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

 EXERCISE 9B

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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

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.

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.

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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.

$$\begin{gathered} \mathrm{i}.\mathrm{e}., \angle CBA+\angle CBD=180° \\ \Rightarrow 60°+\angle CBD=180° \\ \Rightarrow \angle CBD=120° \end{gathered}$$

Triangle BCD is isosceles and BC = BD.
Let $\angle BCD=\angle BDC = x°$.
$$\begin{gathered} \mathrm{In} △CBD, \mathrm{we} \mathrm{have}: \\ \Rightarrow \angle BCD+\angle CBD+\angle CDB=180° \\ \Rightarrow x+120°+x=180 \\ \Rightarrow 2x=60° \\ \Rightarrow x=30° \\ \therefore \angle BCD=\angle BDC=30° \end{gathered}$$

In triangle ADC, $\angle C=\angle ACB + \angle BCD = 50°+30°=80°$
$$\begin{gathered} \angle A=70° \\ and \angle D=30° \end{gathered}$$

$$\begin{gathered} \therefore \angle C>\angle A \\ \Rightarrow AD>CD ...\left(1\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Also}, \angle C>\angle D \\ \Rightarrow AD>AC ...\left(2\right) \end{gathered}$$

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: 

$$\begin{gathered} \mathrm{In} ∆AOB, \\ \angle B<\angle A \\ \Rightarrow AO<BO \left(\mathrm{Side} \mathrm{opposite} \mathrm{to} \mathrm{the} \mathrm{greater} \mathrm{angle} \mathrm{is} \mathrm{longer}\right) .....\left(1\right) \end{gathered}$$

$$\begin{gathered} \mathrm{In} ∆COD, \\ \angle C<\angle D \\ \Rightarrow OD<OC \left(\mathrm{Side} \mathrm{opposite} \mathrm{to} \mathrm{the} \mathrm{greater} \mathrm{angle} \mathrm{is} \mathrm{longer}\right) .....\left(2\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Adding} \left(1\right) \mathrm{and} \left(2\right), \mathrm{we} \mathrm{get} \\ AO+OD<BO+OC \\ \therefore AD<BC \end{gathered}$$

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:

$$\begin{gathered} \mathrm{In} ∆ABC, \\ \because BC>AB \left(\mathrm{Given}, AB \mathrm{is} \mathrm{the} \mathrm{smallest} \mathrm{side}\right) \\ \therefore \angle 1>\angle 2 ...\left(1\right) \end{gathered}$$

$$\begin{gathered} \mathrm{In} ∆ADC, \\ \because CD>AD \left(\mathrm{Given}, CD \mathrm{is} \mathrm{the} \mathrm{largest} \mathrm{side}\right) \\ \therefore \angle 3>\angle 4 ...\left(2\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Adding} \left(1\right) \mathrm{and} \left(2\right), \mathrm{we} \mathrm{get} \\ \angle 1+\angle 3>\angle 2+\angle 4 \\ \therefore \angle A>\angle C \end{gathered}$$


(ii)

Construction: Join BD.

Proof:

$$\begin{gathered} \mathrm{In} ∆ABD, \\ \because AD>AB \left(\mathrm{Given}, AB \mathrm{is} \mathrm{the} \mathrm{smallest} \mathrm{side}.\right) \\ \therefore \angle 5>\angle 6 ...\left(3\right) \end{gathered}$$

$$\begin{gathered} \mathrm{In} ∆CBD, \\ \because CD>BC \left(\mathrm{Given}, CD \mathrm{is} \mathrm{the} \mathrm{greatest} \mathrm{side}.\right) \\ \therefore \angle 7>\angle 8 ...\left(4\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Adding} \left(3\right) \mathrm{and} \left(4\right), \mathrm{we} \mathrm{get} \\ \angle 5+\angle 7>\angle 6+\angle 8 \\ \therefore \angle B>\angle D \end{gathered}$$

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).

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
$$\begin{gathered} 2\left(OA+OB+OC+OD\right)>\left(AB+BC+CD+DA\right) \\ \Rightarrow 2\left(OB+OD+OA+OC\right)>\left(AB+BC+CD+DA\right) \\ \Rightarrow 2\left(BD+AC\right)>\left(AB+BC+CD+DA\right) \end{gathered}$$

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.

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

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

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

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:

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

∴$SQ>SR$

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)

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

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.

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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)
$$\begin{gathered} \mathrm{In} ∆ACD, \\ CD+DA>CA ...\left(1\right) \\ \mathrm{In} ∆ABC, \\ AB+CA>BC ...\left(2\right) \end{gathered}$$
$$\begin{gathered} \mathrm{Adding} \left(1\right) \mathrm{and} \left(2\right), \mathrm{we} \mathrm{get} \\ CD+DA+AB+CA>CA+BC \\ \therefore AB+CD+DA>BC \end{gathered}$$

(ii)
 $$\begin{gathered} \mathrm{In} ∆CDA, \\ CD+DA>CA ...\left(3\right) \\ \mathrm{In} ∆BCA, \\ BC + AB > CA ...\left(4\right) \end{gathered}$$
$$\begin{gathered} \mathrm{Adding} \left(3\right) \mathrm{and} \left(4\right), \mathrm{we} \mathrm{get} \\ CD+AD+BC+AB>CA+CA \\ \therefore CD+AD+BC+AB>2CA \end{gathered}$$

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.​
$$\begin{gathered} AB>OB ...\left(1\right) \\ AC>OC ...\left(2\right) \\ \mathrm{Thus}, AB+AC>OB+OC ...[\mathrm{Adding} (1) \mathrm{and}(2\left)\right] \end{gathered}$$

$$\begin{gathered} \left(ii\right) AB>OA......\left(3\right) \\ BC>OB.....\left(4\right) \\ CA>OC.....\left(5\right) \end{gathered}$$
Adding the above three equations, we get:
$\mathrm{Thus}, AB+BC+CA>OA+OB+OC$              ...(6)

OA cannot be longer than both AB and CA.​
$$\begin{gathered} AB>OB.....\left(5\right) \\ AC>OC.....\left(6\right) \\ AB+AC>OB+OC..........[\mathrm{On} \mathrm{adding} (5) \mathrm{and} (6\left)\right] \end{gathered}$$
Thus, the first equation to be proved is shown correct.

(iii) Now, consider the triangles OAC, OBA and OBC.
We have:
$$\begin{gathered} OA+OC>AC \\ OA+OB>AB \\ OB+OC>BC \\ \mathrm{Adding} \mathrm{the} \mathrm{above} \mathrm{three} \mathrm{equations}, \mathrm{we} \mathrm{get}: \\ OA+OC+OA+OB+OB+OC>AB+AC+BC \\ \Rightarrow 2\left(OA+OB+OC\right)>AB+AC+BC \\ \mathrm{Thus}, OA+OB+OC>\frac{1}{2}\left(AB+BC+CA\right) \end{gathered}$$

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, $AC>AB$.

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