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प्रश्न
For any quadrilateral, prove that its perimeter is greater than the sum of its diagonals.
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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.
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.
In the given figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR >∠PSQ.
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
From the following figure, prove that: AB > CD.
Name the greatest and the smallest sides in the following triangles:
ΔDEF, ∠D = 32°, ∠E = 56° and ∠F = 92°.
Prove that the hypotenuse is the longest side in a right-angled triangle.
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.
In ΔPQR, PS ⊥ QR ; prove that: PQ > QS and PR > PS
