Q10 of 39 Page 158

Let’s prove that the hypotenuse of a right angled triangle is the greatest side.

Given: a right - angled triangle


To prove: the hypotenuse is the greatest side


Let ΔABC be the right angled triangle, right angled at B as shown below,



Now Consider the ΔABC,


We know in a triangle the sum of all three interior angles is equal to 180°.


So in this case,


∠ABC + ∠BAC + ∠ACB = 180°


But the given triangle is right angled, right angle at ∠ABC


So when ∠ABC = 90°, then the other two angles of the triangle together will be equal to 90°, as the sum of all three angles of a triangle is equal to 180°.


Hence ∠ABC > ∠BAC or ∠ABC > ∠ACB


But in a triangle we know the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle


Hence AC > BC or AC > AB


Therefore the hypotenuse of a right angled triangle is the greatest side.


Hence proved


More from this chapter

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8

In ΔABC, AB > AC; the bisector of ∠BAC meets BC at D. Let’s cut off a line segment AE equal in length of AC from the side AB. Let’s join D and E. Let’s prove that

(i) ΔACD≅ΔAED


(ii) ∠ACB > ∠ABC.


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In the figure AB = CD, ∠OCD > ∠COD and ∠OAB < ∠AOB. Let’s prove that OB < OD


 11

Let’s prove that the measurement of the opposite side of the obtuse angle of an obtuse angled triangle is the greatest side.

 12

In ΔABC, the bisectors of ∠ABC and ∠ACB meet at I. If AB > AC then let’s prove that IB > IC.