Therefore, by Pythagoras theorem
XY = YZ + XZ⇒ 200x = 10000 + 3600
⇒ 200x = 13600
⇒ x = 13600/200
Therefore, distance between X and Y = 68 meters.
Therefore, length of each side is 8 cm.
Using the formula solve more word problems on Pythagorean Theorem.
3. Find the perimeter of a rectangle whose length is 150 m and the diagonal is 170 m.
In a rectangle, each angle measures 90°.
Therefore PSR is right angled at S
Using Pythagoras theorem, we get
⇒ PS = √6400
Therefore perimeter of the rectangle PQRS = 2 (length + width)
= 2 (150 + 80) m
= 2 (230) m
= 460 m
4. A ladder 13 m long is placed on the ground in such a way that it touches the top of a vertical wall 12 m high. Find the distance of the foot of the ladder from the bottom of the wall.
Let the required distance be x meters. Here, the ladder, the wall and the ground from a right-angled triangle. The ladder is the hypotenuse of that triangle.
According to Pythagorean Theorem,
Therefore, distance of the foot of the ladder from the bottom of the wall = 5 meters.
5. The height of two building is 34 m and 29 m respectively. If the distance between the two building is 12 m, find the distance between their tops.
The vertical buildings AB and CD are 34 m and 29 m respectively.
Draw DE ┴ AB
Then AE = AB – EB but EB = BC
Therefore AE = 34 m - 29 m = 5 m
Now, AED is right angled triangle and right angled at E.
⇒ AD = √169
Therefore the distance between their tops = 13 m.
The examples will help us to solve various types of word problems on Pythagorean Theorem.
Congruent Shapes
Congruent Line-segments
Congruent Angles
Congruent Triangles
Conditions for the Congruence of Triangles
Side Side Side Congruence
Side Angle Side Congruence
Angle Side Angle Congruence
Angle Angle Side Congruence
Right Angle Hypotenuse Side congruence
Pythagorean Theorem
Proof of Pythagorean Theorem
Converse of Pythagorean Theorem
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The pythagorean theorem with examples.
The Pythagorean theorem is a way of relating the leg lengths of a right triangle to the length of the hypotenuse, which is the side opposite the right angle. Even though it is written in these terms, it can be used to find any of the side as long as you know the lengths of the other two sides. In this lesson, we will look at several different types of examples of applying this theorem.
Table of Contents
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In the examples below, we will see how to apply this rule to find any side of a right triangle triangle. As in the formula below, we will let a and b be the lengths of the legs and c be the length of the hypotenuse. Remember though, that you could use any variables to represent these lengths.
In each example, pay close attention to the information given and what we are trying to find. This helps you determine the correct values to use in the different parts of the formula.
Find the value of \(x\).
The side opposite the right angle is the side labelled \(x\). This is the hypotenuse. When applying the Pythagorean theorem, this squared is equal to the sum of the other two sides squared. Mathematically, this means:
\(6^2 + 8^2 = x^2\)
Which is the same as:
\(100 = x^2\)
Therefore, we can write:
\(\begin{align}x &= \sqrt{100}\\ &= \bbox[border: 1px solid black; padding: 2px]{10}\end{align}\)
Maybe you remember that in an equation like this, \(x\) could also be –10, since –10 squared is also 100. But, the length of any side of a triangle can never be negative and therefore we only consider the positive square root.
In other situations, you will be trying to find the length of one of the legs of a right triangle. You can still use the Pythagorean theorem in these types of problems, but you will need to be careful about the order you use the values in the formula.
Find the value of \(y\).
The side opposite the right angle has a length of 12. Therefore, we will write:
\(8^2 + y^2 = 12^2\)
This is the same as:
\(64 + y^2 = 144\)
Subtracting 64 from both sides:
\(y^2 = 80\)
\(\begin{align}y &= \sqrt{80} \\ &= \sqrt{16 \times 5} \\ &= \bbox[border: 1px solid black; padding: 2px]{4\sqrt{5}}\end{align}\)
In this last example, we left the answer in exact form instead of finding a decimal approximation. This is common unless you are working on an applied problem.
There are many different kinds of real-life problems that can be solved using the Pythagorean theorem. The easiest way to see that you should be applying this theorem is by drawing a picture of whatever situation is described.
Two hikers leave a cabin at the same time, one heading due south and the other headed due west. After one hour, the hiker walking south has covered 2.8 miles and the hiker walking west has covered 3.1 miles. At that moment, what is the shortest distance between the two hikers?
