The area of a triangle can be calculated as:

A = w*h/2

where w is the width of the base, and h is the height.

then we have that:

w*h/2 = 45 in^2

and the height is 2 less than 4 times the width.

h = 4*w - 2in

then we have a system with 2 equations:

w*h/2 = 45 in^2

h = 4*w - 2in

We can replace the second one in the first one and solve it for w.

w*(4*2 - 2in)/2 = 45 in^2

2w^2 - w = 45in^2

2w^2 - w - 45in = 0

then we have to solve the quadratic equation:

the solutions are:

w = (1 +- √(1^2 - 4*2*(-45))/(2*2) = (1 + - √361)/4

where we have two solutions for w, one for each sign of the square root.

we obviously need to take the positive solution (because we can not have a negative length)

w = (1 + √361)/4 = 5in

now, we can replace it in the equation "h = 4*w - 2in" to get the height.

h = 4*5in - 2in = 18in