Q2.State and prove Pythagoras theorem.
Answers
Answer:
Hypotenuse² = Perpendicular² + Base²
c² = a² + b²
Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides“. The sides of this triangle have been named as Perpendicular, Base and Hypotenuse. Here, the hypotenuse is the longest side, as it is opposite to the angle 90°.
Step-by-step explanation:
Proof of Pythagoras Theorem
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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 = b.
Verification of Pythagorean Theorem
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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
Proof of Pythagorean Theorem using Algebra:
Proof of Pythagorean TheoremGiven: A ∆ XYZ in which ∠XYZ = 90°.
To prove: XZ2 = XY2 + YZ2
Construction: Draw YO ⊥ XZ
Proof: In ∆XOY and ∆XYZ, we have,
∠X = ∠X → common
∠XOY = ∠XYZ → each equal to 90°
Therefore, ∆ XOY ~ ∆ XYZ → by AA-similarity
⇒ XO/XY = XY/XZ
⇒ XO × XZ = XY2 ----------------- (i)
In ∆YOZ and ∆XYZ, we have,
∠Z = ∠Z → common
∠YOZ = ∠XYZ → each equal to 90°
Therefore, ∆ YOZ ~ ∆ XYZ → by AA-similarity
⇒ OZ/YZ = YZ/XZ
⇒ OZ × XZ = YZ2 ----------------- (ii)
From (i) and (ii) we get,
XO × XZ + OZ × XZ = (XY2 + YZ2)
⇒ (XO + OZ) × XZ = (XY2 + YZ2)
⇒ XZ × XZ = (XY2 + YZ2)
⇒ XZ 2 = (XY2 + YZ2)
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Answer:
Pythagoras theorem states that “ In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”. The sides of the right-angled triangle are called base, perpendicular and hypotenuse .
refer to the image for proof
