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.
RELATED QUESTIONS
In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

In a huge park people are concentrated at three points (see the given figure):

A: where there are different slides and swings for children,
B: near which a man-made lake is situated,
C: which is near to a large parking and exit.
Where should an ice-cream parlour be set up so that maximum number of persons can approach it?
(Hint: The parlor should be equidistant from A, B and C)
In a triangle PQR; QR = PR and ∠P = 36o. Which is the largest side of the triangle?
Name the greatest and the smallest sides in the following triangles:
ΔABC, ∠ = 56°, ∠B = 64° and ∠C = 60°.
Arrange the sides of the following triangles in an ascending order:
ΔABC, ∠A = 45°, ∠B = 65°.
Name the smallest angle in each of these triangles:
In ΔABC, AB = 6.2cm, BC = 5.6cm and AC = 4.2cm
Name the smallest angle in each of these triangles:
In ΔXYZ, XY = 6.2cm, XY = 6.8cm and YZ = 5cm
In the given figure, ∠QPR = 50° and ∠PQR = 60°. Show that : PN < RN
In ΔPQR is a triangle and S is any point in its interior. Prove that SQ + SR < PQ + PR.
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.
