RD Sharma 2017 Solutions for Class 7 Math Chapter 16 Congruence are provided here with simple step-by-step explanations. These solutions for Congruence are extremely popular among class 7 students for Math Congruence Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2017 Book of class 7 Math Chapter 16 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2017 Solutions. All RD Sharma 2017 Solutions for class 7 Math are prepared by experts and are 100% accurate.

#### Page No 16.10:

#### Question 10:

Triangles *ABC* and *DBC* have side *BC* common, *AB* = *BD* and *AC* = *CD*. Are the two triangles congruent? State in symbolic form. Which congruence condition do you use? Does ∠*ABD* equal ∠*ACD*? Why or why not?

#### Answer:

Yes.

$\text{In\u2206ABCand\u2206DBC}\phantom{\rule{0ex}{0ex}}\text{AB=DB(Given)}\phantom{\rule{0ex}{0ex}}\text{AC=DC(Given)}\phantom{\rule{0ex}{0ex}}\text{BC=BC(Common)}\phantom{\rule{0ex}{0ex}}\text{BySSScriterionofcongruency,\u2206ABC\u2245\u2206DBC}$

No, $\text{\u2220ABDand\u2220ACDarenotequal}$

because AB $\ne $ AC.

#### Page No 16.14:

#### Question 1:

By applying SAS congruence condition, state which of the following pairs (Fig. 28) of triangles are congruent. State the result in symbolic form

#### Answer:

1) We have OA = OC and OB = OD and$$

Therefore by SAS condition,

2) We have BD = DC

$$

and AD = AD

Therefore by SAS condition,

3) We have AB = DC

$\angle $

T

4) We have BC = QR

$$

and AB = PQ

Therefore by SAS condition,

#### Page No 16.14:

#### Question 2:

State the condition by which the following pairs of triangles are congruent.

#### Answer:

1) AB = AD

BC = CD

and AC = CA

Therefore by SSS condition, $\u25b3ABC\cong \u25b3ADC$.

2) AC = BD

AD = BC and AB = BA

Therefore by SSS condition, $\u25b3ABD\cong \u25b3BAC$.

3) AB = AD

$\angle BAC=\angle DAC$

and AC = AC

Therefore by SAS condition, $\u25b3BAC\cong \u25b3DAC$.

4) AD = BC

$\angle $ DAC = $\angle $BCA

and AC = CA

Therefore by SAS condition, $\u25b3ABC\cong \u25b3ADC$.

#### Page No 16.15:

#### Question 3:

In Fig. 30, line segments *AB* and *CD* bisect each other at *O*. Which of the following statements is true?

(i) ∆ *AOC *≅ ∆ *DOB*

(ii) ∆ *AOC* ≅ ∆ *BOD*

(iii) ∆ *AOC* ≅ ∆ *ODB*.

State the three pairs of matching parts, yut have used to arive at the answer.

#### Answer:

We have AO = OB.

And CO = OD

Also

T

Therefore, statement (ii) is true.

#### Page No 16.15:

#### Question 4:

Line-segments *AB* and *CD* bisect each other at *O*. *AC* and *BD* are joined forming triangles *AOC* and *BOD*. State the three equality relations between the parts of the two triangles, that are given or otherwise known. Are the two triangles congruent? State in symbolic form. Which congruence condition do you use?

#### Answer:

We have AO = OB and CO = OD since AB and CD bisect each other at O.

Also $\angle AOC=\angle BOD$ since they are opposite angles on the same vertex.

Therefore by SAS congruence condition, $\u25b3AOC\cong \u25b3BOD\phantom{\rule{0ex}{0ex}}$.

#### Page No 16.15:

#### Question 5:

∆ *ABC* is isosceles with *AB* = *AC*. Line segment *AD* bisects ∠*A* and meets the base *BC* in *D*.

(i) Is ∆ *ADB* ≅ ∆ *ADC*?

(ii) State the three pairs of matching parts used to answer (i).

(iii) Is it true to say that *BD* = *DC*?

#### Answer:

(i) We have AB = AC (given)

$\angle BAD=\angle CAD$ (AD bisects $\angle BAC$)

and AD = AD (common)

Therefore by SAS condition of congruence, $\u25b3ABD\cong \u25b3ACD$.

