# Math Help - Length and area #3

1. ## Length and area #3

2. A circle is inscribed in square. One point of a circle is 1 and 2 away from the two closest sides of the square. Find the area of the square.

3. A certain city block is in the form of a parallelogram. Two of it's sides measure 32ft and 41ft. If the area of the land in the block is 656ft^2, what is the length if it's longer diagonal?

Thanks for helping!

2. Hello, ejaykasai!

2. A circle is inscribed in square. One point on the circle is 1 unit and 2 units away
from the two closest sides of the square. Find the area of the square.
Code:
O         r
- * - - - - - - - * -
| *             *
r-2 |   * r         | r-2
|     *        *|
|       *       |
|         * A * |
B + - - - - - * - + -
|    r-1    :   |
2 |        *  :   | 2
- * -*- - - - + - * -
:    r-1    : 1 :

The circle has center $\,O$ and radius $\,r.$
Point $\,A$ is 1 unit and 2 units from the circumscribed square.

In right triangle $OBA\!:\;\;(r-1)^2 + (r-2)^2 \:=\:r^2$

. . $r^2 - 6r + 5 \:=\:0 \quad\Rightarrow\quad (r-5)(r-1) \:=\:0$

Hence: . $r = 5,\;\rlap{/////}r = 1$

The side of the square is 10 units.

Therefore, the area of the square is: . $10^2 = 100$ units $^2.$

3. Hello again, ejaykasai!

3. A certain city block is in the form of a parallelogram.
Two of it's sides measure 32 ft and 41 ft.
If the area of the block is 656 ft^2, what is the length if its longer diagonal?
Code:
A *--------------------* D
/                    /|
/                    / |
32 /                    /  |
/                  32/   | h
/                    /    |
/                    /     |
*--------------------*- - - *
B - - - - 41 - - - - C  - - E

We have parallelogram $ABCD\!:\;\;AB = 32,\:BC = 41.$
Let the height be: $h \:=\:DE.$

Since $\text{(Area)} \:=\:\text{(base)} \times \text{(height)}$
. . we have: . $41h \:=\:656 \quad\Rightarrow\quad h \:=\:16$

Code:
A *--------------------* D
/                 *  /|
/              *     / |
32 /           *        /  |
/        *         32/   | 16
/     *              /    |
/  *                 /     |
*--------------------*- - - *
B - - - - 41 - - - - C  - - E

In right triangle $DEC\!:\;CE^2 + 16^2 \:=\:32^2 \quad\Rightarrow\quad CE^2 = 768$

Hence: . $CE = 16\sqrt{3} \quad\Rightarrow\quad BE \:=\:41 + 16\sqrt{3}$

In right triangle $DEB\!:\;\;BD^2 \:=\:(41+16\sqrt{3})^2 + (16)^2$

. . $BD^2 \;=\;1681 + 1312\sqrt{3} + 768 + 256 \;=\;2705 + 1312\sqrt{3}$

Therefore: . $BD \;=\;\sqrt{2705 + 1312\sqrt{3}} \;\approx\;70.55\text{ ft}$