Statement: Ina right-angled triangles, the square described on the hypotenuse is equal to the sum of the squares described on the sides containing the right angle.
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In a right-angled triangle the square described on the hypotenuse is equal tothe sum of the squares described on the other two sides containing the right angle.
i.e h2 = p2 + b2, where h is hypotenuse and p & b are the other two sides of right angled triangle.
Given: A right-angled triangle ABC having angle ACB a right angle.
To Prove: AB2 = AC2 + CB2
Construction: On, AB, describe the square ADEB; on AC the square ACGF and on CB the square CBKH. Through C draw CI parallel to AD or BE cutting AB at j and DE at I. Join CD, FB.
1. BC and CG are in same straight line. (As each of the angle s ACB, ACG is a right angle)
2. angle BAD = angle FAC (As both are right angles)
3. The whole angle CAD = the whole angle FAB. (As adding to each the angled CAB)
4. In the triangle s CAD, FAB
a) CA = FA (As side of square ACGF)
b) AD = AB and the (Side of square ADEB)
c) included angle CAD = included angle FAB
5. The triangle CAD and FAB are congruent and equal in area. (S.A.S property)
6.The area of rectangle ADIJ is double that of the area triangle CAD.(As being the same base AD between the same parallels AD and CI)
7. The area of square GFAC is double of the area triangle FAB.( As being onthe same base FA and between same parallels FA and GB)
8. Therefore the rectangle ADIJ = the square GFAC
9. Similarly, by joining CE, AK, the rectangle BJIE = the square HCBK.
10. Therefore the whole square ADEB = the square GFAC + square HCBK.
11. Therefore AB2 = AC2 + CB2
In general approach:
AC2 = BC2 + AB2
or, (Opposite side)2 + (Adjacent side )2 = (Hypotenuse side)2
AC = `sqrt((BC)^2 + (AB)^2)`
This is called as Pythagoras theorem.
ThePythagoras theorem can be proved with the help of similarity of figures. The converse of Pythagoras theorem is also true. That is if thesquare on one side of a triangle is equal to the sum of the squares on the other two sides, then these sides contain a right angle.
If AC2 = BC2 + AB2 Angle B = 90°.
If X is an acute angle in a right triangle, we have the following Pythagorean identities:
cos 2 X + sin 2 X = 1
sec 2 X - tan 2 X = 1
csc 2 X - cot 2 X = 1
We can write the relations as :
cos 2 X + sin 2 X = 1, cos 2 X = 1- sin 2 X and sin 2 X = 1- cos 2 X
sec 2 X - tan 2 X = 1, sec 2 X = 1+ tan 2 X and tan 2 X = sec 2 X -1
csc 2 X - cot 2 X = 1, csc 2 X = 1 + cot 2 X and cot 2 X = csc 2 X -1
Ex1: Find the length of the unknown side in the following figure?
Sol: By the Pythagoras theorem;
x2 + 122 = 132
x2 = 132 - 122 = 169 - 144
x2 = 25
So, x = 5 cm
Ex2: Find the length of the unknown side in the following figure.
Sol: Step1: x2 = 72 + 242
= 49 + 576 = 625
Step2: x = 25cm = Hypotenuse side.
1.Find the hypotenuse side of a triangle when a="6cm" and b="8cm?
2. If the hypotenuse of a triangle is 5cm and one of the side is 4cm. Find the other side of a triangle.