(ii) We have used AB, AC; $\angle BAD=\angle CAD$; AD, DA.

(iii) Now$\u25b3ABD\cong \u25b3ACD$ therefore by c.p.c.t BD = DC.

#### Page No 16.15:

#### Question 6:

In Fig. 31, *AB* = *AD* and ∠*BAC* = ∠*DAC*.

(i) State in symbolic form the congruence of two triangles *ABC* and *ADC* that is true.

(ii) Complete each of the following, so as to make it true:

(a) ∠*ABC* = ........

(b) ∠*ACD* = ........

(c) Line segment *AC* bisects ..... and .....

#### Answer:

i) AB = AD (given)

$\angle BAC=\angle DAC$ (given)

AC = CA (common)

Therefore by SAS conditionof congruency, $\u25b3ABC\cong \u25b3ADC$.

ii) $\angle ABC=\angle ADC$ (c.p.c.t)

$\angle ACD=\angle ACB$ (c.p.c.t)

#### Page No 16.15:

#### Question 7:

In Fig. 32, *AB* || *DC* and *AB* = *DC*.

(i) Is ∆ *ACD* ≅ ∆ *CAB*?

(ii) State the three pairs of matching parts used to answer (i).

(iii) Which angle is equal to ∠*CAD*?

(iv) Does it follow from (iii) that *AD* || *BC*?

#### Answer:

(i) Yes $\u25b3ACD\cong \u25b3CAB\phantom{\rule{0ex}{0ex}}$ by SAS condition of congruency.

(ii) We have used AB = DC, AC = CA and $\angle DCA=\angle BAC$.

(iii) $\angle CAD=\angle ACB$ since the two triangles are congruent.

(iv) Yes, this follows from AD$\parallel $BC as alternate angles are equal.If alternate angles are equal the lines are parallel.

#### Page No 16.19:

#### Question 1:

Which of the following pairs of triangles are congruent by ASA condition?

#### Answer:

1) We have

$\text{Since\u2220ABO=\u2220CDO}=45\xb0\text{andbotharealternateangles},\phantom{\rule{0ex}{0ex}}\text{AB}\parallel \text{DC}\phantom{\rule{0ex}{0ex}}\text{\u2220BAO}=\text{\u2220DCO}(\text{alternateangle},\text{AB}\parallel \text{CD}\text{andACisatransversalline)}\phantom{\rule{0ex}{0ex}}\text{\u2220ABO}=\text{\u2220CDO}=45\xb0\text{(giveninthefigure)}\phantom{\rule{0ex}{0ex}}\text{Also,AB}=\text{DC}(\text{Giveninthefigure})\phantom{\rule{0ex}{0ex}}\text{Therefore},\text{byASA}\u25b3\text{AOB}\cong \u25b3\text{DOC}\phantom{\rule{0ex}{0ex}}$

2)

$\text{In\u25b3ABC,}\phantom{\rule{0ex}{0ex}}\text{NowAB=AC(Given)}\phantom{\rule{0ex}{0ex}}\text{\u2220ABD=\u2220ACD=40\xb0(Anglesoppositetoequalsides)}\phantom{\rule{0ex}{0ex}}\text{\u2220ABD+\u2220ACD+\u2220BAC=180\xb0(Anglesumproperty)}\phantom{\rule{0ex}{0ex}}\text{40\xb0+40\xb0+\u2220BAC=180\xb0}\phantom{\rule{0ex}{0ex}}\text{\u2220BAC=180\xb0-80\xb0=100\xb0}\phantom{\rule{0ex}{0ex}}\text{\u2220BAD+\u2220DAC=\u2220BAC}\phantom{\rule{0ex}{0ex}}\text{\u2220BAD=\u2220BAC-\u2220DAC=100\xb0-50\xb0=50\xb0}\phantom{\rule{0ex}{0ex}}\text{\u2220BAD=\u2220CAD=50\xb0}\phantom{\rule{0ex}{0ex}}\text{Therefore,byASA,\u25b3ABD\u2245\u25b3ADC}\phantom{\rule{0ex}{0ex}}$

3)

