Julibella's Blog Post

Review:
How do we prove the triangles are the same?


-if the angles are the same
AAA (angle angle angle law)

-if the sizes are the same
ASA (angle side angle law)

-if the sides and angles are the same.
SAS (side angle side law)

-if the side and side and side equal
SSS (side side side law)

2/4 = 1/2 = 3/6 = 0.5 or 1/2
or you could write it the opposite way:
6/3 = 2/1 = 4/2
It will still equal the same thing:
0.5 or 1/2

*As long as you are comparing "this" to "that".

Is this triangle the same as well?

No, because not all the angles are the same. The sides are not equivalent to the original shape.



Next Unit: Polygons

Definition of a polygon: 2D shape that is a closed figure made of 3 or more line segments.

This unit will be focusing on identify similar polygons and explain why they are similar. Drawing similar polygons. Solving problems using the properties of similar polygons.



Don't forget to check out COOL MATH SITES on the blog.
Also remember to revisit Similarity and Proportions

Homework:
-Go to Mr. Backé 's site <!--> Normal 0 false false false EN-US X-NONE X-NONE

-Textbook: Chapter 4.4
Practice #3, 5, 6
All Apply
Extend any 3 #13-17
Homework Book Chapter 4.4
Extra Practice 4.4

-Mangahigh

Julibella's Math Question

Chapter 2.4

Question 18)
A frame measures 30 cm by 20 cm. Can you mount a square picture with an area of 500cm^2 in the frame? Explain.


So you're frame is 20cm x 30cm which gives you an area of 600cm^2. You're trying to put a picture with an area of 500cm^2 which dimensions are 22.36cm x 22.36cm. The picture will not fit because it is too big.

Question 19)
A square picture with an area of 100 cm^2 is mounted on a square piece of matting. The matting has 2.5 times the area of the picture. Is the picture is centered on the matting, what width of the matting is visible around the outside of the picture? Give your answer to the nearest tenth of a centimeter.

First find the area of the matting:
100cm^2 x 2.5 = 250cm^2

Fing the side length of the picture:
sqrt 100cm^2 = 10cm

Find the side length of the matting:
sqrt 250cm^2 = 15.81cm

Subtract side lengths to find visible matting:
15.81cm - 10cm = 5.81cm

Make equal on both sides of the picture to find width of visible matting:
5.81 ÷ 2 = 2.9cm

The width of the matting visible around the outside of the picture is 2.9cm.

Julibella's Math Question

Chapter 2.3
22) Taj has three scoops for measuring flour. The largest scoop holds 2 1/2 times as much as the smallest one. The middle scoop holds 1 3/4 times as much as the smallest one.
Describe two different ways in which Taj could measure each of the following quantities. He can use full scoops only.
a) 3 1/4 times as much as the smallest scoop holds
b) 1/2 as much as the smallest scoop holds

First you would have to find the smallest scoop. You think to yourself, " the largest scoop (2 1/2) multiplied by what gives you the largest scoop (2 1/2)?" Then you think to yourself, "the middle scoop (1 3/4) multiplied by what gives you the middle scoop (1 3/4)?"

2 1/2 x _?_= 2 1/2
2 1/2 x _1_= 2 1/2

1 3/4 x _?_= 1 3/4
1 3/4 x _1_= 1 3/4

**What we know**

Can only measure with full scoops.
Smallest scoop equals = 1
Middle scoop= 1 3/4
Largest scoop= 2 1/2

a) One way: 3 1/4 = three middle scoops, take away one small scoop
3 1/4 = 3 x 1 3/4 - 1
3 1/4 = 3/1 x 7/4 - 1
3 1/4 = 21/4 - 1
3 1/4 = 4 1/4 -1
3 1/4 = 3 1/4

Second way: 3 1/4 = one large scoop, plus one middle scoop, take away one small scoop
3 1/4 = 2 1/2 + 1 3/4 - 1
3 1/4 = 5/2 + 7/4 - 1
3 1/4 = 10/4 + 7/4 -1
3 1/4 = 17/4 - 1
3 1/4 = 4 1/4 -1
3 1/4 = 3 1/4

b) One way: 1/2 = one large scoop, take away two small scoops
1/2 = 2 1/2 - 2 x 1
1/2 = 2 1/2 - 2
1/2 = 1/2

Second way: 1/2 = two middle scoops, take away three small scoops
1/2 = 2 x 1 3/4 - 3 x 1
1/2 = 2/1 x 7/4 - 3 x 1
1/2 = 14/4 - 3 x 1
1/2 = 3 2/4 -3
1/2 = 2/4
1/2 = 1/2