Chapter 9: Triangles — Exercise 9.4 | ML Aggarwal Solutions Understanding ICSE Mathematics Class 9

Exercise 9.4

Question 1

In △PQR, ∠P = 70° and ∠R = 30°. Which side of this triangle is longest? Give reason for your answer.

Answer

We know that,

Sum of angles of triangle = 180°.

∴ ∠P + ∠Q + ∠R = 180°

⇒ 70° + ∠Q + 30° = 180°

⇒ ∠Q + 100° = 180°

⇒ ∠Q = 80°.

We know that side opposite to the greatest angle is greatest side.

∴ PR is the greatest side, which is opposite to ∠Q.

Hence, PR is the greatest side because ∠Q = 80° and side opposite to the greatest angle is greatest side.

Question 2

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

Answer

In △ABC,

Show that in a right angled triangle, the hypotenuse is the longest side. Triangles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

∠B = 90°

∠A and ∠C are acute angles i.e. less that 90°.

∴ ∠B is the greatest angle.

∴ ∠B > ∠A and ∠B > ∠C

⇒ AC > BC and AC > AB.

Hence, proved that hypotenuse is the longest side in a right angled triangle.

Question 3

PQR is a right angle triangle at Q and PQ : QR = 3 : 2. Which is the least angle?

Answer

Given,

PQR is a right angle triangle at Q and PQ : QR = 3 : 2.

PQ = 3x and QR = 2x.

Thus, QR is the smallest side and the angle opposite to the smallest side is smallest angle.

∴ ∠P is the least angle.

Hence, ∠P is the least angle.

Question 4

In △ABC, AB = 8 cm, BC = 5.6 cm and CA = 6.5 cm. Which is

(i) the greatest angle?

(ii) the smallest angle?

Answer

In △ABC, AB = 8 cm, BC = 5.6 cm and CA = 6.5 cm. Which is the greatest angle? Triangles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

(i) In △ABC,

AB is the greatest side, the greatest side has greatest angle opposite to it.

∴ ∠C is the greatest angle.

Hence, ∠C is the greatest angle.

(ii) In △ABC,

BC is the smallest side, the smallest side has smallest angle opposite to it.

∴ ∠A is the smallest angle.

Hence, ∠A is the smallest angle.

Question 5

In a △ABC, ∠A = 50°, ∠B = 60°. Arrange the sides of the triangle in ascending order.

Answer

Sum of angles of triangle = 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ 50° + 60° + ∠C = 180°

⇒ 110° + ∠C = 180°

⇒ ∠C = 70°.

So we get,

∠C > ∠B > ∠A.

We know that side opposite to greatest angle is greatest.

∴ AB > CA > BC or BC < CA < AB.

Hence, sides of the triangle in ascending order are BC < CA < AB.

Question 6

In figure given alongside, ∠B = 30°, ∠C = 40° and the bisector of ∠A meets BC at D. Show that

(i) BD > AD

(ii) DC > AD

(iii) AC > DC

(iv) AB > BD.

In figure given alongside, ∠B = 30°, ∠C = 40° and the bisector of ∠A meets BC at D. Show that BD > AD, DC > AD. Triangles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Answer

In △ABC,

∠B = 30° and ∠C = 40°

∠BAC = 180° - (30° + 40°) = 180° - 70° = 110°.

As, AD is bisector of ∠A,

∴ ∠BAD = ∠CAD = $\dfrac{110°}{2} = 55°$ .

(i) Now in △ABD,

∠BAD > ∠B

∴ BD > AD.

Hence, proved that BD > AD.

(ii) Now in △ACD,

∠CAD > ∠C

∴ DC > AD.

Hence, proved that DC > AD.

(iii) Now in △ACD,

∠ADC = 180° - (40° + 55°) = 180° - 95° = 85°.

In △ACD,

∠ADC > ∠CAD

∴ AC > DC.

Hence, proved that AC > DC.

(iv) In △ADB,

∠ADB = 180° - ∠ADC = 180° - 85° = 95°.

Thus, ∠ADB > ∠BAD

∴ AB > BD.

Hence, proved that AB > BD.

Question 7

In the adjoining figure, AD bisects ∠A. Arrange AB, BD and DC in the descending order of their lengths.

Answer

∠A = 180° - 60° - 40° = 80°.

Since, AD bisects ∠A,

∠BAD = ∠DAC = $\dfrac{80°}{2}$ = 40°.

∴ ∠ADB = ∠DAC + ∠C (As exterior angle is equal to sum of two opposite interior angles.)

∠ADB = 40° + 40° = 80°.

∴ In △ABD,

∠BAD < ∠ABD < ∠ADB

⇒ BD < AD < AB (side opp. to smaller angle is smaller) .......(i)

Also in △ADC, ∠DAC = 40° = ∠C ⇒ AD = DC .......(ii)

∴ BD < DC < AB or AB > DC > BD.

Hence, sides in descending order are AB > DC > BD.

Question 8(a)

In the figure (1) given below, prove that (i) CF > AF (ii) DC > DF.

Answer

(i) In △ABC,

⇒ ∠A + ∠B + ∠C = 180°

⇒ 60° + 65° + ∠C = 180°

⇒ ∠C = 180° - 125° = 55°.

In △BEC,

⇒ ∠B + ∠E + ∠C = 180°

⇒ 65° + 90° + ∠C = 180°

⇒ ∠C = 180° - 155° = 25°.

In △DFC,

⇒ ∠D + ∠C + ∠F = 180°

⇒ 90° + 25° + ∠F = 180°

⇒ ∠F = 180° - 115° = 65°.