First, sketch a picture of the information given. Label any unknown value with a variable name, like x.
Due south and due west form a right angle, and the shortest distance between any two points is a straight line. Therefore, we can apply the Pythagorean theorem and write:
\(3.1^2 + 2.8^2 = x^2\)
Here, you will need to use a calculator to simplify the left-hand side:
\(17.45 = x^2\)
Now use your calculator to take the square root. You will likely need to round your answer.
\(\begin{align}x &= \sqrt{17.45} \\ &\approx 4.18 \text{ miles}\end{align}\)
As you can see, it will be up to you to determine that a right angle is part of the situation given in the word problem. If it isn’t, then you can’t use the Pythagorean theorem.
There is one last type of problem you might run into where you use the Pythagorean theorem to write some type of algebraic expression. This is something that you will not need to do in every course, but it does come up.
A right triangle has a hypotenuse of length \(2x\), a leg of length \(x\), and a leg of length y. Write an expression that shows the value of \(y\) in terms of \(x\).
Since no figure was given, your first step should be to draw one. The order of the legs isn’t important, but remember that the hypotenuse is opposite the right angle.
Now you can apply the Pythagorean theorem to write:
\(x^2 + y^2 = (2x)^2\)
Squaring the right-hand side:
\(x^2 + y^2 = 4x^2\)
When the problem says “the value of \(y\)”, it means you must solve for \(y\). Therefore, we will write:
\(y^2 = 4x^2 – x^2\)
Combining like terms:
\(y^2 = 3x^2\)
Now, use the square root to write:
\(y = \sqrt{3x^2}\)
Finally, this simplifies to give us the expression we are looking for:
\(y = \bbox[border: 1px solid black; padding: 2px]{x\sqrt{3x}}\)
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The Pythagorean theorem allows you to find the length of any of the three sides of a right triangle. It is one of those things that you should memorize, as it comes up in all areas of math, and therefore in many different math courses you will probably take. Remember to avoid the common mistake of mixing up where the legs go in the formula vs. the hypotenuse and to always draw a picture when one isn’t given.
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48 pythagorean theorem worksheet with answers [word + pdf].
The simplicity of the Pythagorean Theorem worksheet is the best thing about it. What is the Pythagorean Theorem? Formulated in the 6th Century BC by Greek Philosopher and mathematician Pythagoras of Samos, Pythagorean Theorem is a mathematic equation used for a variety of purposes. Over the years, many engineers and architects have used Pythagorean Theorem worksheet to complete their projects .
Table of Contents
A simple equation, Pythagorean Theorem states that the square of the hypotenuse (the side opposite to the right angle triangle) is equal to the sum of the other two sides . Following is how the Pythagorean equation is written:
In the aforementioned equation, c is the length of the hypotenuse while the length of the other two sides of the triangle are represented by b and a. Though the knowledge of the Pythagorean Theorem predates the Greek Philosopher, Pythagoras is generally credited for bringing the equation to the fore. This is the reason the Pythagorean equation is named after him. Before we discuss the Pythagorean Theorem and the Pythagorean Theorem worksheet in detail, let’s take a look at who Pythagoras of Samos was and how he came up with the Pythagorean equation.
A 6 th century BC Greek philosopher and mathematician, Pythagoras of Samos is widely credited for bringing the Pythagorean equation to the fore. Though others used the relationship long before his time, Pythagoras is the first one who made the relationship between the lengths of the sides on a right-angled triangle public. This is why he’s regarded as the inventor of the Pythagorean equation.
Apart from being a philosopher and mathematician, Pythagoras founded the Pythagoreanism movement. Born in Croton, Italy, Pythagoras travelled to many different countries including Greece, Egypt, and India. After moving back to Croton in 530 BC, Pythagoras established some kind of school. He returned to Samos in 520 BC. It was in late 6 th Century BC that Pythagoras started to make important contributions to philosophy and math. The Pythagorean equation was one of those contributions.
Though he revealed the Pythagorean equation to the world in the late 6 th Century BC while living in Samos, many historians believe that Pythagoras first thought about the equation during his time in Egypt. In fact, according to many historians, Pythagoras learned geometry, the Phoenicians arithmetic and other branches of mathematics from the Egyptians.
Though he has made many important contributions to philosophy, Pythagoras is widely known as the founder of the Pythagorean Theorem. As previously mentioned, the Pythagorean Theorem is a mathematical equation that states that the square of the hypotenuse (the side opposite to the right angle triangle) is equal to the sum of the other two sides .
