Q. 25 A4.3( 4 Votes )

Show that in a ri

Answer :

Given: ∆ABC is right angled triangle at B, where AC = hypotenuse, AB = base and BC = perpendicular

To prove: (hypotenuse)2 = (perpendicular)2 + (base)2

(AC)2 = (BC)2 + (AB)2

Construction: Draw BD perpendicular to AC.

Proof: Since BD is perpendicular to AC,

A theorem is applicable here, which says that if a perpendicular is drawn from the vertex of a right angle of a right triangle to the hypotenuse, then the triangles formed on either sides of the perpendicular are similar to the whole triangle as well as to each other.

This theorem simply means that, ∆ADB ∆ABC and ∆BDC ∆ABC as well as ∆ADB ∆BDC.

Thus, if ∆ADB ∆ABC, by similar triangle property we can write,

AD.AC = AB2 …(i)

Similarly, if ∆BDC ∆ABC

DC.AC = BC2 …(ii)

Now we need to add equations (i) and (ii), we get

AD.AC + DC.AC = AB2 + BC2

AC × (AD + DC) = AB2 + BC2[ AC is common in AD.AC + DC.AC, AC is taken common]

AC × AC = AB2 + BC2 [ AD + DC is clearly AC from the diagram]

AC2 = AB2 + BC2

Hence, it is proved.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses

In Fig. 3,<Mathematics - Board Papers

In an equilateralRD Sharma - Mathematics