Advertisements
Advertisements
प्रश्न
Prove that in a quadrilateral the sum of all the sides is greater than the sum of its diagonals.
Advertisements
उत्तर
We have to prove that the sum of four sides of quadrilateral is greater than sum of diagonal.

Since the sum of two sides of triangle is greater than third side.
In ΔPQR we have
PQ + QR > PR ..........(1)
In ΔRSPwe have
RS + SP >PR ..........(2)
In ΔPQS we have
PQ + SP > QS ........(3)
In ΔQRSwe have
QR + RS > QS .........(4)
Adding (1) & (2) & (3) and (4) we get
2(PQ + QR + RS + SQ ) >2 (PR + QS)
Hence (PQ + QR + RS + SQ > PR + QS)Proved.
APPEARS IN
संबंधित प्रश्न
ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the given figure). If AD is extended to intersect BC at P, show that
- ΔABD ≅ ΔACD
- ΔABP ≅ ΔACP
- AP bisects ∠A as well as ∠D.
- AP is the perpendicular bisector of BC.

AD is an altitude of an isosceles triangles ABC in which AB = AC. Show that
- AD bisects BC
- AD bisects ∠A
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see the given figure). Show that:
- ΔABM ≅ ΔPQN
- ΔABC ≅ ΔPQR

ABC is an isosceles triangle with AB = AC. Drawn AP ⊥ BC to show that ∠B = ∠C.
In two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.
In the following figure, BA ⊥ AC, DE ⊥ DF such that BA = DE and BF = EC. Show that ∆ABC ≅ ∆DEF.

Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side.
In a right triangle, prove that the line-segment joining the mid-point of the hypotenuse to the opposite vertex is half the hypotenuse.
Line segment joining the mid-points M and N of parallel sides AB and DC, respectively of a trapezium ABCD is perpendicular to both the sides AB and DC. Prove that AD = BC.
ABCD is a quadrilateral such that diagonal AC bisects the angles A and C. Prove that AB = AD and CB = CD.