$\text{In\u2206ABC,}\phantom{\rule{0ex}{0ex}}\text{\u2220A+\u2220B+\u2220C=180\xb0(Anglesumproperty)}\phantom{\rule{0ex}{0ex}}\text{\u2220C=180\xb0-\u2220A-\u2220B}\phantom{\rule{0ex}{0ex}}\text{\u2220C=180\xb0-30\xb0-90\xb0=60\xb0}\phantom{\rule{0ex}{0ex}}\text{In\u2206PQR,}\phantom{\rule{0ex}{0ex}}\text{\u2220P+\u2220Q+\u2220R=180\xb0(Anglesumproperty)}\phantom{\rule{0ex}{0ex}}\text{\u2220P=180\xb0-\u2220Q-\u2220R}\phantom{\rule{0ex}{0ex}}\text{\u2220P=180\xb0-60\xb0-90\xb0=30\xb0}\phantom{\rule{0ex}{0ex}}\text{\u2220BAC=\u2220QPR=30\xb0}\phantom{\rule{0ex}{0ex}}\text{\u2220BCA=\u2220PRQ=60\xb0}\phantom{\rule{0ex}{0ex}}\text{andAC=PR(Given)}\phantom{\rule{0ex}{0ex}}\text{Therefore,byASA,\u25b3ABC\u2245\u25b3PQR}$

4)

$\text{WehaveonlyBC=QRbutnoneoftheanglesof\u25b3ABCAND\u25b3PQRareequal.}\phantom{\rule{0ex}{0ex}}\text{Therefore,\u25b3ABC\u2247\u25b3PRQ}$

#### Page No 16.19:

#### Question 2:

In Fig. 37, *AD* bisects ∠*A* and *AD* ⊥ *BC*.

(i) Is ∆ *ADB* ≅ ∆ *ADC*?

(ii) State the three pairs of matching parts you have used in (i).

(iii) Is it true to say that *BD* = *DC*?

#### Answer:

$\text{(i)Yes,\u25b3ADB}\cong \text{\u25b3ADC,byASAcriterionofcongruency}\phantom{\rule{0ex}{0ex}}\text{(ii)}\text{Wehaveused}\mathit{\text{}}\text{\u2220BAD}\mathit{}\mathit{=}\mathit{\text{\u2220}}\text{CAD}\phantom{\rule{0ex}{0ex}}\text{\u2220ADB=\u2220ADC=90\xb0sinceAD\u22a5BC}\phantom{\rule{0ex}{0ex}}\text{andAD=DA}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{(iii)Yes,BD\hspace{0.17em}=DCsince,\u25b3ADB\u2245\u25b3ADC}\phantom{\rule{0ex}{0ex}}$

#### Page No 16.20:

#### Question 3:

Draw any triangle *ABC*. Use ASA condition to construct another triangle congruent to it.

#### Answer:

We have drawn

$\u25b3\mathrm{ABC}\mathrm{with}\angle \mathrm{ABC}=60\xb0\mathrm{and}\angle \mathrm{ACB}=70\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{now}\mathrm{construct}\u25b3\mathrm{PQR}\cong \u25b3\mathrm{ABC}\phantom{\rule{0ex}{0ex}}\u25b3\mathrm{PQR}\mathrm{has}\angle \mathrm{PQR}=60\xb0\mathrm{and}\angle \mathrm{PRQ}=70\xb0\phantom{\rule{0ex}{0ex}}\mathrm{Also}\mathrm{we}\mathrm{construct}\u25b3\mathrm{PQR}\mathrm{such}\mathrm{that}\mathrm{BC}=\mathrm{QR}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore}\mathrm{by}\mathrm{ASA}\mathrm{the}\mathrm{two}\mathrm{triangles}\mathrm{are}\mathrm{congruent}$

#### Page No 16.20:

#### Question 4:

In ∆ *ABC*, it is known that ∠*B* = ∠*C*. Imagine you have another copy of ∆ *ABC*

(i) Is ∆ *ABC* ≅ ∆ *ACB*?

(ii) State the three pairs of matching parts you have used to answer (i).

(iii) Is it true to say that *AB* = *AC*?

#### Answer:

(i) Yes $\u25b3ABC\cong \u25b3ACB\phantom{\rule{0ex}{0ex}}$.

