ML Aggarwal Solution Class 9 Chapter 10 Triangles Exercise 10.4

Exercise 10.4

Question 1

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

Sol :
In ∆PQR, ∠P=70° , ∠R=30°
But ∠P+∠Q+∠R=180°
⇒70°+30°+∠Q=180°
⇒100°+∠Q=180°








∴∠Q=180°-100°=80°
∴∠Q=80° the greatest angle
∴It opposite side PR is the longest side
(Side opposite to greatest angle is longest)

Question 2

Show that in a right angled triangle, the hypotenuse is the longest side.
Sol :
Given  : In right angled ΔABC, ∠B=90°











To prove : AC is the longest side
Proof : In ΔABC,
∵∠B=90°
∴∠A and ∠C are acute angles
i.e. less than 90°
∴∠B is the greatest angle
or ∠B>∠C and ∠B>∠A
∴AC>AB and AC>BC
Hence AC is the longest side

Question 3

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









Here , PAR is right angle triangle at Q. Also, given that
⇒PQ : QR=3 : 2
Let PQ=3x, then QR=2x
It is clear that QR is the least side
Then , we know that the least angle has least side opposite to it.
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 ?
Sol :









Given that ABΔ=8 cm , BC=5.6 cm ,CA=6.5 cm
Here AB is the greatest side
Then ∠C is the greatest angle
(∴The greater side has greater angle opposite to it)
Also, BC is the least side
then ∠A is the least angle
(∵The least side has least angle opposite to it)

Question 5

In ∆ABC, ∠A = 50°, ∠B= 60°, Arrange the sides of the triangle in ascending order.
Sol :









Given in a ΔABC
⇒∠A=50°
⇒∠B= 60°
⇒∠C=180°-(∠A+∠B)
[Sum of all angles in a triangle  is 180°]
⇒∠C=180°-(50°+60°)
⇒∠C=180°-110°=70°

Now , ∠A＜∠B＜∠C
⇒BC<CA<AB
(∵Greater angles has greater side opposite to it)
Hence, side of ΔABC in ascending order as BC, CA, AB

Question 6

In figure given alongside, ∠B = 30°, ∠C = 40° and the bisector of ∠A meets BC at D. Show
(i) BD > AD
(ii) DC > AD
(iii) AC > DC
(iv) AB > BD







Sol :
Given : In ΔABC, ∠B=30°
, ∠C=40° and bisector of ∠A meets BC at D
To prove :
(i) BD>AD
(ii) DC>AD
(iii) AC>DC
(iv) AB>BD
Proof : In ΔABC,
⇒∠B=30° and ∠C=40°
⇒∴∠BAC=180°-(30°+40°)
=180°-70°=110°
∵AD is bisector of ∠A
∴∠BAD=∠CAD$=\frac{110^{\circ}}{2}=55^{\circ}$
 
(i) Now in ΔABD
∵∠BAD>∠B
∴BD>AD

(ii) In ΔACD,
∵∠CAD>∠C
∴DC>AD

(iii) ∠ADC=180°-(40°+55°)
=180°-95°=85°
In ΔADC,
∵∠ADC>∠CAD
∴AC>DC

(iv) Similarly,
∠ADB=180°-∠ADC
=180°-85°=95°
∴In ΔADB
AB>BD 
Hence proved

Question 7

(a) In the figure (1) given below, AD bisects ∠A. Arrange AB, BD and DC in the descending order of their lengths.
(b) In the figure (2) given below, ∠ABD = 65°, ∠DAC = 22° and AD = BD. Calculate ∠ ACD and state (giving reasons) which is greater : BD or DC ?
Sol :








(a) Given : In ΔABC, AD bisects ∠A, ∠B=60° and ∠C=40°
To arrange : AD, BD and DC in the descending order
In ΔABC
⇒∠BAC+∠B+∠C=180°
[Sum of all angles in a triangle is 180°]

⇒∠BAC+60°+40°=180°
[From given , ∠B=60°, ∠C=40°]
⇒∠BAC=180°-100°
⇒∠BAC=80°
∴AD bisects ∠A
∴∠BAD$=\angle \mathrm{DAC}=\frac{1}{2} \times \angle \mathrm{BAC}$
⇒$\angle \mathrm{BAD}=\angle \mathrm{DAC}=\frac{1}{2} \times 80^{\circ}$
∠BAD=∠DAC=40°..(1)

