can any one help me to prove the PGT theorm
Answers
Answer:
: In any right angled triangles, the square of the hypotenuse of that triangle will always be equal to the sum of the squares of other two sides.
XYZ is a rt. angled triangle with <Y a right angle and its hypotenuse (XZ) is "a", height (XY) is "b" and base (YZ) is "c"
a² = b² + c²
A square PQRS is constructed with all its sides equal to (b+c). Inside it, a tilted square EFGH is drawn whose all sides are equal to 'a'.
Area of square EFGH = a²
Area of all four right angled triangles formed = 4 × 1/2 × b × c = 2 bc
Area of square PQRS = (b+c)²
Area of square PQRS = Area of square EFGH + Area of four rt. angled triangles
= a² + 2 bc
By condition,
(b+c)² = a² + 2bc
Or, b²+2bc+c² = a²+2bc
Or, a² = b²+c² [PROVED]
THIS IS THE MOST EASY WAY TO PROVE PYTHAGORAS THEOREM.
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Answer:
Step-by-step explanation:
Here is your answer mate..
Pythagoras' theorem :-
→ In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.
Step-by-step explanation:
It's prove :-
➡️ Given :-
➡️A △ABC in which ∠ABC = 90° .
➡️To prove :-
➡️AC² = AB² + BC² .
➡️Construction :-
➡️Draw BD ⊥ AC .
➡️ Proof :-
In △ADB and △ABC , we have
∠A = ∠A ( common ) .
∠ADB = ∠ABC [ each equal to 90° ] .
∴ △ADB ∼ △ABC [ By AA-similarity ] .
➡️AD/AB = AB/AC .
➡️AB² = AD × AC ............(1) .
In △BDC and △ABC , we have
∠C = ∠C ( common ) .
∠BDC = ∠ABC [ each equal to 90° ] .
∴ △BDC ∼ △ABC [ By AA-similarity ]
➡️ DC/BC = BC/AC .
➡️ BC² = DC × AC. ............(2) .
Add in equation (1) and (2) , we get
➡️AB² + BC² = AD × AC + DC × AC .
➡️AB² + BC² = AC( AD + DC ) .
➡️AB² + BC² = AC × AC .
Hope it helps u..