(ii) We have used $\angle ABC=\angle ACBand\angle ACB=\angle ABCagain\phantom{\rule{0ex}{0ex}}$.

Also BC = CB

(iii) Yes, it is true to say that AB = AC since $\angle ABC=\angle ACB$.

#### Page No 16.20:

#### Question 5:

In Fig. 38, AX bisects ∠*BAC* as well as ∠*BDC*. State the three facts needed to ensure that ∆ *ABD *≅ ∆ *ACD*.

#### Answer:

$\text{Asperthegivenconditions,\u2220CAD=\u2220BAD}\phantom{\rule{0ex}{0ex}}\text{and\u2220CDA=\u2220BDA(becauseAXbisects\u2220BAC)}\phantom{\rule{0ex}{0ex}}\text{AD=DA(common)}\phantom{\rule{0ex}{0ex}}\text{Therefore,byASA,\u25b3ACD\u2245\u25b3ABD}$

#### Page No 16.20:

#### Question 6:

In Fig. 39, *AO* = *OB* and ∠*A* = ∠*B*.

(i) Is ∆ *AOC* ≅ ∆ *BOD*?

(ii) State the matching pair you have used, which is not given in the question.

(iii) Is it true to say that ∠*ACO* = ∠*BDO*?

#### Answer:

We have

$\text{\u2220OAC=\u2220OBD,AO=OB}\phantom{\rule{0ex}{0ex}}\text{Also,\u2220AOC=\u2220BOD(Oppositeanglesonsamevertex)}\phantom{\rule{0ex}{0ex}}\text{Therefore,byASA\u25b3AOC\u2245\u25b3BOD}$

#### Page No 16.23:

#### Question 1:

In each of the following pairs of right triangles, the measures of some parts are indicated along side. State by the application of RHS congruence condition which are congruent. State each result in symbolic form. (Fig. 46)

#### Answer:

i) $\text{\u2220ADC}=\text{\u2220BCA}=90\xb0\phantom{\rule{0ex}{0ex}}\text{AD}=\text{BC}\phantom{\rule{0ex}{0ex}}\text{andhypAB}=\text{hypAB}\phantom{\rule{0ex}{0ex}}\text{Therefore,byRHS,}\u25b3\text{ADB}\cong \u25b3\text{ACB.}\phantom{\rule{0ex}{0ex}}$

ii)

$\text{AD=AD}\text{(Common)}\phantom{\rule{0ex}{0ex}}\text{hypAC}=\text{hypAB}\text{(Given)}\phantom{\rule{0ex}{0ex}}\text{\u2220ADB}+\text{\u2220ADC}=180\xb0\text{(Linearpair)}\phantom{\rule{0ex}{0ex}}\angle \text{ADB}+90\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\angle \text{ADB}=180\xb0-90\xb0=90\xb0\phantom{\rule{0ex}{0ex}}\angle \text{ADB}=\angle \text{ADC}=90\xb0\phantom{\rule{0ex}{0ex}}\text{Therefore,}\text{byRHS,}\text{\u25b3ADB}=\text{\u25b3ADC}$

iii)

$\text{hypAO}=\text{hypDO}\phantom{\rule{0ex}{0ex}}\text{BO}=\text{CO}\phantom{\rule{0ex}{0ex}}\angle \text{B}=\angle \text{C}=90\xb0\phantom{\rule{0ex}{0ex}}\text{Therefore,}\text{byRHS,}\text{\u25b3AOB}\cong \text{\u25b3DOC}$

iv)

$\text{HypAC}=\text{HypCA}\phantom{\rule{0ex}{0ex}}\text{BC}\hspace{0.17em}=\text{DC}\phantom{\rule{0ex}{0ex}}\text{\u2220ABC}=\text{\u2220ADC}=90\xb0\phantom{\rule{0ex}{0ex}}\text{Therefore,}\text{byRHS,}\text{\u25b3ABC}\cong \text{\u25b3ADC}\phantom{\rule{0ex}{0ex}}$

v)

$\text{BD}=\text{DB}\phantom{\rule{0ex}{0ex}}\text{HypAB}=\text{HypBC,}\text{asperthegivenfigure.}\phantom{\rule{0ex}{0ex}}\text{\u2220BDA}+\text{\u2220BDC}=180\xb0\phantom{\rule{0ex}{0ex}}\text{\u2220BDA}+90\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\text{\u2220BDA}=180\xb0-90\xb0=90\xb0\phantom{\rule{0ex}{0ex}}\text{\u2220BDA}=\text{\u2220BDC}=90\xb0\phantom{\rule{0ex}{0ex}}\text{Therefore,}\text{byRHS,}\text{\u25b3ABD}\cong \text{\u25b3CBD}$

#### Page No 16.24:

#### Question 2:

∆ *ABC* is isosceles with *AB* = *AC*. *AD* is the altitude from *A* on *BC*.

