Advertisements
Advertisements
प्रश्न
D is a point on the side of the BC of ΔABC. Prove that the perimeter of ΔABC is greater than twice of AD.
Advertisements
उत्तर

Construction: Join AD
In triangle ACD,
AC + CD > AD ...(i)
(Sum of two of a triangle greater than the third side)
Similarly, in triangle ADB,
AB + BD > AD ...(ii)
Adding (i) and (ii),
AC + CD + AB + BD > AD
AB + BC + AC > 2AD. ...(Since, CD + BD = BC)
APPEARS IN
संबंधित प्रश्न
Show that in a right angled triangle, the hypotenuse is the longest side.
In the given figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR >∠PSQ.

From the following figure, prove that: AB > CD.

If two sides of a triangle are 8 cm and 13 cm, then the length of the third side is between a cm and b cm. Find the values of a and b such that a is less than b.
In the following figure, ∠BAC = 60o and ∠ABC = 65o.

Prove that:
(i) CF > AF
(ii) DC > DF
Name the greatest and the smallest sides in the following triangles:
ΔABC, ∠ = 56°, ∠B = 64° and ∠C = 60°.
In ΔABC, the exterior ∠PBC > exterior ∠QCB. Prove that AB > AC.
Prove that the perimeter of a triangle is greater than the sum of its three medians.
ABCD is a trapezium. Prove that:
CD + DA + AB + BC > 2AC.
In ΔPQR is a triangle and S is any point in its interior. Prove that SQ + SR < PQ + PR.
