Prove that in a right angled triangle hypotenuse is equal to the sum of square of other two sides
Answers
Answer:
Step-by-step explanation:
Answer:
To prove:AB²+BC²=AC²
Step-by-step explanation:
- Let the triangle be ΔABC,∠B=90°
- now,draw BD⊥AC.
By theorem 6.7(in textbook);i.e;if a perpendicular is drawn form the vertex of the right angle of a right triangle to the hypotenuse, then the triangles on both sides of the perpendicular are similar to the whole triangle and to each other,
ΔADB≈ΔABC ('≈' means similar )
so, AD/AB=AB/AC.................[REASON:In similar triangles,sides are
proportional ]
by cross multiplying, AB²= AC.BD ....................eq.1
Similarly, by the same theorem ,ΔBCD≈ΔABC
so,CD/BC=BC/AC...........................[same reason]
by cross-multiplying, BC²=AC.CD.....................eq.2
Adding eq.1 and eq.2;
AB²+BC²=AC.BD+AC.CD
AB²+BC²=AC(BD+CD)
AB²+BC²=AC.AC.................................since BD+CD=AC
AB²+BC²=AC²
Hence, proved.
Hope you would like the answer.
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!!!