# Thread: Help With A Grid

1. ## Help With A Grid

2. Originally Posted by dgenerationx2
Looks like this to me:

Rectangular portion below first roof angle is 5 X 8 minus the doors (6) = 34 squares

From first roof angle to second roof angle we have a trapezoid with bases 8 and 4 with height 1.

Using $A=\frac{1}{2}h(b_1+b_2)$, we can find the area of the trapezoid to be 6 squares

The remaining portion is a triangle with base 4 and height 4. Using $A=\frac{1}{2}bh$, we find the area of the triangle to be 8 squares.

That makes a total of 48 squares X 9 square feet per square = 432 square ft.

3. Hello, dgenerationx2!

I must assume you are familiar with basic area formulas:
. . area of a triangle, area of a rectangle.
Then exactly where is your difficulty?

The front of the building below need to be repainted.

How many square feet will need to be repainted, excluding the door?
Code:
      -                 *
:                *|*
:               */|/*
:              *//|//*
1             *///|///*
:            *////|12//*
:           */////|/////*
:          *//////|//////*
-       . *-------*-------* .
2   . *///////////|3//////////* .
- *-------*-------*-------*-------*
: |///6///////6///////6///////6///|
: |///////////////////////////////|
: |///////////////////////////////|
3 |///////////////////////////////| 15
: |///////////* - - - *///////////|
: |///////////|   6   |///////////|
: |///////////|9      |///////////|
: |///////////|       |///////////|
- *-----------*-------*-----------*
: - - - - - -  24 - - - - - - - :

Level 1: An isosceles triangle with base 12 and height 12.
. . Its area is: . $\tfrac{1}{2}(12)(12) \:=\:72$ ft².

Level 2: A trapezoid with bases 12 and 24, and height 3.
. . Its area is: . $\tfrac{3}{2}(12 + 24) \:=\:54$ ft².

Level 3: A 24×15 rectangle minus a 6×9 rectangle.
. . Its area is: . $(24\cdot15) - (6\cdot9) \:=\:360 - 54 \:=\:306$ ft².

The total area is: . $72 + 54 + 306 \;=\;{\color{blue}432\text{ ft}^2}$

Ha! . . . masters beat me to it!
.