prove that in a right triangle the square on the hypoteneuse is equal to the sum of squares on the other two sides
Answers
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aim : to Prove that in a right triangle the square on the hypoteneuse is equal to the sum of squares on the other two sides
procedure : construct a right triange right angled at B.
construction: construct a perpendicular BD on side of AC
given: angleB= 90 degrees
to prove: AB2+BC2=AC2
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=AB2 eq1.
Similarly, tri BDC is similar to tri ABC
BC/DC=AB/BC (sides are in proportion)
which implies AC*DC=BC2 eq-2
add 1 and 2
AD.AC=AB2
AD.AC+AC.DC=AB2+BC2
=AC(AD+DC)=AB2+BC2
=AC2=AB2+BC2
hence it is proved
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AC²=AB²+BC²
Given: A right angled ∆ABC, right angled at B
To Prove: AC²=AB²+BC²
Construction: Draw perpendicular BD onto the side AC .
Proof:
We know that if a perpendicular is drawn from the vertex of a right angle of a right angled triangle to the hypotenuse, than triangles on both sides of the perpendicular are similar to the whole triangle and to each other.
We have
△ADB∼△ABC. (by AA similarity)
Therefore, AD/ AB=AB/AC
(In similar Triangles corresponding sides are proportional)
AB²=AD×AC……..(1)
Also, △BDC∼△ABC
Therefore, CD/BC=BC/AC
(in similar Triangles corresponding sides are proportional)
Or, BC²=CD×AC……..(2)
Adding the equations (1) and (2) we get,
AB²+BC²=AD×AC+CD×AC
AB²+BC²=AC(AD+CD)
( From the figure AD + CD = AC)
AB²+BC²=AC . AC
Therefore, AC²=AB²+BC²
Hope this helps you!!!