Question

David has a rectangle and a right triangle.

Answer 1

Answer:

2w^2+10w

Step-by-step explanation:

Let's start with the rectangle.

The length of the rectangle is:

L = 5 + w

The width of the rectangle is:

W

So, the equation of the area for the rectangle would be:

(w)(5+w) due to the area of a rectangle being length x width

Now, onto the triangle.

The length of the shorter side is equal to the width of the rectangle:

L(1) = w

The length of the longer side of the triangle is twice the length of the rectangle, so going back to what we have the rectangles length being:

L(2) = (2)(5+w) or 10 + 2w

Now to find the area of the triangle, use the equation 1/2(b)(height)

So plug it in:

1/2(w)(10 + 2w)

Which equals 1/2(10w + 2w^2)

Now, we times it by 1/2, so we will put it into a fraction:

1 x (10w + 2w^2)/2

10w + 2w^2/2

Now factor it:

2w(5 + w)/2

2 cancels out 2, so we are left with

w(w+5) = w^2 + 5w

So, the equation says what is the combined area of the rectangle and triangle

Rectangle area: w(w+5) = w^2 + 5w

Triangle area: w^2 + 5w

So, the combined area is 2w^2 + 10w

hope that helps!! I understand this question is hard, so I tried to explain as best as I can :)

for me, this is an EOC question, so to all those taking the test good luck! study hard!!