RS AGGARWAL CLASS 9 CHAPTER 9 CONGRUENCE OF TRIANGLES AND INEQUALITIES IN A TRIANGLE EXERCISE 9B

 EXERCISE 9B

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Q1 | Ex-9B |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 1:

Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.
(i) 5 cm, 4 cm, 9 cm
(ii) 8 cm, 7 cm, 4 cm
(iii) 10 cm, 5 cm, 6 cm
(iv) 2.5 cm, 5 cm, 7 cm
(v) 3 cm, 4 cm, 8 cm

Answer 1:

(i) No, because the sum of two sides of a triangle is not greater than the third side.
5 + 4 = 9

(ii) Yes, because the sum of two sides of a triangle is greater than the third side.
7 + 4 > 8; 8 + 7 > 4; 8 + 4 > 7

(iii) Yes, because the sum of two sides of a triangle is greater than the third side.
5 + 6 > 10; 10 + 6 > 5; 5 + 10 > 6

(iv) Yes, because the sum of two sides of a triangle is greater than the third side.
2.5 + 5 > 7; 5 + 7 > 2.5; 2.5 + 7 > 5

(v) No, because the sum of two sides of a triangle is not greater than the third side.
3 + 4 < 8

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Question 2:

In ΔABC, ∠A = 50° and ∠B = 60°. Determine the longest and shortest sides of the triangle.

Answer 2:

Given: In ΔABC, ∠A = 50° and ∠B = 60°

In ΔABC,
∠A + ∠B + ∠C = 180°           (Angle sum property of a triangle)
$\Rightarrow$50° + 60° + ∠C = 180°
$\Rightarrow$110° + ∠C = 180°
$\Rightarrow$∠C = 180° $-$ 110°
$\Rightarrow$∠C = 70°

Hence, the longest side will be opposite to the largest angle (∠C = 70°)  i.e. AB.
And, the shortest side will be opposite to the smallest angle (∠A = 50° ) i.e. BC.

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Question 3:

(i) In ∆ABC, ∠A = 90°. Which is its longest side?
(ii) In ∆ABC, ∠A = ∠B = 45°. Which is its longest side?
(iii) In ∆ABC, ∠A = 100° and ∠C = 50°. Which is its shortest side?

Answer 3:

(i) Given: In ∆ABC, ∠A = 90°

So, sum of the other two angles in triangle ∠B + ∠C = 90°

i.e. ∠B, ∠C < 90°

Since, ∠A is the greatest angle.

So, the longest side is BC.

(ii) Given: ∠A = ∠B = 45°

Using angle sum property of triangle,

∠C = 90°

Since, ∠C is the greatest angle.

So, the longest side is AB.

(iii) Given: ∠A = 100° and ∠C = 50°

Using angle sum property of triangle,

∠B = 30°

Since, ∠A is the greatest angle.

So, the shortest side is BC.

PAGE NO-297

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Question 4:

In ∆ABC, side AB is produced to D such that BD = BC. If ∠B = 60° and ∠A = 70°, prove that (i) AD > CD and (ii) AD > AC.

Answer 4:

In triangle CBA, CBD is an exterior angle.

i.e., ∠CBA+∠CBD=180°

⇒60°+∠CBD=180°

⇒∠CBD=120°


Triangle BCD is isosceles and BC = BD.
Let ∠BCD=∠BDC = x°.
In △CBD, we have:

⇒∠BCD+∠CBD+∠CDB=180°

⇒x+120°+x=180

⇒2x=60°

⇒x=30°

∴∠BCD=∠BDC=30°

In triangle ADC, ∠C=∠ACB + ∠BCD = 50°+30°=80°
∠A=70°and ∠D=30°

∴∠C>∠A

⇒AD>CD ...(1)

Also, ∠C>∠D

⇒AD>AC ...(2)

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Question 5:

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

Answer 5:

Given: ∠B < ∠A and ∠C < ∠D

To prove: AD > BC

Proof:

In ΔAOB,∠B<∠A

⇒AO<BO (Side opposite to the greater angle is longer) .....(1)

In ΔCOD,∠C<∠D

⇒OD<OC (Side opposite to the greater angle is longer) .....(2)

Adding (1) and (2), we get

AO+OD<BO+OC

∴AD<BC

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Question 6:

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

Answer 6:

Given: In quadrilateral ABCD, AB and CD are respectively the smallest and largest sides.

To prove:
​(i) ∠A > ∠C
(ii) ∠B > ∠D


Construction: Join AC.

Proof:

In ΔABC,∵ BC>AB (Given, AB is the smallest side)

∴ ∠1>∠2 ...(1)

In ΔADC,∵ CD>AD (Given, CD is the largest side)

∴ ∠3>∠4 ...(2)

Adding (1) and (2), we get

∠1+∠3>∠2+∠4

∴ ∠A>∠C


(ii)

Construction: Join BD.