In ΔABD, ext ∠ADC=∠B+∠BAD
[In triangle exterior angle is equal to sum of opposite interior angles]

∴∠ADC=60°+40°=100°
⇒∠ADC=100°..(2)
Similarly, In ΔACD, ΔADB
=40°+40°=80°..(3)

Now , ∠ADB=80°  [From (3)]
⇒∠BAD=40° [From (2)]
⇒∠DAC=40° [From (1)]

Now , ∠ADB>∠DAC=∠BAD
[∵80°>40°=40°]
Hence, AB, DC, BD in the descending order of their lengths
(Note : It can also written as AB, BD , DC in the descending order ∵DC=BD)


(b) Given : In ΔABC , ∠ABD=65°
∠DAC=22° , and AD=BD

To calculate the ∠ACD and say which greater , BD or DC

Now , in ΔABD

∴AD=BD (given)
∴∠ABD=∠BAD..(1)
(In a triangle , equal sides have equal angles opposite to them)

Also, ∠ABD=65°..(2)
From (1) and (2) , we get
∠BAD=65°..(3)

In ΔABC,
⇒∠A+∠B+∠C=180°
[Sum of all angles in a triangle is 180°]

⇒(∠BAD+∠DAC)+∠B+∠C=180°
[∴∠A=∠BAD+∠DAC]

⇒∠BAD+∠DAC+∠B+∠ACD=180° [∴∠C=ACD]

⇒65°+22°+65°+∠ACD=180°
(Substituting the value of ∠BAD, ∠DAC and ∠B)

⇒152°+∠ACD=180°
⇒∠ACD=180°-152°
⇒∠ACD=28°

Now ∠BAD=65°  [From (3)]
and ∠CAD=22° (Given)

∴∠BAD>∠CAD
∴BD>DC
[Greater angle has greater opposite side]
Hence , BD is greater than DC

Question 8

(a) In the figure (1) given below, prove that (i) CF> AF (ii) DC>DF.
(b) In the figure (2) given below, AB = AC.
Prove that AB > CD.
(c) In the figure (3) given below, AC = CD. Prove that BC < CD.

















Sol :
(a) Given : In ΔABC, AD⊥BC, CF⊥AB and AD and E intersects at F
∠BAC=60° , ∠ABC=65°

To prove : (i) CF>AF 
(ii) DC>DF

Proof : (i) In ΔAEC,
⇒∠B+∠BEC+∠BCE=180°...(1)
(sum of angle of a triangle=180°)

⇒∠B=65°  (given)..(2)
⇒∠BEC=90° (CE⊥AB)..(3)

Putting these value in equation (1),  we get
⇒65°+90°+∠BEC=180°
⇒155°+∠BCE=180°
⇒∠BCE=25°
⇒∠DCF=25° [BCE=∠DCF]..(4)

Now in ΔCDF,
⇒∠DCF+∠FDC+∠CFD=180°
[sum of all angles in a triangle is 180°]
⇒25°+90°+∠CFD=180°

[From (4) ∠DCF=25° and AD⊥BC, ∠FDC=90°]
⇒115°+∠CFD=180°
⇒∠CFD=180°-115°
⇒∠CFD=65°...(5)

Also, ∠AFC+∠CFD=180°
[AFD is a straight line]

⇒∠AFC+65°=180°
⇒∠AFC=180-65 (∠CFD=65°)
⇒∠AFC=115°..(6)

Now in ACE,
⇒∠ACE+∠CEA+∠BAC=180°
[sum of all angles in a triangle is 180°]

⇒∠ACE+90°+60°=180°
[∵∠CEA=90°, ∠BAC=60°]

⇒∠ACE+150°=180°
⇒∠ACE=30°..(7)

Now, in ΔAFC,
⇒∠AFC+∠ACF+∠FAC=180°
[sum of all angles in a triangle is 180°]

⇒115°+30°+∠FAC=180°  {by (6) and (7)}
⇒145°+∠FAC=180°
⇒∠FAC=180°-145°
⇒∠FAC=35°..(8)

Now in ΔAFC
⇒∠FAC=35°  [From equation (8)]
⇒∠ACF=30°  [From equation (7)]
∴∠FAC>∠ACF (∵35°>30°)
∴CF>AF
[Greater angle has greater side opposite to it]

