Advertisements
Advertisements
प्रश्न
In ΔPQR is a triangle and S is any point in its interior. Prove that SQ + SR < PQ + PR.
उत्तर
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
संबंधित प्रश्न
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.
Arrange the sides of ∆BOC in descending order of their lengths. BO and CO are bisectors of angles ABC and ACB respectively.
In the following figure, ∠BAC = 60o and ∠ABC = 65o.
Prove that:
(i) CF > AF
(ii) DC > DF
Arrange the sides of the following triangles in an ascending order:
ΔABC, ∠A = 45°, ∠B = 65°.
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 ABC, P, Q and R are points on AB, BC and AC respectively. Prove that AB + BC + AC > PQ + QR + PR.
ABCD is a trapezium. Prove that:
CD + DA + AB + BC > 2AC.
ABCD is a trapezium. Prove that:
CD + DA + AB > BC.
In ΔABC, AE is the bisector of ∠BAC. D is a point on AC such that AB = AD. Prove that BE = DE and ∠ABD > ∠C.