Today the aforementioned equation bears Pythagoras’s name but it’s important to know that he wasn’t the first one to use the equation. Before Pythagoras’s time, the Indians and the Babylonians utilized the Pythagorean Theorem or equation. Since they constructed the first proof of the theorem, Pythagoras and his disciples are regarded as the inventors of the equation.
Many historians say that Pythagoras worked in a very secretive manner. This is the reason little evidence is available that the Greek Philosopher/ mathematician himself worked on and proved the Pythagorean Theorem. It is important to note that the first time Pythagoras was given credit for the Theorem was five centuries after his death. This makes Pythagoras’s contribution to the Theorem even more debatable. Nonetheless, since Pythagoras is the only one connected to the Pythagorean Theorem known today, we have to give him due credit. Now that we’ve discussed who Pythagoras of Samos was and how he came up with the Pythagorean equation, it’s time to take a detailed look at the Pythagorean Theorem and the Pythagorean Theorem worksheet.
According to Pythagorean Theorem, the sum of the squares on the right-angled triangle’s two smaller sides is equal to the side opposite to the right angle triangle (the square on hypotenuse). Using a Pythagorean Theorem worksheet is a good way to prove the aforementioned equation. An amazing discovery about triangles made over two thousand years ago, Pythagorean Theorem says that when a triangle has a 90° angle and squares are made on each of the triangle’s three sides, the size of the biggest square is equal to the size of the other two squares put together! A short equation, Pythagorean Theorem can be written in the following manner:
In Pythagorean Theorem, c is the triangle’s longest side while b and a make up the other two sides. The longest side of the triangle in the Pythagorean Theorem is referred to as the ‘hypotenuse’. Many people ask why Pythagorean Theorem is important. The answer to this is simple: you’ll be able to find the length of a right-angled triangle’s third side if you know the length of the other two sides. This equation works like magic and can be used to find any missing value. Following is an example that uses the Pythagorean Theorem to solve a triangle.
In this equation, the longest side of the triangle ‘c’ is missing. By finding out the sum of the squares of the two other sides, we were able to find the missing value. The most famous mathematical contribution of Pythagoras, the Pythagoras Theorem was one of the earliest documented theorems. Though Pythagoras is given most of the credit for the theorem, a major contribution to the theorem was made by his students.
When you look at a Pythagoras Theorem worksheet, you’ll notice that the theorem enables you to find the length of any right angle triangle side provided you know the length of the other two sides. Also, using the theorem, you can check whether a triangle is a right triangle. The Pythagoras Theorem is extremely useful in solving many math problems. Further, you can use it in many real life situations. This is illustrated by a Pythagoras Theorem worksheet.
A good way to review the Pythagoras Theorem and expand the mathematical equation is using a Pythagoras Theorem worksheet. By using the worksheet, you’ll be able to get a good understanding of geometry. Additionally, the worksheet will give you an opportunity to review the knowledge related to the different types of triangles. Finally and most importantly, you’ll be able to practice the ancient equation invented by the Greek mathematician and philosopher, Pythagoras. Before you start using the Pythagoras Theorem worksheet, just remember that ‘c’ is the hypotenuse while the shorter sides of the triangle are represented by ‘a’ and ‘b’.
A Pythagoras Theorem worksheet presents students with triangles of various orientations and asks them to identify the longest side of the triangle i.e. the hypotenuse. As you know by now, the formula used in Pythagoras Theorem is a²+b²=c². Regardless of what the worksheet asks the students to identify, the formula or equation of the theorem always remain the same. Though, the students could be presented with different challenges including solving triangles:
The symbols used in the Pythagoras Theorem are something students will find on their calculators. Figuring out how to use these functions is what students need to establish. There is involvement of the Babylonians and the Egyptians in the invention of the Pythagoras Theorem but the earliest known proof of the theorem was produced by the school of Pythagoras.
Many Pythagorean triples were known to the Babylonians while the Egyptians knew and used the (3, 4, 5) triple. The Chinese and Indians also played a role in the invention of the Pythagoras Theorem. The first diagrammatic proof of the theorem was produced by the Chinese while the Indians discovered many triples. In 1995, the theorem became part of the Guinness Book of Records as the most proved theorem of all time.
