YOU MUST SHOW WORK!
1. A banner is in the shape of a right triangle of area 63in^2 The height of the banner is 4in less than twice the width of the banner. Find the height and width of the banner.
2. A rectangular poster has an area of 190in^2 The height of the poster is 1in less than twice it's width. Find the dimensions of the poster.
1. A banner is in the shape of a right triangle of area 63in^2 The height of the banner is 4in less than twice the width of the banner. Find the height and width of the banner.
2. A rectangular poster has an area of 190in^2 The height of the poster is 1in less than twice it's width. Find the dimensions of the poster.
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1. Area = 63 in^2
h = 2w-4
you know the equation for area of right triangle is 1/2 base * height
(base meaning width)
So now lets plug in our value of height
1/2 * (2w-4) * w = 63
lets distribute:
(w-2)*w = 63
distribute again:
w^2-2w = 63
So when we solve for W we get w = 9 in
lets check
9^2 - 9*2 = 63
81-18 = 63 Correct.
2. Area = 190 in ^2
area of a rectangle is length x width (height x with in this case)
the height = 2 times the width - 1
so h = 2w-1
now plug in:
area = l * w ====> area = w * (2w-1)
Now distribute:
2w^2 - w = 190
In this case w = 10
Lets check:
2 * 10^2 - 10 = 190
2*100 - 10 = 190
200-10 = 190 CORRECT
h = 2w-4
you know the equation for area of right triangle is 1/2 base * height
(base meaning width)
So now lets plug in our value of height
1/2 * (2w-4) * w = 63
lets distribute:
(w-2)*w = 63
distribute again:
w^2-2w = 63
So when we solve for W we get w = 9 in
lets check
9^2 - 9*2 = 63
81-18 = 63 Correct.
2. Area = 190 in ^2
area of a rectangle is length x width (height x with in this case)
the height = 2 times the width - 1
so h = 2w-1
now plug in:
area = l * w ====> area = w * (2w-1)
Now distribute:
2w^2 - w = 190
In this case w = 10
Lets check:
2 * 10^2 - 10 = 190
2*100 - 10 = 190
200-10 = 190 CORRECT