He was the 1^{st} individual to come up with this theorem. So, what does it entail? Simple – the relation between the three sides of a right-angle triangle. If we revise the definition once, Pythagorean or Pythagoras Theorem states that the summation of squares of 2 sides’ (base and height) lengths of a triangle is equivalent to the square of the 3^{rd} side, which is the longest. This side is termed as the hypotenuse. Well, learning the **Pythagoras Theorem formula** and different ways to prove it are equally simple if you understand it thoroughly.

Here, at Vedantu, we aim to help you precisely with the same so that you can excel in exams not only with better marks but also with a stronger concept.

Here, ‘a’ is the perpendicular, ‘b’ is the base and ‘c’ is the hypotenuse. You can term ‘a’ and ‘b’ as the legs of that triangle which meet each other at 90°.

Hence, the formula will be a^{2} + b^{2} = c^{2}.

There’s another way of stating this theorem.

In this case, ‘a’ is the base, ‘b’ is the perpendicular and ‘c’ is the hypotenuse.

So, the formula will be the area of a + area of b = area of c.

If a = 6 and b = 8, then what is the value of hypotenuse, x?

According to the **Pythagoras Theorem formula**, it is x^{2 }= 6^{2 }+ 8^{2}.

Now, consider it this way, x^{2} = 100, because 6^{2} is 36 and 8^{2} is 64.

So, x = , i.e., 10.

Vedantu guides thoroughly with various **Pythagorean Theorem formula and examples** so that students get a grip and can solve mathematical problems effortlessly. Not only students in schools but those appearing in competitive exams can also find our formulae list handy and easy to learn.

Besides, students can also learn about **Pythagorean Theorem formula proof** and solve simple to complex numerical calculations.

Interestingly, there are various applications of Pythagoras Theorem in real life. Some of those are as follows:

So, if you opt for such careers, rest assured you’ll need in-depth knowledge of the Pythagoras Theorem.

There are many applications of Pythagorean theorem some are listed below:

i.e., \[A{C^2} = A{B^2} + B{C^2}\]

\[\angle ADB = \angle ABC\] [Each equal to \[90^\circ \]]

and, \[\angle A = \angle A\] [Common]

So, by AA-similarity criterion, we have

\[\Delta ADB \sim \Delta ABC\]

\[ \Rightarrow \,\,\frac{{AD}}{{AB}} = \frac{{AB}}{{AC}}\] [\[\because \] In similar triangles corresponding sides are proportional]\[ \Rightarrow A{B^2} = AD \times AC\] ……. (i)

In triangles BDC and ABC, we have

\[\angle CDB = \angle ABC\] [Each equal to \[90^\circ \]]

and, \[\angle C = \angle C\]

So, by AA-similarity criterion, we have

\[\Delta BDC \sim \Delta ABC\]

\[ \Rightarrow \frac{{DC}}{{BC}} = \frac{{BC}}{{AC}}\] [\[\because \] In similar triangles corresponding sides are proportional]

\[ \Rightarrow B{C^2} = AC \times DC\] ……… (ii)

Adding equations (i) and (ii), we get

\[A{B^2} + B{C^2} = AD \times AC + AC \times DC\]

\[ \Rightarrow \,\,A{B^2} + B{C^2} = AC\left( {AD + DC} \right)\]

\[ \Rightarrow \,\,A{B^2} + B{C^2} = AC \times AC\]

\[ \Rightarrow A{B^2} + B{C^2} = A{C^2}\]

Hence, \[A{C^2} = A{B^2} + B{C^2}\].

Let AB be the width of the street and C be the foot of the ladder. Let D and E be the windows at heights of 9 m and 12 m respectively from the ground. Then, CD and EF are the two positions of the ladder.

Clearly, AD = 9 m, BE = 12 m, CD = CE = 15 m.

In \[\Delta \,ACD,\] we have

\[C{D^2} = A{C^2} + A{D^2}\]

\[ \Rightarrow \,\,{15^2} = A{C^2} + {9^2}\]

\[ \Rightarrow \,\,A{C^2} = 225 - 81 = 144\]

\[ \Rightarrow \,\,AC = 12\,m\]

In \[\Delta \,BCE,\] we have

\[C{E^2} = B{C^2} + B{E^2}\]

\[ \Rightarrow \,\,{15^2} = B{C^2} + {12^2}\]

\[ \Rightarrow \,\,B{C^2} = 225 - 144 = 81\]

\[ \Rightarrow \,\,BC = 9\,m\]

Hence, width of the street \[ = AB = AC + CB = \left( {12 + 9} \right)m = 21\,m\].

(a) \[\sqrt {q + 1} \]

(b) \[\sqrt {q - 1} \]

(c) \[\sqrt {2q + 1} \]

(d) \[\sqrt {2q - 1} \]

Let the third side be

\[{p^2} = {q^2} + {x^2}\]

\[ \Rightarrow {x^2} = {p^2} - {q^2} = \left( {p - q} \right)\left( {p + q} \right) = p + q\] \[\left[ {\because \,\,p - q = 1} \right]\]

\[ \Rightarrow x = \sqrt {p + q} = \sqrt {2q + 1} \] \[\left[ {\because \,\,p - q = 1\,\,\,\therefore \,\,p = q + 1} \right]\]

Hence, the length of the third side is \[\sqrt {2q + 1} \] cm.

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