Advertisements
Advertisements
Question
In ΔPQR is a triangle and S is any point in its interior. Prove that SQ + SR < PQ + PR.
Advertisements
Solution
In ΔPQR,
PQ + PR > QR ....(∵ Sum of two sides of a triangle is always greater than the third side.) ....(i)
Also, in ΔSQR,
SQ + SR > QR ....(∵ Sum of two sides of a triangle is always greater than the third side.) ....(ii)
Dividing (i) by (ii),
`"PQ + PR"/"SQ + SR" > "QR"/"QR"`
`"PQ + PR"/"SQ + SR" > 1`
PQ + PR > SQ + SR
i.e. SQ + SR < PQ + PR.
APPEARS IN
RELATED QUESTIONS
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.
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
In the following figure, write BC, AC, and CD in ascending order of their lengths.
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°.
Arrange the sides of the following triangles in an ascending order:
ΔDEF, ∠D = 38°, ∠E = 58°.
Name the smallest angle in each of these triangles:
In ΔABC, AB = 6.2cm, BC = 5.6cm and AC = 4.2cm
Prove that the perimeter of a triangle is greater than the sum of its three medians.