(i) Is ∆ *ABD* ≅ *ACD*?

(ii) State the pairs of matching parts you have used to answer (i).

(ii) Is it true to say that *BD* = *DC*?

#### Answer:

(i)Yes, $\u25b3ABD\cong \u25b3ACD$ by RHS congruence condition.

(ii) We have used Hyp AB = Hyp AC

AD = DA

and $\angle ADB=\angle ADC=90\xb0$ (AD$\perp $BC at point D)

(iii)Yes, it is true to say that BD = DC (c.p.c.t) since we have already proved that the two triangles are congruent.

#### Page No 16.24:

#### Question 3:

∆ *ABC* is isoseles with *AB* = *AC*. Also, *AD* ⊥ *BC* meeting *BC* in *D*. Are the two triangles *ABD* and *ACD* congruent? State in symbolic form. Which congruence condtion do you use? Which side of ∆ *ADC* equls *BD*? Which angle of ∆ *ADC* equals ∠*B*?

#### Answer:

We have AB = AC ......(1)

AD = DA (common)........(2)

and $\angle ADC=\angle ADB$ (AD$\perp $BC at point D)........(3)

Therefore from 1, 2 and 3, by RHS congruence condition,

$\text{\u25b3ABD}\cong \text{\u25b3ACD}\phantom{\rule{0ex}{0ex}}\text{Now,}\text{thetrianglesarecongruent}.\text{Therefore,}\text{BD}=\text{CD.}\phantom{\rule{0ex}{0ex}}\text{And}\text{\u2220ABD}=\text{\u2220ACD}\text{(c.p.c.t).}$

#### Page No 16.24:

#### Question 4:

Draw a right triangle *ABC*. Use *RHS* condition to construct another triangle congruent to it.

#### Answer:

Consider

$\text{\u25b3ABC}\text{with}\text{\u2220B}\text{asrightangle.}\phantom{\rule{0ex}{0ex}}\text{WenowconstructanotherrighttriangleonbaseBC,}\text{suchthat}\phantom{\rule{0ex}{0ex}}\text{\u2220C}\text{isarightangleand}\text{AB}=\text{DC}\phantom{\rule{0ex}{0ex}}\text{Also,}\text{BC}=\text{CB}\phantom{\rule{0ex}{0ex}}\text{Therefore,}\text{byRHS,}\text{\u25b3ABC}\cong \text{\u25b3DCB}$

#### Page No 16.24:

#### Question 5:

In Fig. 47, *BD* and *CE* are altitudes of ∆ *ABC* and *BD* = *CE*.

(i) Is ∆ *BCD* ≅ ∆ *CBE*?

(ii) State the three pairs of matching parts you have used to answer (i).

#### Answer:

(i) Yes, $\u25b3BCD\cong \u25b3CBE$ by RHS congruence condition.

(ii) We have used hyp BC = hyp CB

BD = CE (given in question)

and $\angle BDC=\angle CEB=90\xb0$.

#### Page No 16.3:

#### Question 1:

Explain the concept of congruence of figures with the help of certain examples.

#### Answer:

Congruent objects or figures are exact copies of each other or we can say mirror images of each other. The relation of two objects being congruent is called congruence.

Consider Ball A and Ball B. These two balls are congruent.

Now consider the two stars below. Star A and Star B are exactly the same in size, colour and shape. These are congruent stars.

Let us look at the triangles below, Here we have triangle PQR and triangle DEF.

These two triangles have corresponding angles equal and corresponding sides equal. Thus these triangles are congruent to each other.

#### Page No 16.3:

#### Question 2:

Fill in the blanks:

(i) Two line segments are congruent if .......