The triples used in the Pythagoras Theorem include (3, 4, 5), (6, 8, 10), (5, 12, 13), (8,15,17), (7,24,25), (20,21,29), (12,35,37), (9,40,41), (28,45,53), (11,60,61), (16,63,65), (33,56,65) and (48,55,73). The aforementioned triples aren’t multiples of a smaller triple and the name given to them is ‘primitive’ triples. To solve a particular problem, the Pythagoras Theorem can be arranged. For example, if you’re asked to find b which is one of the two smaller sides of the right-angled triangle, you can rearrange the theorem to b²=c²-a². By doing this, you’ll be able to easily find the missing value.
The Pythagoras Theorem has many different proofs. However, when checking your answers, following are the two things that you must always remember:
To understand this better, take a look at a Pythagoras Theorem worksheet. Today, you can get easy access to Pythagorean Theorem worksheet with answers. Nonetheless, we’re going to try and understand the Pythagoras Theorem as much as we can.
As mentioned earlier, if you know the size of the other two sides, you will be able to find out the length of the third side of the right angle triangle. Also, after being squared, the shorter length is subtracted from the square of the hypotenuse when the hypotenuse is one of the two known lengths. As seen earlier, the lengths of each side of the triangle in the Pythagoras Theorem are whole numbers. Such triangles are known as Pythagorean triangles.
Though there are many different proofs of the Pythagoras Theorem, only three of them can be constructed by students and other people on their own. The first proof starts off as rectangle and is then divided into three triangles that individually contain a right angle. To see first proof, you can use a computer or something as straight forward as an index card cut up into right triangles.
Beginning with a rectangle, the second proof of the Pythagoras Theorem starts off by constructing rectangle CADE with BA=DA. This is followed by the construction of the <BAD’s angle bisector. Once constructed, the bisector is allowed to intersect ED at point F. This makes <BAF and <DAF congruent, BA=DA, and AF=AF. This in turn makes the triangle DAF equal to triangle BAF which means that since ADF is a right angle, ABF will also be a right angle. The third and final proof of the Pythagorean Theorem that we’re going to discuss is the proof that starts off with a right angle. In this proof, triangle ABC is right angle and its right side is angle C.
The three proofs stated above are just few of the many Pythagoras Theorem. You’ll come across these proofs when you take a look at the Pythagorean Theorem worksheet with answers. Learning and understanding the Pythagorean concept is extremely important for students and other people who’ll use this theorem in their practical life.
It is important that you understand the algebraic representation of the Pythagoras Theorem as well as the geometric concepts behind it. You can accomplish this by using proofs, manipulatives, and computer technology. By using these methods to learn Pythagorean Theorem, you’ll be able to see the connections and benefit greatly.
Formulated in the 6th Century BC by Pythagoras of Samos, Pythagoras Theorem is widely used today. If you want to practice Pythagoras Theorem then you can do that easily. Pythagoras Theorem worksheets with answers are easily available and you can use these worksheets to get a good grip of the Theorem.
Number of problems found: 1366.
Problem 1 :
A man goes 18 m due east and then 24 m due north. Find the distance of his current position from the starting point?
the distance of his current position from the starting point = √18 2 + 24 2
= √(324 + 576)
= √900
= 30 m
So, the required distance is 30 m.
Problem 2 :
There are two paths that one can choose to go from Sarah’s house to James house. One way is to take C street, and the other way requires to take A street and then B street. How much shorter is the direct path along C street? (Using figure).
By choosing the C street, he has to cover the distance,
= √2 2 + 1.5 2
= √(4 + 2.25)
= √6.25
= 2.5 miles
By choosing the alternative way, he has to cover the distance = 2 + 1.5
= 3.5 miles
The difference between these two paths = 3.5 - 2.5
= 1 mile
So, by choosing the direct path, he may save 1 miles faster than other way.
Problem 3 :
To get from point A to point B you must avoid walking through a pond. You must walk 34 m south and 41 m east. To the nearest meter, how many meters would be saved if it were possible to make a way through the pond?
By drawing the rough picture using the given information, we get
AC = √34 2 + 41 2
= √1156 + 1681
= √2837
= 53.26
Miles saved = (34 + 41) - 53.26
= 75 - 53.26
= 21.74 m
Problem 4 :
In the rectangle WXYZ, XY + YZ = 17 cm, and XZ + YW = 26 cm. Calculate the length and breadth of the rectangle?
XY + YZ = 17 cm
XZ + YW = 26 cm
To calculate : - Length and breadth of the rectangle.
We know that,
Diagonals of a rectangle are equal.