From figure,

∠AFE = ∠DFC = 65° (Vertically opposite angles are equal.)

An exterior angle is equal to the sum of two opposite interior angles.

⇒ ∠AFE = ∠FAC + ∠FCA

⇒ 65° = ∠FAC + 30°

⇒ ∠FAC = 65° - 30° = 35°.

In △AFC,

∠FAC = 35°

∠FCA = 30°

∴ ∠FAC > ∠FCA

∴ CF > AF (As side opposite to greater angle is greater.)

Hence, proved that CF > AF.

(ii) In △DFC,

∠DFC = 65°

∠DCF = 25°

∴ ∠DFC > ∠DCF

∴ DC > DF (As side opposite to greater angle is greater.)

Hence, proved that DC > DF.

Question 8(b)

In the figure (2) given below, AB = AC. Prove that AB > CD.

Answer

Since, AB = AC.

∴ ∠ABC = ∠ACB = 70° (As angles opposite to equal sides of an isosceles triangle are equal.)

From figure,

⇒ ∠ACB + ∠ACD = 180°

⇒ 70° + ∠ACD = 180°

⇒ ∠ACD = 110°.

In △ACD,

⇒ ∠CAD + ∠ADC + ∠ACD = 180°

⇒ ∠CAD + 40° + 110° = 180°

⇒ ∠CAD + 150° = 180°

⇒ ∠CAD = 30°.

In △ACD,

∠ADC = 40°

∠CAD = 30°

∴ ∠ADC > ∠CAD

∴ AC > CD (As side opposite to greater angle is greater.)

Since, AB = AC,

∴ AB > CD.

Hence, proved that AB > CD.

Question 8(c)

In the figure (3) given below, AC = CD. Prove that BC < CD.

Answer

From figure,

⇒ ∠ACB + ∠ACD = 180°

⇒ 70° + ∠ACD = 180°

⇒ ∠ACD = 110°.

In △ACD,

AC = CD

∴ ∠CAD = ∠CDA (As angles opposite to equal sides are equal.)

Let ∠CAD = ∠CDA = x.

In △ACD,

⇒ ∠CAD + ∠CDA + ∠ACD = 180°

⇒ x + x + 110° = 180°

⇒ 2x + 110° = 180°

⇒ 2x = 70°

⇒ x = 35°.

⇒ ∠CAD = 35°.

From figure,

⇒ ∠CAD + ∠CAB = 70°

⇒ 35° + ∠CAB = 70°

⇒ ∠CAB = 35°

In △ABC,

⇒ ∠ABC + ∠ACB + ∠BAC = 180°

⇒ ∠ABC + 70° + 35° = 180°

⇒ ∠ABC + 105° = 180°

⇒ ∠ABC = 75°.

In △ABC,

∠ABC > ∠BAC

⇒ AC > BC (As side opposite to greater angle is greater)

∵ AC = CD

∴ CD > BC or BC < CD.

Hence, proved that BC < CD.

Question 9(a)

In the figure (i) given below, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Answer

Given,

∠B < ∠A

∴ AO < BO (As side opposite to smaller angle is smaller.) ......(i)

∠C < ∠D

∴ OD < OC (As side opposite to smaller angle is smaller.) .......(ii)

From (i) and (ii) it can be concluded,

⇒ AO + OD < BO + OC

⇒ AD < BC.

Hence, proved that AD < BC.

Question 9(b)

In the figure (ii) given below, D is any point on the side BC of △ABC. If AB > AC, show that AB > AD.

In the figure (ii), D is any point on the side BC of △ABC. If AB > AC, show that AB > AD. Triangles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Answer

Given,

AB > AC

∴ ∠ACB > ∠ABC (As angle opposite to greater side is greater.)

From figure,

∠ADB = ∠ACB + ∠DAC (As exterior angle is equal to sum of two opposite interior angles.)

⇒ ∠ADB > ∠ACB

⇒ ∠ADB > ∠ABC [∵ ∠ACB > ∠ABC]

∴ AB > AD (As side opposite to greater angle is greater.)

Hence, proved that AB > AD.

Question 10(i)

Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm? Give reason for your answer.

Answer

We know that sum of any two sides is greater than the third side.

Here,

4 cm + 3 cm = 7 cm, which is equal to third side.

Hence, construction of triangle with sides 4 cm, 3 cm and 7 cm is not possible.

Question 10(ii)

Is it possible to construct a triangle with lengths of its sides as 9 cm, 7 cm and 17 cm? Give reason for your answer.

Answer

We know that sum of any two sides is greater than the third side.

Here,

9 cm + 7 cm = 16 cm, which is less than third side (i.e. 17 cm).

Hence, construction of triangle with sides 9 cm, 7 cm and 17 cm is not possible.

Question 10(iii)

Is it possible to construct a triangle with lengths of its sides as 8 cm, 7 cm and 4 cm? Give reason for your answer.

Answer

Here,

⇒ 8 cm + 7 cm = 15 cm, which is 15 cm > 4cm,

⇒ 7 cm + 4 cm = 11 cm, which is 11 cm > 8 cm,

⇒ 8 cm + 4 cm = 12 cm, which is 12 cm > 7 cm.

We know that sum of any two sides is greater than the third side.

Hence, construction of triangle with sides 8 cm, 7 cm and 4 cm is possible.

Chapter 9

Triangles — Exercise 9.4

Class - 9 ML Aggarwal Understanding ICSE Mathematics

Exercise 9.4

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8(a)

Question 8(b)

Question 8(c)

Question 9(a)

Question 9(b)

Question 10(i)

Question 10(ii)

Question 10(iii)