Advertisements
Advertisements
प्रश्न
For any quadrilateral, prove that its perimeter is greater than the sum of its diagonals.
उत्तर
Given: PQRS is a quadrilateral.PR and QS are its diagonals.
To Prove: PQ + QR + SR + PS > PR + QS
Proof: In ΔPQR
PQ + QR > PR ...(Sum of two sides of triangle is greater than the third side)
Similarly, In ΔPSR, PS + SR > PR
In ΔPQS, PS + PQ > QS and in QRS we have QR + SR > QS
Now we have
PQ +QR > PR
PS + SR > PR
PS + PQ > QS
QR + SR > QS
After adding above inequalities we get
2(PQ + QR + PS + SR) > 2(PR + QS)
⇒ PQ + QR + PS + SR > PR +QS.
APPEARS IN
संबंधित प्रश्न
In the given figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR >∠PSQ.
ABC is a triangle. Locate a point in the interior of ΔABC which is equidistant from all the vertices of ΔABC.
In a triangle PQR; QR = PR and ∠P = 36o. Which is the largest side of the triangle?
Arrange the sides of ∆BOC in descending order of their lengths. BO and CO are bisectors of angles ABC and ACB respectively.
D is a point in side BC of triangle ABC. If AD > AC, show that AB > AC.
In the following figure ; AC = CD; ∠BAD = 110o and ∠ACB = 74o.
Prove that: BC > CD.
Name the greatest and the smallest sides in the following triangles:
ΔXYZ, ∠X = 76°, ∠Y = 84°.
Name the smallest angle in each of these triangles:
In ΔPQR, PQ = 8.3cm, QR = 5.4cm and PR = 7.2cm
In ABC, P, Q and R are points on AB, BC and AC respectively. Prove that AB + BC + AC > PQ + QR + PR.
In ΔPQR is a triangle and S is any point in its interior. Prove that SQ + SR < PQ + PR.