Advertisements
Advertisements
प्रश्न
In ΔPQR is a triangle and S is any point in its interior. Prove that SQ + SR < PQ + PR.
Advertisements
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
Show that in a right angled triangle, the hypotenuse is the longest side.
In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.
In the given figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR >∠PSQ.
D is a point in side BC of triangle ABC. If AD > AC, show that AB > AC.
ABCD is a quadrilateral in which the diagonals AC and BD intersect at O. Prove that AB + BC + CD + AD < 2(AC + BC).
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 ΔPQR, PS ⊥ QR ; prove that: PQ > QS and PR > PS
In ΔABC, D is a point in the interior of the triangle. Prove that DB + DC < AB + AC.