XZ = YW = 26/2 = 13 cm
In ∆XYZ, let YZ = P. Then
XY = 17 - P
Then, by Pythagoras theorem,
(P) 2 + (17 - P) 2 = (13) 2
P 2 + 289 - 34P + P 2 = 169
2P 2 - 34P = 169 - 289
2(P 2 - 17P) = - 120
P 2 - 17P = - 120/2
P 2 - 17P = - 60
P 2 - 17P + 60 = 0
P 2 - 12P - 5P + 60 = 0
P(P - 12) - 5(P - 12) = 0
(P - 12)(P - 5) = 0
P - 12 = 0 or P = 12
P = 12 cm or P = 5 cm
YZ = P = 12 cm [Because , YZ is the length of the rectangle ,so we will assign it the greatest value of P]
Again, XY = (17 - P) = (17 - 12) cm = 5 cm
[Because , XY is thee breadth]
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Aug 02, 24 11:32 AM
Aug 02, 24 11:23 AM
Ollie Lovell
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On my last placement I taught a unit on Pythagoras' theorem to my year 9 class. I found it deceptively difficult. Why? The challenge was twofold: First, students had seen the theorem in previous years so it was a bit old hat to them. Second, I wondered how to stretch students and provide them with variety and challenge beyond the old trick of calling the hypotenuse a ladder?
I was thinking to myself ‘ WHY does Pythagoras’ theorem actually matter, and how can I make that tangible and relevant to students? I came up with the following slideshow.
I had this first slide on the screen as students walked into the classroom, it created a bit of a buzz in the room…
I got students to vote (hands up) on which corner they thought was square (which do you think it is?), then revealed the answer.
I've presented the remainder of the slideshow in the following video with examples of the kind of teacher questions and the script that I used. Seemed easier and clearer than writing it all out…
(If you want to use any of this you can download the slideshow file here: Pythagoras Slides_Oliver Lovell_www.ollielovell.com )
Clarifying points…
The general structure of lessons in this unit from then on was a bit of book work (using the ‘points' system that I'll write about in another post) with a challenge question for those students who got all of their points done. Lessons were wrapped up with students sharing on the board their approach to the challenge question and what they had learned from attempting it. I also book-ended most lessons with a micro-revision, 3 questions on the basics as a mini-test at the start, and an exit card at the end to help me to keep track of how students were progressing.
Below are the questions that I used as challenge questions. I wrote all of the following questions, excepting the final question, which I found here . I'd also like to acknowledge Dr Max Stephens who suggested that I create questions that encourage students to solve for exact values rather than untidy decimals. I've left out answers on purpose in the knowledge that future students will possibly visit this post.
You’ve always got to include a ladder question don’t you? (this one is a little bit different).
A Cat is stuck in a tall tree and needs to be saved by the Firefighters. The fire truck has a 35m ladder. The closest that it can get to the tree (because there are parked cars in the way) is 12m. The cat is in the tree, 39m up. Comment on the likelihood of the cat getting saved and justify your suggestion mathematically.
Q: A cube has side length of 10cm. what is the length of the diagonal through its centre? (could accompany this with the picture below).
Alternatively you could ask, Q: ‘What is the length of d in the following if the side lengths are 10cm?' (referring to the above image)
Prior to giving students the above question we’d had a discussion about the formula of the diagonal of a box and students had worked out that it’s just the square root of the sum of the squares of the box’s dimensions. This helped them to solve this question. One of the students actually proposed this ‘pythagoras in 3 dimensions' formula, so I encouraged them all to check it and see if the proposal seemed plausible.
An additional question that I also posed to students was, Q: “Show that the formula for the diagonal of a box is √a²+b²+c².” A discussion of this also helped to scaffold the students.
Q: A rectangular prism has sides of length a, 4a/3, and 4a. It has an internal diagonal of 26cm. What are the dimensions of the rectangular prism?
Or, you could word it in a more tricky way: Q: A rectangular prism has a width one quarter of its length and a height one third of its width. It has an internal diagonal of 26cm. What are the dimensions of the rectangular prism?
Another question:
In the 6 sided pyramid pictured below, what is the height of the apex ‘V’ above the point ‘O’ ? (You can assume that O is in the centre of the base, the base is a regular hexagon, and V is directly above O).
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Pythagorean identities are fundamental trigonometric identities derived from the Pythagorean theorem. They establish essential relationships between the sine, cosine, and tangent functions. These identities are crucial in simplifying complex trigonometric expressions and solving various mathematical problems.
In this article, we will learn what are Pythagorean identities, derivations of other identities some more important trigonometry identities, and some practice questions on Pythagorean identities.
