AB and CD are the smallest and largest sides of a quadrilateral ABCD. Out of ∠B and ∠D decide which is greater.
Solution:
Given, ABCD is a quadrilateral.
AB and CD are the smallest and largest sides of a quadrilateral.
We have to find the greater angle among ∠B and ∠D.
Join the diagonal BD of the quadrilateral.
In triangle ABD,
Given, AB is the smallest side of ABCD.
So, AD > AB
We know that in a triangle angle opposite to the larger side is greater.
So, ∠1 > ∠3 ------------ (1)
Given, CD is the largest side of ABCD
So, CD > BC
We know that in a triangle angle opposite to the larger side is greater.
So, ∠2 > ∠4 ----------------- (2)
Adding (1) and (2),
∠1 + ∠2 > ∠3 + ∠4
From the figure,
∠B = ∠1 + ∠2
∠D = ∠3 + ∠4
Now, ∠B > ∠D
Therefore, it is proven that angle B is greater than angle D.
✦ Try This: Diagonals of a quadrilateral ABCD bisect each other. If ∠A=45 degrees, then ∠B is equal to
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 7
NCERT Exemplar Class 9 Maths Exercise 7.4 Problem 19
AB and CD are the smallest and largest sides of a quadrilateral ABCD. Out of ∠B and ∠D decide which is greater
Summary:
AB and CD are the smallest and largest sides of a quadrilateral ABCD. Out of ∠B and ∠D, angle B is greater than angle D
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