Question

"Question 4 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.

Class 9 - Math - Triangles Page 132"

Answer 1

Solution:

Given:
In quadrilateral ABCD, AB smallest & CD is longest sides.

To Prove: ∠A>∠C
& ∠B>∠D

Construction: Join AC.
Mark the angles as shown in the figure..

Proof:
In △ABC , AB is the shortest side.

BC > AB
∠2>∠4 …(i)
[Angle opposite to longer side is greater]

In △ADC , CD is the longest side

CD > AD
∠1>∠3 …(ii)
[Angle opposite to longer side is greater]

Adding (i) and (ii), we have


∠2+∠1>∠4+∠3
⇒∠A>∠C

Similarly, by joining BD, we can prove that
∠B>∠D

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Hope this will help you......

Answer 2

Hi friend☝

ABCD is a quadrilateral such that AB is it's smallest side and CD is it largest side.


Join AC and BD

In ∆ABC ,we have

Bc>Ab

angle 8 >angle 3--------(1)

{Angle opposite to longer side is greater.}


Since CD is the longest side of quadrilateral ABCD

In ∆ACD ,I have

Cd>Ad

angle 7>angle 4---------(2)

{{{{{{. Same reason }}}}}}}}

Adding equ.(1) and(2)

angle 8 +angle 7>angle 3 +angle 4

Angle A> angle C

Again ,I have ......

Ad>Ab {AB I the shortest side}

=angle 1 >angle 6....--------(3)

In ∆BCD,I have

CD>BC

Angle 2 and angle 5....-----(4)


Adding (3) and (4)

angle 1+angle2 >angle 5+angle 6

=angle B>angle D

So,,,Then.....angle A>angleC

Angle B >angle D .

Hope it is helpful.....