Pythagorean identities are composed of these six trigonometric functions present in mathematics: sine, cosine, tan, sec, cosec, and cot. The trigonometric functions sine (sin), cosine (cos), and tangent (tan) are involved in the three main Pythagorean identities , which are obtained from the Pythagorean theses. These identities are:
sin²θ + cos²θ = 1 1 + tan²θ = sec²θ 1 + cot²θ = cosec²θ
Problem 1: Verify the identity: sin 2 θ + cos 2 θ = 1 for θ = 45 o .
Sin45 o = √2/2 cos 45 o = √2/2 sin 2 45 o + cos 2 45 o = (√2/2) 2 + (√2/2) 2 = 2/4 + 2/4 = 4/4 = 1
Problem 2: If cosθ = 5/13 , then find sinθ.
Using the identity: sin 2 θ+ cos 2 θ = 1 sin 2 θ + (5/13) 2 = 1 sin 2 θ + 25/169 = 1 sin 2 θ = 1 – 25/169 sin 2 θ = 144/169 sinθ = ± 12/13
Problem 3: Solve for θ if sec 2 θ = 5.
Using the identity: sec 2 θ = 1 + tan 2 θ 5 = 1 + tan 2 θ tan 2 θ = 4 tanθ= ±2 θ = tan -1 (2)
Problem 4: Prove the identity: cos 2 θ = 1 – sin 2 θ.
Using the basic identity: sin 2 θ + cos 2 θ = 1 cos 2 θ = 1 – sin 2 θ
Problem 5: If sinθ = 3/5 , then find the value of cos θ using Pythagorean identity.
Using the identity: sin 2 θ+ cos 2 θ = 1 (3/5) 2 + cos 2 θ = 1 9/25 + cos 2 θ = 1 cos 2 θ = 1 – 9/25 cos 2 θ = 16/25 cosθ = ± 4/5
Problem 6: Find cosec θ if cot θ = 2.
Using the identity: cosec 2 θ – cot 2 θ = 1 1 + cot 2 θ = cosec 2 θ 1 + 2 2 = cosec 2 θ 1 + 4 = cosec 2 θ cosec 2 θ = 5 cosecθ = ±√5
Problem 7: Express cot 2 θ in terms of cosec 2 θ.
We know by identity that: 1 + cot 2 θ = cosec 2 θ Rearranging above equation we get: cot 2 θ = cosec 2 θ – 1
Problem 8: Find sinθ if secθ = 2.
secθ = 1/cosθ secθ = 2 cosθ = 1/2 sin 2 θ + cos 2 θ = 1 sin 2 θ + (1/2) 2 = 1 sin 2 θ= 1 – 1/4 sin 2 θ = 3/4 sinθ = ± √3/2
Problem 9: If sinθ = 1/2, find tanθ.
Using the identity: sin 2 θ + cos 2 θ = 1 (1/2) 2 + cos 2 θ = 1 cos 2 θ = 1 – 1/4 cos 2 θ = 3/4 cosθ = √3/2 tanθ = sinθ/cosθ tanθ = 1/2/√3/2 tanθ = 1/√3
Problem 10: Simplify the expression using Pythagorean identities: sin 2 θ + cos 2 θ + 1.
By Pythagorean identity, we know that: sin 2 θ + cos 2 θ = 1 So, the value of sin 2 θ + cos 2 θ + 1 = 1 + 1 = 2
Practice Questions on Pythagorean Identities
You can download these Practice problems and their Solution from – Pythagorean Identities Practice Problems
Pythagorean identities, which in real life connect the basic trigonometric functions, are an essential part of trigonometry. It defines all three main formulas of trigonometry. After understanding these Pythagorean identities one has a deeper understanding of trigonometric identities and will be able to solve questions. The identities and practice questions given in this article will help you to understand it in a better way.
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Pythagorean Identities Trigonometric Identities Trigonometric Identities Class 10
What are pythagorean identities.
Trigonometric identities also known as Pythagorean identities, which express relationships between the sine, cosine, and tangent functions, are derived using the famous Pythagorean theorem.
They are important for deriving solutions to trigonometric equations and serve as the basic understanding for different calculus and physics ideas.
They are derived by applying the Pythagorean theorem to a right triangle and then changing the fundamental identities(by exhanging LHS and RHS side of equation) to obtain other identities.
Yes, they are important for solving a variety of calculus problems, such as those involving the differentiation and integration of trigonometric functions.
The main relationships that are directly derived from the Pythagorean theorem are known as basic identities, and these basic relationships can be changed to create derived identities.
