Advertisements
Advertisements
प्रश्न
Prove that the hypotenuse is the longest side in a right-angled triangle.
उत्तर
Let us consider a right angled triangle ABC, right angle at B.
In ΔABC
∠A + ∠B +∠C = 180° ...(angle sum property of a triangle)
∠A + 90° + ∠C = 180°
∠A +∠C = 90°
Hence, the other two angles have to be acute (i.e. less than 90°).
∴ ∠B is the largest angle in ΔABC.
⇒ ∠B > ∠A and ∠B > ∠C
⇒ AC > BC and AC > AB
[In ant triangle, the side opposite to the larger (greater) angle is longer]
So, Ac is the largest side in ΔABC.
But AC is the hypotenuse of ΔABC. Therefore, hypotenuse is the longest side in a right angled triangle.
APPEARS IN
संबंधित प्रश्न
In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.
In the following figure ; AC = CD; ∠BAD = 110o and ∠ACB = 74o.
Prove that: BC > CD.
Name the greatest and the smallest sides in the following triangles:
ΔDEF, ∠D = 32°, ∠E = 56° and ∠F = 92°.
Arrange the sides of the following triangles in an ascending order:
ΔABC, ∠A = 45°, ∠B = 65°.
In ΔABC, the exterior ∠PBC > exterior ∠QCB. Prove that AB > AC.
For any quadrilateral, prove that its perimeter is greater than the sum of its diagonals.
In ΔPQR, PR > PQ and T is a point on PR such that PT = PQ. Prove that QR > TR.
In the given figure, ∠QPR = 50° and ∠PQR = 60°. Show that : PN < RN
In ΔABC, BC produced to D, such that, AC = CD; ∠BAD = 125° and ∠ACD = 105°. Show that BC > CD.
Prove that in an isosceles triangle any of its equal sides is greater than the straight line joining the vertex to any point on the base of the triangle.