Proof:

In ΔABD,∵ AD>AB (Given, AB is the smallest side.)

∴ ∠5>∠6 ...(3)

In ΔCBD,∵ CD>BC (Given, CD is the greatest side.)

∴ ∠7>∠8 ...(4)

Adding (3) and (4), we get
∠5+∠7>∠6+∠8
∴∠B>∠D

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Question 7:

In a quadrilateral ABCD, show that (AB + BC + CD + DA) > (AC + BD).

Answer 7:

Given: Quadrilateral ABCD

To prove: (AB + BC + CD + DA) > (AC + BD)

Proof:

$$\begin{gathered} \mathrm{In} ∆ABC, \\ AB+BC>AC ...\left(\mathrm{i}\right) \\ \mathrm{In} ∆CAD, \\ CD+AD>AC ...\left(\mathrm{ii}\right) \\ \mathrm{In} ∆BAD, \\ AB+AD>BD ...\left(\mathrm{iii}\right) \\ \mathrm{In} ∆BCD, \\ BC+CD>BD ...\left(\mathrm{iv}\right) \end{gathered}$$

Adding (i), (ii), (iii) and (iv), we get

2(AB + BC + CD + DA) < 2( AC + BD)

Hence, (AB + BC + CD + DA) < (AC + BD).

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Question 8:

In a quadrilateral ABCD, show that
$\left(AB+BC+CD+DA\right)<2\left(BD+AC\right)$

Answer 8:

Given: Quadrilateral ABCD
To prove: (AB + BC + CD + DA) < 2(BD + AC).

Proof:

In ∆AOB, 
$OA+OB>AB .....\left(i\right)$
In ∆BOC, 
$OB+OC>BC .....\left(ii\right)$
In ∆COD, 
$OC+OD>CD .....\left(iii\right)$
In ∆AOD, 
$OD+OA>AD .....\left(iv\right)$
Adding $\left(i\right),\left(ii\right),\left(iii\right) \mathrm{and} \left(iv\right)$, we get
$$\begin{gathered} 2\left(OA+OB+OC+OD\right)>\left(AB+BC+CD+DA\right) \\ \Rightarrow 2\left(OB+OD+OA+OC\right)>\left(AB+BC+CD+DA\right) \\ \Rightarrow 2\left(BD+AC\right)>\left(AB+BC+CD+DA\right) \end{gathered}$$

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Question 9:

In ΔABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in X. Arrange AX, BX and CX in descending order.

Answer 9:

Given: In ΔABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in X.

$$\begin{gathered} \mathrm{In} ∆ABX, \\ \because \angle BAX>\angle ABX \\ \therefore BX>AX ...\left(\mathrm{i}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Similarly}, \mathrm{in} ∆ACX, \\ \because \angle ACX>\angle XAC \\ \therefore AX>CX ...\left(\mathrm{ii}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{From} \left(\mathrm{i}\right) \mathrm{and} \left(ii\right), \mathrm{we} \mathrm{get} \\ BX>AX>CX \end{gathered}$$

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Question 10:

In the given figure, PQ > PR and QS and RS are the bisectors of ∠Q and ∠R respectively. Show that SQ > SR.

Answer 10:

Since the angle opposite to the longer side is greater, we have:

$$\begin{gathered} PQ>PR \\ \Rightarrow \angle R>\angle Q \\ \Rightarrow \frac{1}{2}\angle R>\frac{1}{2}\angle Q \\ \Rightarrow \angle SRQ>\angle RQS \\ \Rightarrow QS>SR \end{gathered}$$

∴$SQ>SR$

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Question 11:

D is any point on the side AC of ΔABC with AB = AC. Show that CD < BD.

Answer 11:

Given: In $∆$ABC, AB = AC

To prove: CD < BD

Proof:

In $∆$ABC,

Since, AB = AC       (Given)

So, $\angle ABC=\angle ACB$      ...(i)

$$\begin{gathered} \mathrm{In} ∆ABC \mathrm{and} ∆DBC, \\ \angle ABC>\angle DBC \\ \Rightarrow \angle ACB>\angle DBC \left[\mathrm{From} \left(\mathrm{i}\right)\right] \\ \Rightarrow BD>CD \left(\mathrm{Side} \mathrm{opposite} \mathrm{to} \mathrm{greater} \mathrm{angle} \mathrm{is} \mathrm{longer}.\right) \\ \therefore CD<BD \end{gathered}$$

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Question 12:

Prove that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater than $\frac{2}{3}$ of a right angle.

Answer 12:

Given: In $∆$ABC, BC is the longest side.