Similar reads.
Rules, Formula and more
The sum of the squares of the lengths of the legs equals the square of the length of the hypotenuse .
Usually, this theorem is expressed as $$ A^2 + B^2 = C^2 $$ .
A right triangle has one $$ 90^{\circ} $$ angle ($$ \angle $$ B in the picture on the left) and a variety of often-studied formulas such as:
SOHCAHTOA only applies to right triangles ( more here ) .
The hypotenuse is the largest side in a right triangle and is always opposite the right angle.
In the triangle above, the hypotenuse is the side AB which is opposite the right angle, $$ \angle C $$.
Online tool calculates the hypotenuse (or a leg) using the Pythagorean theorem.
Below are several practice problems involving the Pythagorean theorem, you can also get more detailed lesson on how to use the Pythagorean theorem here .
Find the length of side t in the triangle on the left.
Substitute the two known sides into the Pythagorean theorem's formula : A² + B² = C²
What is the value of x in the picture on the left?
Set up the Pythagorean Theorem : 14 2 + 48 2 = x 2 2,500 = X 2
$$ x = \sqrt{2500} = 50 $$
$$ x^2 = 21^2 + 72^2 \\ x^2= 5625 \\ x = \sqrt{5625} \\ x =75 $$
Find the length of side X in the triangle on on the left?
Substitue the two known sides into the pythagorean theorem's formula : $$ A^2 + B^2 = C^2 \\ 8^2 + 6^2 = x^2 \\ x = \sqrt{100}=10 $$
What is x in the triangle on the left?
x 2 + 4 2 = 5 2 x 2 + 16 = 25 x 2 = 25 - 16 = 9 x = 3
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Here's the Pythagorean Theorem formula for your quick reference. The longer leg is twice the shorter leg. Find the hypotenuse. If the longest leg is half the hypotenuse, what is the length of the shortest leg? Here are eight (8) Pythagorean Theorem problems for you to solve. You might need to find either the leg or the hypotenuse of the right ...
How to answer Pythagorean Theorem questions. 1 - Label the sides of the triangle a, b, and c. Note that the hypotenuse, the longest side of a right triangle, is opposite the right angle and will always be labeled. 2 - Write down the formula and substitute the values>. a^2+b^2=c^2 a2 +b2 = c2. 3 - Calculate the answer.
Practice Questions on Pythagoras Theorem. 1. Find the area of a right-angled triangle whose hypotenuse is 13 cm and one of the perpendicular sides is 5 cm. 2. Find the Pythagorean triplet whose one member is 15. 3. Find the perimeter of a rectangle whose diagonal is 5 cm and one of its sides is 4 cm.
a) d) 8) A right triangle has legs of 52.6 cm and 35.7 cm. Determine the length of the triangle's hypotenuse. 9) A right triangle has a hypotenuse of 152.6 m. The length of one of the other sides is 89.4 m. Determine the length of the third side. 10) For each of the following, the side lengths of a triangle are given.
The following questions involve using Pythagoras' theorem to solve a range of word problems involving 'real-life' type questions. On the first sheet, only the hypotenuse needs to be found, given the measurements of the other sides. Illustrations have been provided to support students solving these word problems.
What is the Pythagorean Theorem? The Pythagorean Theorem states that the square of the longest side of a right triangle (called the hypotenuse) is equal to the sum of the squares of the other two sides.. Pythagorean Theorem formula shown with triangle ABC is:. a^2+b^2=c^2 . Side c is known as the hypotenuse.The hypotenuse is the longest side of a right triangle.
kramberol 2022-07-11. pythagorean theorem extensions. are there for a given integer N solutions to the equations. ∑ n = 1 N x i 2 = z 2. for integers x i and zan easier equation given an integer number 'a' can be there solutions to the equation. ∑ n = 1 N x i 2 = a 2. for N=2 this is pythagorean theorem.
Pythagorean theorem word problems. Google Classroom. Microsoft Teams. You might need: Calculator. Steve is turning half of his backyard into a chicken pen. His backyard is a 24 meter by 45 meter rectangle. He wants to put a chicken wire fence that stretches diagonally from one corner to the opposite corner. How many meters of fencing will Steve ...
Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere. ... Multi-step word problem with Pythagorean theorem . Report a problem
Pythagorean Theorem: where a and b are lengths of the legs of a fight triangle and c is the length of the hypotenuse "sum of the squares of the legs is equal to the square of the hypotenuse" Example: 49 _ 65 c fight triangle acute triangle obtuse triangle AV Identifying triangles by their sides: a a a Distance Formula mustrates Pythagorean Theorem!