Now in ΔCDF,
⇒∠DCF=25°
⇒∠CFD=65°
∴∠CFD>DCF (∵65°>25°)
∴DC>DF
[greater angle has greater side opposite to it]
[Q.E.D]

(b) Given : In ΔABD, AC meets BD in C
∠B=70° , ∠D=40°, AB=AC

To prove : AB>CD
Proof : In ΔABC,
AB=AC  (given)
∴∠ACB=∠B..(1)
(In a triangle , equal sides have equal angles opposite to them)

Also, ∠B=70°  [Given ]...(2)
From (i) and (ii) , we get
⇒∠ACB+∠ACD=180°  [BCD is a straight line]
⇒70°+∠ACD=180°  [From equation (3)]
⇒∠ACD=180°-70°=110°..(4)

Now , in ΔACD,
⇒∠CAD+∠ACD+∠D=180° 
[sum of all angles in a triangle is 180°]

⇒∠CAD+110°+40°=180°  [From (4)]
⇒∠ACD=110° and ∠D=40°  (given)
⇒∠CAD+150°=180°
⇒∠CAD=180°-150°
⇒∠CAD=30°...(5)

Now, in ΔACD,
⇒∠ACD=110°  [From equation (4)]
⇒∠CAD=30°  [From equation (5)]
⇒∠D=40°  (given)
∴∠D>∠CAD  (40°>30°)
∴AC>CD
[greater angle has greater side opposite to it]
⇒AB>CD  [∵AB=AC given]
[Q.E.D]


(c) Given : In ΔACD, AC=CD, ∠BAD=60° ,∠ACB=70°
To prove : BC>CD
Proof : In ΔACD
∴AC=CD  (Given)

∴∠CAD=∠CDA..(1)
[In a triangle if two sides are equal , then angles opposite to them are also equal]

Also, ∠ACB=70°...(2)

Now, ∠ACB=∠CAD+∠CDA
[exterior angle is equal to sum of two interior opposite angles]
⇒70°=∠CAD+∠CAD
[From (1) and (2)]

⇒70°=2∠CAD
⇒2∠CAD=70°
⇒∠CAD$=\frac{70^{\circ}}{2}$=35°
∵∠BAD=60°  (given)
∴∠BAC=∠BAD-∠CAD
=60°-35°=25°
∴∠BAC<∠CAD  [∵25°<35°]
∴BC>CD

[greater angles has greater side opposite to it]
[Q.E.D]

Question 9

(a) In the figure (i) given below, ∠B < ∠A and ∠C < ∠D. Show that AD < BC. (b) In the figure (ii) given below, D is any point on the side BC of ∆ABC. If AB > AC, show that AB > AD.
Sol :
(a) In the given figure,
∠B<∠A and ∠C<∠D

To prove : AD>BC
Proof : In ΔABO
∠B<∠A 
∴AO<BO...(i)












Similarly in ΔOCD
∠C<∠D  (given)
∴OD>OC
Adding (i) and (ii)
⇒AO+OD<BO+OC
⇒AD<BC
Hence AD<BC

(b) In the given figure,
D is any point on BC of ΔABC
⇒AB>AC
To prove : AB>AD
Proof : In ΔABC
⇒AB>AC
⇒∠C>∠B












In ΔABD
Ext.∠ADC>∠B  (∵∠C>∠B)
∴AC>AD ..(i)

But  AB>AC
(Given )..(ii)

∴From (i) and (ii)
⇒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,
(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.
(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.
Sol :
(i) Length of sides of a triangle are 4 cm, 3 cm and 7 cm
We know that sum of any two sides of a triangle is greater than its third side But 4 + 3 = 7 cm
Which is not possible
Hence to construction of a triangle with sides 4 cm, 3 cm and 7 cm is not possible.
(ii) Length of sides of a triangle are 9 cm, 7 cm and 17 cm
We know that sum of any two sides of a triangle is greater than its third side Now 9 + 7 = 16 < 17 ∴ It is not possible to construct a triangle with these sides.
(iii) Length of sides of a triangle are 8 cm, 7 cm and 4 cm We know that sum of any two sides of a triangle is greater than its third side Now 7 + 4 = 11 > 8
Yes, It is possible to construct a triangle with these sides.

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