State and prove Pythagoras Theorem
Answers
Answer:
Hey mate,here is your answer:-)
Step-by-step explanation:
Statement: In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Given: ABC is a triangle in which \angle ABC={90}^{\circ}∠ABC=90
∘
Construction: Draw BD\bot ACBD⊥AC.
Proof:
In \triangle ADB△ADB and \triangle ABC△ABC
\angle A=\angle A∠A=∠A [Common angle]
\angle ADB=\angle ABC∠ADB=∠ABC [Each {90}^{\circ}90
∘
]
\triangle ADB \sim \triangle ABC△ADB∼△ABC [A-AA−A Criteria]
So, \dfrac {AD}{AB}=\dfrac{AB}{AC}
AB
AD
=
AC
AB
Now, {AB}^{2}=AD\times ACAB
2
=AD×AC ..........(1)
Similarly,
{BC}^{2}=CD\times ACBC
2
=CD×AC ..........(2)
Adding equations (1) and (2) we get,
{AB}^{2}+{BC}^{2}=AD\times AC+CD\times ACAB
2
+BC
2
=AD×AC+CD×AC
=AC(AD+CD)=AC(AD+CD)
=AC\times AC=AC×AC
\therefore{AB}^{2}+{BC}^{2}={AC}^{2}∴AB
2
+BC
2
=AC
2
[hence \, proved]
Step-by-step explanation:
Step-by-step explanation:hope this will help u
Step-by-step explanation:hope this will help u PLEASE..........MARK. ME. AS. BRILLIANTLIST