Pythagorean theorem intro problems. ... Practice using the Pythagorean theorem to solve for missing side lengths on right triangles. Each question is slightly more challenging than the previous. Pythagorean theorem. ... Just a quick question, so in a test it could say find the distance of the other 2 sides of the triangle. ...
The Corbettmaths Practice Questions on Pythagoras. Next: Direct and Inverse Proportion Practice Questions
Right Triangle Questions - using the theorem. The Theorem helps us in: Finding Sides: If two sides are known, we can find the third side. Determining if a triangle is right-angled: If the sides of a triangle are known and satisfy the Pythagoras Formula, it is a right-angled triangle. There is a proof of this theorem by a US president.
Calculate the total area of the plot. Answer: square kilometers. It is given the length of the diagonal of the square. It divides it into two equal triangles. In addition, the two triangles are right and the legs of the same length. be the length of square side and by the Pythagorean theorem we get: \displaystyle x^ {2} +x^ {2}=\left (2\sqrt {2 ...
Word problems using the Pythagorean Theorem: 1. A person has to walk 100 m to go from position X in the north of east direction to the position B and then to the west of Y to reach finally at position Z. The position Z is situated at the north of X and at a distance of 60 m from X. Find the distance between X and Y. Solution:
Solution. The side opposite the right angle is the side labelled \ (x\). This is the hypotenuse. When applying the Pythagorean theorem, this squared is equal to the sum of the other two sides squared. Mathematically, this means: \ (6^2 + 8^2 = x^2\) Which is the same as: \ (100 = x^2\) Therefore, we can write:
Pythagoras' Theorem and Trigonometry - Short Problems. This is part of our collection of Short Problems. You may also be interested in our longer problems on Pythagoras' Theorem and Trigonometry. Printable worksheets containing selections of these problems are available here: Pythagoras Stage 4 ★. Sheet 1.
A simple equation, Pythagorean Theorem states that the square of the hypotenuse (the side opposite to the right angle triangle) is equal to the sum of the other two sides. Following is how the Pythagorean equation is written: a²+b²=c². In the aforementioned equation, c is the length of the hypotenuse while the length of the other two sides ...
Pythagorean theorem - practice problems. The Pythagorean Theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. This can be written as: where c is the length of the hypotenuse, and a and b are the ...
Example 2 (solving for a Leg) Use the Pythagorean theorem to determine the length of X. Step 1. Identify the legs and the hypotenuse of the right triangle . The legs have length 24 and X X are the legs. The hypotenuse is 26. Step 2. Substitute values into the formula (remember 'C' is the hypotenuse). A2 + B2 = C2 x2 + 242 = 262 A 2 + B 2 = C 2 ...
By choosing the alternative way, he has to cover the distance = 2 + 1.5. = 3.5 miles. The difference between these two paths = 3.5 - 2.5. = 1 mile. So, by choosing the direct path, he may save 1 miles faster than other way. Problem 3 : To get from point A to point B you must avoid walking through a pond. You must walk 34 m south and 41 m east.
An additional question that I also posed to students was, Q: "Show that the formula for the diagonal of a box is √a²+b²+c².". A discussion of this also helped to scaffold the students. Q: A rectangular prism has sides of length a, 4a/3, and 4a. It has an internal diagonal of 26cm.
Pythagorean identities are fundamental trigonometric identities derived from the Pythagorean theorem. They establish essential relationships between the sine, cosine, and tangent functions. ... these Pythagorean identities one has a deeper understanding of trigonometric identities and will be able to solve questions. The identities and practice ...
A Right Triangle's Hypotenuse. The hypotenuse is the largest side in a right triangle and is always opposite the right angle. (Only right triangles have a hypotenuse ). The other two sides of the triangle, AC and CB are referred to as the 'legs'. In the triangle above, the hypotenuse is the side AB which is opposite the right angle, ∠C ∠ C .
The Pythagorean Theorem [tex]\bullet\,\,\,\boldsymbol{a^2+b^2=c^2.}\bullet[/tex]. in which a and b are the two legs of the right triangle, and c is the hypotenuse, the longest side of the right triangle. ... The question is incomplete but I will try to help you the much I can. ... Solving Problems Involving Inverse Functions With solutions ...
A.I.'s math problem reflects how much the new technology is a break with computing's past. By Steve Lohr In the school year that ended recently, one class of learners stood out as a seeming ...