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Prove that: In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Prove that: In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Grade:12th pass

1 Answers

Pawan Prajapati
askIITians Faculty 60787 Points
2 years ago
Consider a right angled triangle ABC, right angled at B and construct a perpendicular to the side AC passing through B. Use the properties related to similarity of triangles to reach a result which might help you to conclude the mentioned statement. Complete step-by-step answer: First, we will draw a right angled triangle ABC, right angled at B and construct a perpendicular to the side AC passing through B. Now in triangle ADB and triangle ABC , we have Angle A is common between the two triangles, according to our construction ∠ADB=∠ABC=90∘ . So, by AA similarity criteria, we can say that ΔADB∼ΔABC . Now we know that the ratio of corresponding sides of two similar triangles is constant. ∴ABAD=ACAB ⇒(AB)2=AC×AD..........(i) Now in triangle BDC and triangle ABC , we have Angle B is common between the two triangles, according to our construction ∠BDC=∠ABC=90∘ . So, by AA similarity criteria, we can say that ΔBDC∼ΔABC . Now we know that the ratio of corresponding sides of two similar triangles is constant. ∴DCBC=BCAC ⇒(BC)2=AC×DC..........(ii) Now, if we add equation (i) and (ii), we get (BC)2+(AB)2=AC×DC+AC×AD (BC)2+(AB)2=AC(DC+AD) Now from the figure, we can see that DC+AD=AC . So, our equation becomes: (BC)2+(AB)2=AC×AC ⇒(BC)2+(AB)2=(AC)2 So, we have proved that, in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Note: The main part in the above solution is the construction of the line BD. Also, it might be difficult to see the similar triangles in case of the above question. However, it is very important that you remember the Pythagoras theorem, as it is used in almost every question related to right angled triangles.

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