To prove: $\angle$BAC > $\frac{2}{3}$ of a right angle, i.e., $\angle$BAC > 60$°$

Construct: Mark a point D on side AC such that AD = AB = BD.

Proof:

In $∆$ABD,

$\because$ AD = AB = BD    (By construction)

$\therefore \angle 1=\angle 3=\angle 4=60°$

$$\begin{gathered} \mathrm{Now}, \\ \angle BAC=\angle 1+\angle 2=60°+\angle 2 \\ \mathrm{but} 60° =\frac{2}{3} \mathrm{of} \mathrm{a} \mathrm{right} \mathrm{angle} \\ \mathrm{So}, \angle BAC=\frac{2}{3} \mathrm{o}\mathrm{f} \mathrm{a} \mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t} \mathrm{a}\mathrm{n}\mathrm{g}\mathrm{l}\mathrm{e} + \angle 2 \end{gathered}$$

Hence, $\angle$BAC > $\frac{2}{3}$ of a right angle.

PAGE NO-298

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Question 13:

In the given figure, prove that
(i) CD + DA + AB > BC
(ii) CD + DA + AB + BC > 2AC.

Answer 13:

Given: Quadrilateral ABCD

To prove:
(i) CD + DA + AB > BC
(ii) CD + DA + AB + BC > 2AC

Proof:
(i)
In ΔACD,CD+DA>CA ...(1)

In ΔABC,AB+CA>BC ...(2)

Adding (1) and (2), we get

CD+DA+AB+CA>CA+BC

∴ AB+CD+DA>BC

(ii)
In ΔCDA,CD+DA>CA ...(3)

In ΔBCA,BC + AB > CA ...(4)

Adding (3) and (4), we get

CD+AD+BC+AB>CA+CA

∴ CD+AD+BC+AB>2CA

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Question 14:

If O is a point within ∆ABC, show that:
(i) AB + AC > OB + OC
(ii) AB + BC + CA > OA + OB + OC
(iii) OA+OB+OC>$\frac{1}{2}$(AB+BC+CA)

Answer 14:

Given:

In triangle ABC, O is any interior point.
We know that any segment from a point O inside a triangle to any vertex of the triangle cannot be longer than the two sides adjacent to the vertex.
Thus, OA cannot be longer than both AB and CA (if this is possible, then O is outside the triangle).

(i) OA cannot be longer than both AB and CA.​
AB>OB ...(1)

AC>OC ...(2)

Thus, AB+AC>OB+OC ...[Adding (1) and(2)]

(ii) AB>OA......(3)

BC>OB.....(4)

CA>OC.....(5)

Adding the above three equations, we get:
Thus, AB+BC+CA>OA+OB+OC ...(6)

OA cannot be longer than both AB and CA.​
AB>OB.....(5)

AC>OC.....(6)

AB+AC>OB+OC..........[On adding (5) and (6)]

Thus, the first equation to be proved is shown correct.

(iii) Now, consider the triangles OAC, OBA and OBC.
We have:
OA+OC>AC

OA+OB>AB

OB+OC>BC

Adding the above three 

equations, we get:

OA+OC+OA+OB+OB+OC>AB+AC+BC

⇒2OA+OB+OC>AB+AC+BC

Thus, 

OA+OB+OC>$\frac{1}{2}$AB+BC+CA

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Question 15:

In the given figure, AD ⊥ BC and CD > BD. Show that AC > AB.

Answer 15:

Given: AD ⊥ BC and CD > BD

To prove: AC > AB

Proof:

$$\begin{gathered} \angle ADB=\angle ADC=90° \left(AD\perp BC\right) ...\left(1\right) \\ \angle BAD<\angle DAC \left(CD>BD\right) ...\left(2\right) \end{gathered}$$

In $∆ABD$,

Using angle sum property of a triangle,

$$\begin{gathered} \angle B=180°-\angle ADB-\angle BAD \\ \angle B=90°-\angle BAD ...\left(3\right) \end{gathered}$$

In $∆ADC$,

Using angle sum property of a triangle,

$\angle ACD=90°-\angle DAC ...\left(4\right)$

From (2), (3) and (4), we get

$\angle B>\angle C$

Therefore, $AC>AB$.

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Question 16:

In the given figure, D is a point on side BC of a ΔABC and E is a point such that CD = DE. Prove that AB + AC > BE.

Answer 16:

Given: CD = DE

To prove: AB + AC > BE

Proof:

In $∆ABC$,

$AB+AC>BC ...\left(1\right)$

In $∆BED$,

$$\begin{gathered} BD+CD>BE \\ \Rightarrow BC>BE ...\left(2\right) \end{gathered}$$

From (1) and (2), we get

AB + AC > BE