state Pythagoras theoram and prove it
Answers
Answer:
The proof of Pythagorean Theorem in mathematics is very important. In a right angle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. States that in a right triangle that, the square of a (a2) plus the square of b (b2) is equal to the square of c (c2).
Step-by-step explanation:
Let QR = a, RP = b and PQ = c. Now, draw a square WXYZ of side (b + c). Take points E, F, G, H on sides WX, XY, YZ and ZW respectively such that WE = XF = YG = ZH = Then, we will get 4 right-angled triangle, hypotenuse of each of them is ‘a’: remaining sides of each of them are band c. Remaining part of the figure is the
square EFGH, each of whose side is a, so area of the square EFGH is a2.
Now, we are sure that square WXYZ = square EFGH + 4 ∆ GYF
or, (b + c)2 = a2 + 4 ∙ 1/2 b ∙ c
or, b2 + c2 + 2bc = a2 + 2bc
or, b2 + c2 = a2
Statement :
›»› For any right angled traingle with sides a and b and hypotenuse, The square of the hypotenuse is equal to the sum of the square of the other two sides.
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Given :
➤ ABC is a triangle in which ∠ABC=90°
To Prove :
➤ AB² + BC² = AC²
Construction :
➤ Draw BD⊥AC.
Proof :
In ∆ADB and ∆ABC
⪼ ∠A = ∠A [Common angle]
⪼ ∠ADB = ∠ABC [Each 90°]
⪼∆ADB ∼ ∆ABC [AA Criteria]
∆ADB and ∆ABC are similar
So ,
⇒ (AB/AD) = (AC/AB)
⇒ AB² = AD × AC ..........❶
∆BDC and ∆ABC are similar
So ,
⇒ (CD/BC) = (BC/AC)
⇒ BC² = CD × AC ..........❷
Adding equations ❶ and ❷ we get,
⇒ AB² + BC² = AD × AC + CD × AC
⇒ AB² + BC² = AC(AD + CD)
⇒ AB² + BC² = AC × AC
⇒ AB² + BC² = AC²