Introduction
The Pythagoras theorem is given by the Greek mathematician Pythagoras. Before Pythagoras, this theorem was discovered by Indian mathematician Boudhayana so this theorem is also known as Boudhayana Theorem. This theorem is also called Pythagorean Theorem.

Statement – According to this theorem, In a right-angled triangle, the square of the hypotenuse side is equal to the sum of the square of the base side and perpendicular side.

Formula – (Hypotenuse)2 = (Base)2 + (Perpendicular)2
We can also write, H2 = B2 + P2
Note – This theorem can be used for right-angled triangles only.
How is it Used, Let us See тАУ
Pythagoras theorem is used to find any side of a right-angled triangle when the remaining two sides are given.
Example тАУ In a right-angled triangle, the base is 4 cm and the perpendicular is 3 cm then find its hypotenuse side.
Solution тАУ Here, Base(B) = 4 cm , Perpendicular(P) = 3cm and Hypotenuse(H) = ?

By Pythagoras theorem, H2 = B2 + P2
H2 = (4)2 + (3)2
H2 = 16 + 9
H2 = 25
H = тИЪ25
H = 5 cm
Therefore, Hypotenuse is 5 cm. Ans.
Proofs of Pythagoras Theorem
(1) In a right-angled triangle, the area of the square made on the hypotenuse is equal to the sum of the areas of the squares made on the other two sides.

Given тАУ In тИЖABC, тИаB = 90┬░ and squares ACDE, ABFG, and BCHK are made on sides AC, AB, and BC respectively.
To Prove тАУ ar(ACDE) = ar(ABFG) + ar(BCHK) (ar = Area)
Construction тАУ Drawn BMтКеDE and Joined AH and BD.
Proof – тИаACD = тИаBCH = 90┬░ (Angles of squares)
Adding тИаACB on both sides
тИаACD + тИаACB = тИаBCH + тИаACB
тИаBCD = тИаACH —————-(1)
In тИЖACH and тИЖDCB
AC = CD (Sides of square ACDE)
CH = BC (Sides of square BCHK)
тИаACH = тИаBCD (From equation (1)
By SAS congruency criterion
тИЖACH тЙЕ тИЖDCB
тИ┤ ar(тИЖACH) = ar(тИЖDCB) (areas of congruent triangles are equal) —————-(2)
тИ╡ тИаABC = тИаCBK = 90┬░
тИ┤ тИаABC + тИаCBK = 180┬░
It means ABK is a straight line.
тИ╡ CH тИе BK
тИ┤ CH тИе AK
тИ╡ тИЖACH and square BCHK are made on the same base CH and between the same parallel lines AK and CH
тИ┤ ar( тИЖACH) = ┬╜ ar (square BCHK) —————(3)
Similarly, тИЖDCB and Rectangle CDML are made on the same base CD and between the same parallel lines CD and BM
тИ┤ ar( тИЖDCB) = ┬╜ ar (rectangle CDML) —————-(4)
From equations (2), (3) and (4) we get
ar(тИЖACH) = ar(тИЖDCB)
┬╜ ar (square BCHK) = ┬╜ ar (rectangle CDML)
ar (square BCHK) = ar (rectangle CDML) —————-(5)
Similarly, ar(square ABFG) = ar(rectangle AEML) —————–(6)
Adding equations (5) and (6),
ar (square BCHK) + ar(square ABFG) = ar (rectangle CDML) + ar(rectangle AEML)
ar (square BCHK) + ar(square ABFG) = ar (square ACDE)
or ar (square ACDE) = ar (square BCHK) + ar(square ABFG) Hence Proved.
(2) In a right-angled triangle, the square of the hypotenuse side is equal to the sum of the squares of the remaining two sides.

Given тАУ In тИЖABC, тИаB = 90┬░
To Prove тАУ AC2 = AB2 + BC2
Construction тАУ Drawn BD тКе AC
Proof тАУ In тИЖABC and тИЖADB
тИаABC = тИаADB (Both are 90┬░)
тИаBAC = тИаDAB (Common angle in both triangles)
By AA similarity criterion, тИЖABC тИ╝ тИЖADB
Therefore, AB тИХ AD = AC тИХ AB (By Thales theorem)
AB2 = ACтипAD —————-(1)
Now in тИЖABC and тИЖBDC
тИаABC = тИаBDC (Both are 90┬░)
тИаACB = тИаBCD (Common angle in both triangles)
By AA similarity criterion,┬а┬а┬а┬а┬а тИЖABC тИ╝┬атИЖBDC
Therefore, BC / DC = AC / BC (By Thales theorem)
BC2 = ACтипCD —————-(2)
By adding equations (1) and (2),
AB2 + BC2 = ACтипAD + ACтипCD
AB2 + BC2 = AC (AD + CD)
AB2 + BC2 = ACтипAC
AB2 + BC2 = AC2
AC2 = AB2 + BC2 Hence Proved.
Some Other Examples of Pythagoras Theorem
Example 1) In a right-angled triangle, Hypotenuse is 7 cm and the perpendicular is 4 cm then find its Base.

Solution – By Pythagoras theorem,
H2 = B2 + P2
(7)2 = B2 + (4)2
49 = B2 + 16
49 тАУ 16 = B2
B2 = 33
B = тИЪ33
B = 5.74 cm Ans.
Example 2) Length and breadth of a rectangle are 5 cm and 7 cm. find its diagonal.

Solution – Because of the diagonal, this Rectangle is divided into two right-angled triangles.
So, by Pythagoras theorem, in тИЖABC
AC2 = AB2 + BC2
AC2 = (5)2 + (7)2
AC2 = 25 + 49
AC2 = 74
AC = тИЪ74
AC = 8.60 cm
Therefore, the length of the diagonal is 8.60 cm. Ans.
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