More Practice on Chapter 7

Today in math class we did different kinds of questions that will be on the test.




First we were given a question to find the area of rectangle with a missing piece using polynomials.





There are two ways to figuring out the area of this shape. First way is to break up the shape into 3 pieces, and add Areas 1 and 2, and subtract area 3.







The second way is to pretend the missing piece is there (area 1) so it makes a whole rectangle, then subtract area 2 (missing piece) to area 1. I chose to do this method because it seemed like a faster way for me.



Show your work:
l = 8x - 4 + 4x + 9 = 12x +5
w = 6x + 2 + 3x - 5 = 9x - 3

A1 = lw
= (12x + 5)(9x -3)
= (12x)(9x) + (12x)(-3) + (5)(9x) + (5)(-3)
= 108x^ - 36x + 45x - 15

A2 = lw
= (8x - 4)(3x -5)
= (8x)(3x) + (8x)(-5) + (-4)(3x) + (-4)(-5)
= 24x^ - 40x -12x + 20
= 24x^ - 52x + 20

A1 - A2
= 108x^ + 9x -15 - (24x^ - 52x + 20)
= 108x^ +9x - 15 -24x^ +52x - 20
= 84x^ + 61x -35

Therefore, the area of the shape is 84x^ + 61x - 35.


The second question was a triangle. This was much simpler because it didn't contain a missing piece.

A = bh/2
= (8x +4)(5x)/2
= (8x)(5x) + (4)(5x)
= 40x^ + 20x/2
= 40x^/2 + 20x/2
= 20x^ + 10x

The area of the triangle is 20x^ + 10x.

The third question was to find the width of the rectangular prism.

V = lwh
48x³ = (2x)(w)(3x)
48x³ = 6x^
w = 48x³/6x^
w = 8x


The width of the rectangular prism is 8x.




The fourth question is similar to the first one, except with different values. Again, I used the A1-A2 method.





Step 1







Step 2






Step 3






The area of the shape is 26x^ - 35x - 12.


The fifth question is finding the ratio of the rectangle to the circle. This won't be on the test but it's better to know it now for future references.

r = d/2
r = 6x/2
r = 3x

rectangle / circle
= lw / pi r^
= 6x(12x) / pi(3x)^
= 72x^/pi 9x^
= 8 / pi



The last question was to find the ratio of the small circle to the large circle.

So/Lo = pi r^/pi r^
= pi(2x)^/pi(4x)^
= 4x^/16x^
= 4/16
= 1/4






Solve:
1. 3(5x + 3) - (10x - 6)
= 15x +9 - 10x + 6
= 5x + 15

2. (1/2t)^ 3t
= (1/2t)(1/2t)(3t)
= 1/4t^(3t)
= 3/4t³


Remember:
Always use the FOIL(First, Outside, Inside, Last) method when multiplying
Only add/subtract like terms
When subtracting polynomials, (-) before the 2nd polynomial means to multiply (-1) to each term. (Change each sign to its opposite)
Study, go on Mangahigh, and practice!

The next person to do the scribe will be Jocelle Garcia!

Marie's Blog Question

The question I was given for 2.4 was #21:

"A baseball diamond is a square area of about 750 m². What is the distance from first base to second base? Give your answer to the nearest tenth of a metre."






I knew that the distance between first base to second base is also equal to the side length of the baseball diamond. Since 750 m² is the area, all I needed to do was find the square root of 750m²















For the answer, I got 27.3861279 m.

I had to round it to the nearest tenth, and since .08 is greater than .05, I changed 27.3 to 27.4 m.

The distance from first base to second base is 27.4 m.

Surface Area

Quiz Answers (Review)

Area of rectangle: lh or lw

Area of circle: πr²

Area of triangle: bh/2
Area of square: s²
Circumference: πd or 2πr
Radius: d/2
Diameter: 2r or c/π
Perimeter of triangle: s + s + s
Perimeter of quadrilateral: s + s + s + s
Perimeter of any n-gon: Add all side lengths
_________________________________________


Parallelogram:





Formula:
a = bh

Trapezoid/Isosceles Trapezoid

Formula:
a = ( b1 + b2/2 )h

Surface Area of nets

Square:








First way:
SA = 6s²
= 6(2²)
= 6(4)
= 24u²

Second way:
SA = 2(s²) + (Pob)h
= 2(2²) + 8(2)
= 2(4) + 16
= 8 + 16
= 24u²

Rectangle:










First way:
SA = 2(s²) + 4(lw)
SA = 2(2²) + 4(3x2)
SA = 2(4) + 4(6)
SA = 8 + 24
SA = 32u²

Second way:
SA = 2(lw) + (Pob)h
SA = 2(3x2) + 10(2)
SA = 2(6) + 20
SA = 12 + 20
SA = 32u²

Homework:
Identify 2 ways to find the surface area of this net.