Theorem of Pythagoras on Wake Flat, Example of Problem and Discussion

Pythagoras Theorem Formula On Breathe - Do you still remember what is meant by getting up flat? Waking flat is a two-dimensional wake where there is only a long and wide side and is bounded by curved lines and straight lines. As you know, wake flat consists of eight types of square, rectangular, parallelogram, trapezoid, triangle, kite, rhombus and the last is a circle. Each of these flat layers has different circumferential and circumferential formulas and sometimes when we compute the formulas, a calculation that involves the Pythagorean theorem formulas
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Do you guys know in what kind of situations pythagoras theorem is used on the flat wake? If you do not know it then you are obliged to read this material to the end because the basic mathematical formula will explain in detail the application of the pythagoras theorem in solving the problems associated with a flat build. So, let's check it out !! [1945907]
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The use of Pythagorean Theorem Formula on Wake Flat


Looking for a diagonal plane on a square and a rectangle

We can use the pythagoras theorem formula to find the diagonal plane on the rectangle if we already know the length and width. While the pythagoras formula we can use to find the diagonal plane on the square if the length of the side has been known. For more details, see the example example below:
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Example Problem 1 [1945907] Given a rectangle having a length of 20 cm and a width of 15 Cm. Then how long is one diagonal on the rectangle? [1945909]

Discussion:
Diagonal = √ (length 2 + 2 )
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Diagonal = √ ( 20 2 )
Diagonal = √400 + 225
Diagonal = √625
Diagonal = 25cm

[194590] Looking for diagonal kites and rhombus

[1945908] The Pythagoras formula can be We use to look for one diagonal on a kite and a rhomb if it has been known the length of the side and one of the diagonal sides. Consider these two examples of the following questions: [1945907]
Example Problem 2

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Compute the extent of the wake of the kites below:
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[1945909] [1945909] Discussion:
Since the diagonal of EG and FH intersect at point M, Then we first find the length of EM:
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EM = ½ x EG [1945907] [1945909]
EM = ½ x 16 [1945907] [1945908]] EM = 8 cm
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After that, use the pythagoras theorem To know the length of FM and HM:
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FM = √ (EF 2 - EM 2 ) [1] 2 - 8 2 )
[1945907] [1945907] [1945907] ] [1945909] FM = √ (225 - 64)
FM = √161
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FM = 12.6 cm
HM = √ (EH 2 - EM 2 [1945907] HM = √ (20 2 - 8 2 ) ]

HM = √ (400 - 64)
HM = √336

HM = 18.3 cm

The diagonal length of FH is:

] FH = FM + HM
FH = 12.6 + 18.3
FH = 30.9 cm

Now we are looking broadly from the kite:
L = ½ x d1 x d2
L = ½ x EG x FH
L = ½ x 16 x 30.9
L = ½ x 494.4
L = 247 , 2 Cm 2

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Example Problem 3
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Notice the following rhombic image:


Pythagoras Theorem Formula on Wake Flat, Sample Problem and Discussion "border =" 0 "src =" https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEig7AER57M-2hSTvXi-KoEKIwjGHI4I5Rtx2UbuDZyIK3_9vWbpaptyvb96f2cipxvsjKZo2JA3TxFs1YW8B_n9VU2Xi7rgDcFP6hQAMyE5FGxFtrEMGaBbw2xLlV5gtKiT_nhIJh5ummU/s1600/Rumus+Teorema % 2BPythagoras% 2Bpada% 2BBangun% 2BDatar.JPG "title =" Pythagoras Theorem Formula on Wake Flat, Sample Problem and Discussion "> </span> </div> <span style=
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If the known length of the PQRS diamond is 15 cm and the length of one diagonal is 24 cm, then what is the width of the rhombus?
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Discussion:
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When the diagonal crossing of PR and QS in the rhombus is present at point X, then:
] PX = ½ x PR
PX = ½ x 24 [1945907] [1945909] [1945909]
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Now we use the formula of the pythagoras theorem to find the length of QX:
QX = √ (PQ 2 - PX

)

[1945907] QX = √ (15 2 - 12 2
QX = √ (225 - 144)
QX = √81 [1945907] [1945907] QX = 9 cm [1945909]
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QS = 2 x QX
QS = 2 x 9
QS = 18 cm [1945907]

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