# Math Help - [SOLVED] Finding areas

1. ## [SOLVED] Finding areas

Hi I need some more help with word problems. I don't know if I can get some help because there are drawings that go with it but I'll explain them as best as I can. Or maybe just some ideas on how to plug it into an equation.

1. Length and area. Find the length y in the figure. The area of the shaded region is given. both pictures are completely shaded. There is an attached picture.

2. Hello, Lucy!

Are you sure that you couldn't figure these out yourself?

Find the length $y$ in the figure.
Code:
                  * - - - - - *
* |           |
*   |           |
*     |           | y    Area: 120 in²
*       |           |
*         |           |
* - - - - - * - - - - - *
y           y

The triangle has area: . $\tfrac{1}{2}y^2$

The square has area: . $y^2$

The total area is 120 in²: . $\tfrac{1}{2}y^2 + y^2 \:=\:120 \quad\Rightarrow\quad\tfrac{3}{2}y^2 \:=\:120 \quad\Rightarrow\quad y^2 \:=\:80$

Therefore: . $y \:=\:\sqrt{80} \:=\:4\sqrt{5}\text{ in.}$

Code:
                  * - - *
* |     |
*   |     |
*     |     | y    Area: 1200 cm²
*       |     |
*         |     |
* - - - - - * - - *
y        1

The triangle has area: . $\tfrac{1}{2}y^2$

The rectangle has area: . $y\cdot1 \:=\:y$

The total area is 1200 cm²: . $\tfrac{1}{2}y^2 + y \:=\:1200 \quad\Rightarrow\quad y^2 + 2y - 2400 \:=\:0$

. . which factors: . $(y-48)(y+50) \:=\:0$

. . and has the positive root: . $y \:=\:48$

3. ## questions and thanks yous for word problem help

Thank you so much for your help. I am extremely bad at word problems I have been working on these for the past three days and nothing came to me. I am also a little confused as to where the 2y and 2400 came from in B. Thank you again.