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Prove the pythagorean theorom i.e. perpendiclar square + base square = hypotenuse square?
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Prove the pythagorean theorom i.e. perpendiclar square + base square =...
Pythagoras theorem:
statement:in a right triangle,the square of the hypotenuse is equal to the sum of the squares of the other two sides.
construct a right triange right angled at B.
construction:construct a perpendicular BD on side AC.
given:angleB=90
to prove:AB^2+BC^2=AC^2
proof: in triangle ABD and tri ABC,
angle A=angle A
angle ADB=angleBDC=90
therefore,triADB is similar to triABC.
which implies AD/AB=AB/AC (sides are in proportion)
which implies AD*AC=AB^2-------1
IIIly, tri BDC is similar to tri ABC
BC/DC=AB/BC (sides are in proportion)
which implies AC*DC=BC^2-----2
add 1 and 2
AD.AC=AB^2
AD.AC+AC.DC=AB^2+BC^2
=AC(AD+DC)=AB^2+BC^2
=AC^2=AB^2+BC^2
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Prove the pythagorean theorom i.e. perpendiclar square + base square = hypotenuse square?
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Prove the pythagorean theorom i.e. perpendiclar square + base square = hypotenuse square? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Prove the pythagorean theorom i.e. perpendiclar square + base square = hypotenuse square? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Prove the pythagorean theorom i.e. perpendiclar square + base square = hypotenuse square?.
Prove the pythagorean theorom i.e. perpendiclar square + base square = hypotenuse square? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Prove the pythagorean theorom i.e. perpendiclar square + base square = hypotenuse square? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Prove the pythagorean theorom i.e. perpendiclar square + base square = hypotenuse square?.
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