(ii) Two angles are congruent if ......

(iii) Two squres are congruent if .......

(iv) Two rectangles are congruent if .......

(v) Two circles are congruent if .......

#### Answer:

1) They have the same length, since they can superpose on each other.

2) Their measures are the same. On superposition, we can see that the angles are equal.

3) Their sides are equal. All the sides of a square are equal and if two squares have equal sides, then all their sides are of the same length. Also angles of a square are 90^{o} which is also the same for both the squares.

4) Their lengths are equal and their breadths are also equal. The opposite sides of a rectangle are equal. So if two rectangles have lengths of the same size and breadths of the same size, then they are congruent to each other.

5) Their radii are of the same length. Then the circles will have the same diameter and thus will be congruent to each other.

#### Page No 16.3:

#### Question 3:

In Fig., ∠*POQ* ≅ ∠*ROS*, can we say that ∠*POR* ≅ ∠*QOS*

#### Answer:

We have,

$\text{\u2220POQ\u2245\u2220ROS(1)}\phantom{\rule{0ex}{0ex}}\text{Also,\u2220ROQ\u2245\u2220ROQ(sameangle)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{Therefore,adding\u2220ROQtobothsidesof(1),weget:}\phantom{\rule{0ex}{0ex}}\text{\u2220POQ+\u2220ROQ\u2245\u2220ROQ+\u2220ROS}\phantom{\rule{0ex}{0ex}}\text{Therefore,\u2220POR\u2245\u2220QOS}\phantom{\rule{0ex}{0ex}}\text{Henceproved.}$

#### Page No 16.4:

#### Question 4:

In Fig., *a* = *b* = *c*, name the angle which is congruent to ∠*AOC*.

#### Answer:

We have,

$\text{\u2220AOB=\u2220BOC=\u2220COD}\phantom{\rule{0ex}{0ex}}\text{Therefore,\u2220AOB=\u2220COD}\phantom{\rule{0ex}{0ex}}\text{Also,\u2220AOB+\u2220BOC=\u2220BOC+\u2220COD}\phantom{\rule{0ex}{0ex}}\text{\u2220AOC=\u2220BOD}\phantom{\rule{0ex}{0ex}}\text{Hence,\u2220AOC\u2245\u2220BOD}\phantom{\rule{0ex}{0ex}}\text{\u2220BOD\hspace{0.17em}iscongruentto\u2220AOC}$

#### Page No 16.4:

#### Question 5:

Is it correct to say that any two right angles are congruent? Give reasons to justify your answer.

#### Answer:

Two right angles are congruent to each other because they both measure 90 degrees.

We know that two angles are congruent if they have the same measure.

#### Page No 16.4:

#### Question 6:

In Fig. 8, ∠*AOC* ≅ ∠*PYR* and ∠*BOC* ≅ ∠*QYR*. Name the angle which is congruent to ∠*AOB*.

#### Answer:

$\angle AOC\cong \angle PYR..........\left(i\right)\phantom{\rule{0ex}{0ex}}Also,\angle BOC\cong \angle QYR\phantom{\rule{0ex}{0ex}}Now,\angle AOC=\angle AOB+\angle BOCand\angle PYR=\angle PYQ+\angle QYR\phantom{\rule{0ex}{0ex}}Byputtingthevalueof\angle AOCand\angle PYRinequation\left(i\right),weget:\phantom{\rule{0ex}{0ex}}\angle AOB+\angle BOC\cong \angle PYQ+\angle QYR\phantom{\rule{0ex}{0ex}}\angle AOB\cong \angle PYQ(\angle BOC\cong \angle QYR)\phantom{\rule{0ex}{0ex}}Hence\angle AOB\cong \angle PYQ$

#### Page No 16.4:

#### Question 7:

Which of the following statements are true and which are false;

(i) All squares are congruent.

(ii) If two squares have equal areas, they are congruent.

(iii) If two rectangles have equal area, they are congruent.

(iv) If two triangles are equal in area, they are congruent.

#### Answer:

i) False. All the sides of a square are of equal length. However, different squares can have sides of different lengths. Hence all squares are not congruent.

ii) True

Area of a square = side $\times $side

Therefore, two squares that have the same area will have sides of the same lengths. Hence they will be congruent.

iii) False

Area of a rectangle = length $\times $ breadth

Two rectangles can have the same area. However, the lengths of their sides can vary and hence they are not congruent.

