Question

NCERT class 9 exercise 7.4 Questions 4,5 solution with figure

Answer 1

Please refer to figures above.

Answer 4:

Let us join AC.
Now in ∆ABC, AB < BC
[∵ AB is the smallest side of Quadrilateral ABCD]
=> BC > AB
∴ [Angle Opposite to BC] < [Angle Opposite to AB]
=> ∠BAC > ∠BCA
Again, in ∆ACD,
CD > AD -----------> (1)
[∵ CD is the longest side of the quadrilateral ABCD]
∴ [Angle Opposite to CD] > [Angle Opposite to AD]
=> ∠CAD > ∠ACD -----------> (2)
Adding (1) and (2), we get
[∠BAC + ∠CAD] > [∠BCA + ∠ACD]
=> ∠A > ∠C
Similarly, by joining BD, we have
∠B > ∠D
----------------------------------------------------

Answer 5:

In ∆PQR, PS biscets ∠QPR [Given]
∴ ∠QPS = ∠RPS
∵ PR > PQ [Given]
∴ [Angle Opposite to PR] > [Angles Opposite to PQ]
=> ∠PQS > ∠PRS
=> [∠PQS + ∠QPS] > [∠PRS + ∠RPS] -----------> (1)
[∵ ∠QPS = ∠RPS]
∵ Exterior ∠PSR = [∠PQS + QPS]
[∵ An exterior angle is equal to the sum of interior opposite angles]
And Exterior ∠PSQ = [∠PRS + ∠RPS]
Now, from (1), we have
∠PSR > ∠PSQ
----------------------------------------------------

Hope it helps!
Mark as branliest.

Answer 2

Answer:

please mark as brainliest