Previous exercise – Triangles – Exercise 7.3 – Class IX

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

Solution:

Let ABC is aright angled triangle, right angle at B i.e, ∟B = 90˚

Now, ∠A + ∠C = 90˚

Since the sum of ∠A and ∠B is equal to 90˚, then each ∠A and ∠B is less than 90˚

Therefore, ∠B > ∠A ⇒ AC > BC ……………..(1)

and ∠B > ∠C ⇒ AC > AB………………….(2)

Hence, (1) and (2), we have,

AC is greater than both BC and AB.

i.e, hypotenuse is greater than other two sides of a right angled triangle.

⇒ hypotenuse is the longest side of a right angled triangle.

**In Fig. 7.48 sides AB and AC of ∆ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB**

Solution:

∠ABC + ∠PBC = 180˚ (linear pair)

⇒∠ABC = 180˚ – ∠PBC ……………(1)

Similarly, ∠ACB = 180˚ – ∠QCB ……………..(2)

It is given that, ∠PBC < ∠QCB

⸫ 180˚ – ∠QCB < 180˚ – ∠PBC

i.e, ∠ACB < ∠ABC [from(1) and (2)]

⇒AB < AC

Hence proved.

**In the fig7.49. ∠B < ∠A and ∠C < ∠D. Show that AD < BC**

Solution:

Given, ∠B < ∠A

i.e, OA < OB ………………………(1) [sides opposite to greater angle is greater]

It is also given that, ∠C < ∠D

i.e, OD < OC……………………….(2) [sides opposite to greater angle is greater]

Adding (i) and (ii),

OA + OD < OB + OC

AD < BC

**AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD. Show that ∠A > ∠C and ∠B > ∠D**

Solution:

Given, ABCD is a quadrilateral, AB is the smallest and CD is the longest side of the quadrilateral ABCD.

We have to show, ∠A > ∠C and ∠B > ∠D

Construction: Draw a line joining A and C .

In ∆ABC , AB < BC [AB is the shortest side & angle opposite to longer side is greater]

∠ACB < ∠BAC ………….(1)

In ∆ADC, AD < CD[CD is the longest side & angle opposite to longer side is greater]

∠ACD < ∠CAD …………….(2)

From (1) and (2),

∠ACB + ∠ACD < ∠BAC + ∠CAD

∠BCD < ∠BAD

Similarly, we can draw a line joining B and D to prove, ∠ADC < ∠ABC.

**In fig PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ**

Solution:

Given, PR > PQ and PS bisects ∠QPR

We have show that, ∠PSR > ∠PSQ

Since, PR > PQ

⇒∠PQR > ∠PRQ …………….(1) [angles opposite to the greater line the greater angle]

Since PS bisects ∠QPR, we have,

∠QPS = ∠RPS ……………..(2)

∠PSR is the exterior angle of ∆PQS, then,

∠PSR = ∠PQR + ∠QPS ………………..(3)

∠PSQ is the exterior angle of ∆PRS, then,

∠PSQ = ∠SPR + ∠PRQ ……………(4)

Adding (1) and (2),

∠PQR + ∠QPS > ∠PRQ + ∠RPS………………..(5)

⇒∠PSR > ∠PSQ *[from (5), (3) and (4)]*

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

Solution:

Let the point be O. Line segments from O are OP, OQ, OR and OS, where OP is perpendicular line segment *l*.

We have to prove, OP is the shortest *l*.

i.e., OP<OQ , OP < OR and OP < OS

In ∆OPQ, ∠P = 90˚

⸫ ∠Q is an acute angle (i.e, ∠Q = 90˚)

⸫ ∠Q < ∠P

Hence, OP < OQ [side opposite to greater angle is longer]

Similarly, we can prove that OP is shorter than OR , OS.

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