Example: Suppose rectangle 1 has sides 8 m and 8 m and area 64 metre square.

Rectangle 2 has sides 16 m and 4 m and area 64 metre square.

Then rectangle 1 and 2 are not congruent.

iv) False

Area of a triangle = $\frac{1}{2}\times \mathrm{base}\times \mathrm{height}$

Two triangles can have the same area but the lengths of their sides can vary and hence they cannot be congruent.

#### Page No 16.8:

#### Question 1:

In the following pairs of triangles (Fig. 12 to 15), the lengths of the sides are indicated along sides. By applying SSS condition, determine which are congruent. State the result in symbolic form.

#### Answer:

1) In $\u2206\text{ABCand\u2206DEF}$

AB = DE = 4.5 cm (Side)

BC = EF = 6 cm (Side)

and AC = DF = 4 cm (Side)

Therefore, by SSS criterion of congruence, $\u25b3\text{ABC}\cong \u25b3\text{DEF}$.

2)

$\text{In}\u25b3\text{ACBand}\u25b3\text{ADB}\phantom{\rule{0ex}{0ex}}\text{AC=AD(Side)}\phantom{\rule{0ex}{0ex}}\text{BC=BD(Side)}\phantom{\rule{0ex}{0ex}}\text{andAB\hspace{0.17em}=AB(Side)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Therefore, by SSS criterion of congruence, $\u25b3\text{ACB}\cong \u25b3\text{ADB}$.

3)

$\text{In}\u25b3\text{ABDand}\u25b3\text{FEC,}\phantom{\rule{0ex}{0ex}}\text{AB=FE(Side)}\phantom{\rule{0ex}{0ex}}\text{AD=FC(Side)}\phantom{\rule{0ex}{0ex}}\text{BD=CE(Side)}\phantom{\rule{0ex}{0ex}}$

Therefore, by SSS criterion of congruence, $\u25b3\text{ABD}\cong \u25b3\text{FEC}$.

#### Page No 16.8:

#### Question 2:

In Fig. 16, *AD* = *DC* and *AB* = *BC*.

(i) Is ∆* ABD* ≅ ∆* CBD*?

(ii) State the three parts of matching pairs you have used to answer (i).

#### Answer:

Yes $\u25b3ABD\cong \u25b3CBD$ by the SSS criterion.

We have used the three conditions in the SSS criterion as follows:

AD = DC

AB = BC

and DB = BD

#### Page No 16.9:

#### Question 3:

In Fig. 17, *AB* = *DC* and *BC* = *AD*.

(i) Is ∆ *ABC* ≅ ∆ *CDA*?

(ii) What congruence condition have you used?

(iii) You have used some fact, not given in the question, what is that?

#### Answer:

We have AB = DC

BC = AD

and AC = AC

Therefore by SSS $\u25b3ABC\cong \u25b3CDA$.

We have used Side Side Side congruence condition with one side common in both the triangles.

Yes, we have used the fact that AC = CA.

#### Page No 16.9:

#### Question 4:

If ∆ *PQR* ≅ ∆ EFD,

(i) Which side of ∆ *PQR* equals *ED*?

(ii) Which angle of ∆ *PQR* equals ∠*E*?

#### Answer:

△PQR $\cong $ △EDF

1) Therefore PR = ED since the corresponding sides of congruent triangles are equal.

2) $\angle QPR=\angle FED$ since the corresponding angles of congruent triangles are equal.

#### Page No 16.9:

#### Question 5:

Triangles *ABC* and *PQR* are both isosceles with *AB* = *AC* and *PQ* = *PR* respectively. If also, *AB* = *PQ* and *BC* = *QR*, are the two triangles congruent? Which condition do you use?

If ∠*B* = 50°, what is the measure of ∠*R*?

#### Answer:

We have AB = AC in isosceles $\u25b3$ABC

and PQ = PR in isosceles $\u25b3$PQR.

Also, we are given that AB = PQ and QR = BC.

Therefore, AC = PR (AB = AC, PQ = PR and AB = PQ)

Hence, $\u25b3\text{ABC}\cong \u25b3\text{PQR}$.

