solve the problem of widespread two dimensional wake mathematics

solve the problem of widespread two dimensional wake mathematics

The development of mathematical concepts needs to be emphasized in a simpler form that is more easily understood by learners. As good as any educator who teaches with thousands of scientific treasures will not be enough to make students understand the concept of mathematics. The focal point of educating mathematics lies in the simplicity of the teaching style and the content taught, of course this is only relevant to the level of junior high school education. A higher level of education may require deep deductive axiomatic standards in mathematical study.

Step educator in simplifying the concept of mathematical applications in the start of doing the discovery activities of the concept of concept in question. At this time I will try to share to the father / mother or younger sister about how to simplify the calculation of Reduction Area two-dimentional figure a straight build using smart solution. My experience during teaching with this technique is more understandable to students.

Consider the Case below.

Look at the following garden sketch drawings!

Given a plot of parallelogram parallelogram EFGH. Inside the garden is made garden with long 
ABCD the rest is the road. Define Street area!
 
Case analysis
Because ABCD and EFGH are similar then EFGH is the result of enlargement of ABCD.
The scale of the magnification (s) = EF / AB
 EF = s. AB and KH = s. DJ
 
Then the Area of ​​EFGH = EF. KH
                           = s.AB. s. DJ
                           = s2 AB. DJ
                              = 〖(EF / AB)〗 ^ 2.Large ABCD
 
Area of ​​Road = Area of ​​EFGH - Area ABCD
 
= 〖(EF / AB)〗 ^ 2. Area ABCD - Area ABCD
 
= ((〖EF〗 ^ 2 / 〖AB〗 ^ 2) - 1) Area ABCD
= ((〖EF〗 ^ 2- 〖AB〗 ^ 2) / 〖AB〗 ^ 2) Area of ​​ABCD
By the same analogy
 
Area of ​​road = ((〖EF〗 ^ 2- 〖AB〗 ^ 2) / 〖EF〗 ^ 2) Area of ​​EFGH
 
 
Example Use of the formula to solve mathematical problems
 
(UN Junior High School 2016)


Look at the following garden sketch drawings!


A plot of parallelogram parallelogram. Inside the garden is a garden with length AB = 20 m and 

length DJ = 15 m. Around the park will be made way. If the gardens and parks are similar, then 
the wide road is ....
A. 66 m2 C. 300 m2
B. 132 m2 D. 360 m2
 
Answer
From the image AB = 20 m and EF = 20 + 4 = 24 m
Then AB: EF = 20: 24 = 5: 6
Area of ​​the Road = ((6 ^ 2-5 ^ 2) / 5 ^ 2) 20 x 15
= (36-25) / 25 x 300
= 11/25 x 300 = 132 m2
2. A photo is affixed to a carton as shown


On the top left part of the photo

There is 5 cm cardboard remaining. If the photo
and cartons of widespread carton
which is not closed photo is ...
a. 750 cm2
b. 850 cm2
c. 1,050 cm2
d. 1.350 cm2


Answer
Photo width = 40-5-5 = 30
Photo: carton = 30: 40 = 3: 4
Uncovered cardboard area of ​​photograph = ((4 ^ 2-3 ^ 2) / 4 ^ 2) .40 x 60
= (16-9) / 16 x 240
= 7/16 x 240 = 105 cm2
 
That's the smart solution to solving the widespread problem of 

two-dimentional figure. Hope to benefit.