Advertisements
Advertisements
Question
In ABC, P, Q and R are points on AB, BC and AC respectively. Prove that AB + BC + AC > PQ + QR + PR.
Advertisements
Solution
In triangle APR,
AP + AR > PR ......(i)
In triangle BPQ,
BQ + PB > PQ .......(ii)
In triangle QCR,
QC + CR > QR .......(iii)
Adding (i), (ii) and (iii)
AP + AR + BQ + PB + QC + CR > PR + PQ + QR
(AP + PB) + (BQ + QC) + (CR + AR) > PR + QR + PQ)
⇒ AB + BC + AC > PQ + QR + PR.
APPEARS IN
RELATED QUESTIONS
Show that in a right angled triangle, the hypotenuse is the longest side.
In the given figure sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see the given figure). Show that ∠A > ∠C and ∠B > ∠D.
Name the greatest and the smallest sides in the following triangles:
ΔABC, ∠ = 56°, ∠B = 64° and ∠C = 60°.
Name the greatest and the smallest sides in the following triangles:
ΔXYZ, ∠X = 76°, ∠Y = 84°.
In ΔABC, the exterior ∠PBC > exterior ∠QCB. Prove that AB > AC.
ΔABC is isosceles with AB = AC. If BC is extended to D, then prove that AD > AB.
Prove that the hypotenuse is the longest side in a right-angled triangle.
D is a point on the side of the BC of ΔABC. Prove that the perimeter of ΔABC is greater than twice of AD.
In the given figure, ∠QPR = 50° and ∠PQR = 60°. Show that : PN < RN
