NCERT Solution for Triangles 7.4

Exercise 7.4 (Page 132)

1.  Show that in a right angled triangle, the hypotenuse is the longest side

Ans. Let us consider ΔABC such that ∠B = 90°

          ∴ ∠A + ∠B + ∠C = 180°

          ∴ [∠A + ∠C] + ∠B = 180°

          ⇒ ∠A + ∠C = ∠B

          ∴ ∠B > ∠A and ∠B > ∠C

          ⇒ Side opposite to ∠B is longer than the side opposite to ∠A.

          i.e. AC > BC

...(1)

          Similarly, AC > AB

...(2)

          From (1) and (2), we get AC is the longest side.

          But AC is the hypotenuse of the triangle.

          Thus, hypotenuse is the longest side.

2.  In the adjoining figure, sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

Ans. ∠ABC + ∠PBC = 180°

[Linear pair]

            ∠ACB + ∠QCB = 180°

[Linear pair]

            ∴ ∠ABC + ∠PBC = ∠ACB + ∠QCB

            But ∠PBC < ∠QCB

[Given]

            ∴ ∠ABC > ∠ACB

            ⇒ [The side opposite to ∠ABC]

> [The side opposite to ∠ACB]

            ⇒ AC > AB

3.  In the figure, ∠B < ∠A and ∠C < ∠D. Show that AD > BC.

Ans. ∵∠B < ∠A

[Given]

                ⇒ ∠A > ∠B

                ∴ OB > OA

[Side opposite to greater angle is longer]

...(1)

                Similarly,

                OC > OD

...(2)

                From (1) and (2), we have

                [OB + OC] > [OA + OD]

                ⇒ BC > AD

                or AD < BC

4.  AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that ∠A > ∠C and ∠B > ∠D.

Ans. Let us join AC.

                Now, in ΔABC,          AB < BC

                [∵AB is the smallest side of quadrilateral ABCD]

                ⇒ BC > AB

∴ [Angle opposite to BC] < [Angle opposite to AB]

                ⇒ ∠BAC > ∠BCA

...(1)

                Again, in ΔACD,

                CD > AD

[∵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

5.  In the figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

Ans. In ΔPAR, PS bisects ∠QPR

[Given]

                ∴ ∠QPS = ∠RPS

                ∴ PR > PQ

[Given]

∴ [Angle opposite to PR] > [Angle 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

6.  Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Ans. Let us consider the ΔPMN such that ∠M = 90°

                Since, ∠M + ∠N + ∠P = 180°

                [Sum of angles of a triangle]

                ∴ ∠M = 90°

                [∴ PM ⊥ l]

                ⇒ ∠N < ∠M

                ⇒ PM < PN

...(1)

                Similarly, Pm < PN1

...(2)

                PM < PN2

...(3)

                From (1), (2) and (3), we have PM is the smallest line segment drawn from P on the line l. Thus, the perpendicular segment is the shortest line segment drawn on a line from a point not on it.