Now

$\angle \mathrm{ABC}=\angle \mathrm{PQR}(\mathrm{Since}\mathrm{triangles}\mathrm{are}\mathrm{congruent})\phantom{\rule{0ex}{0ex}}\text{However,\u25b3PQRisisosceles.}\phantom{\rule{0ex}{0ex}}\text{Therefore,\u2220PRQ=\u2220PQR=\u2220ABC}=50\xb0$

#### Page No 16.9:

#### Question 6:

*ABC* and *DBC* are both isosceles triangles on a common base *BC* such that *A* and *D* lie on the same side of *BC*. Are triangles *ADB* and *ADC* congruent? Which condition do you use? If ∠*BAC* = 40° and∠*BDC* = 100°; then find ∠*ADB*.

#### Answer:

YES $\u25b3\text{ADB}\cong \u25b3\text{ADC(BySSS)}$

AB = AC , DB = DC AND AD= DA

$\text{\u2220BAD=\u2220CAD(c.p.c.t)}\phantom{\rule{0ex}{0ex}}\text{\u2220BAD+\u2220CAD}=40\xb0\phantom{\rule{0ex}{0ex}}2\text{\u2220BAD}=40\xb0\phantom{\rule{0ex}{0ex}}\angle \text{BAD}=\frac{40\xb0}{2}=20\xb0\phantom{\rule{0ex}{0ex}}$

$\text{\u2220ABC+\u2220BCA+\u2220BAC}=180\xb0\left(\text{Anglesumproperty}\right)\phantom{\rule{0ex}{0ex}}\text{Since}\u2206\text{ABCisanisoscelestriangle,}\phantom{\rule{0ex}{0ex}}\angle \text{ABC=\u2220BCA}\phantom{\rule{0ex}{0ex}}\text{\u2220ABC+\u2220ABC}+40\xb0=180\xb0\phantom{\rule{0ex}{0ex}}2\text{\u2220ABC}=180\xb0-40\xb0=140\xb0\phantom{\rule{0ex}{0ex}}\angle \text{ABC}=\frac{140\xb0}{2}=70\xb0$

$\text{\u2220DBC+\u2220BCD+\u2220BDC}=180\xb0\left(\text{Anglesumproperty}\right)\phantom{\rule{0ex}{0ex}}\text{Since\u2206ABCisanisoscelestriangle,}\phantom{\rule{0ex}{0ex}}\text{\u2220DBC=\u2220BCD}\phantom{\rule{0ex}{0ex}}\text{\u2220DBC+\u2220DBC}+100\xb0=180\xb0\phantom{\rule{0ex}{0ex}}2\text{\u2220DBC}=180\xb0-100\xb0=80\xb0\phantom{\rule{0ex}{0ex}}\angle \text{DBC}=\frac{80\xb0}{2}=40\xb0$

$\text{In\u2206BAD,}\phantom{\rule{0ex}{0ex}}\text{\u2220ABD+\u2220BAD+\u2220ADB}=180\xb0\text{(Anglesumproperty)}\phantom{\rule{0ex}{0ex}}30\xb0+20\xb0+\text{\u2220ADB}=180\xb0\left(\text{\u2220ABD=\u2220ABC-\u2220DBC}\right)\phantom{\rule{0ex}{0ex}}\angle \text{ADB}=180\xb0-20\xb0-30\xb0\phantom{\rule{0ex}{0ex}}\angle \text{ADB}=130\xb0\phantom{\rule{0ex}{0ex}}$

$\angle ADB=130\xb0$

#### Page No 16.9:

#### Question 7:

∆ *ABC* and ∆ *ABD* are on a common base *AB*, and *AC* = *BD* and *BC* = *AD* as shown in Fig. 18. Which of the following statements is true?

(i) ∆ *ABC* ≅ ∆ *ABD*

(ii) ∆ *ABC* ≅ ∆ *ADB*

(iii) ∆ *ABC* ≅ ∆ *BAD*

#### Answer:

In $\u25b3$ABC and $\u25b3$BAD we have,

AC = BD (given)

BC = AD (given)

and AB = BA (common)

Therefore by SSS criterion of congruency, $\u25b3$ABC $\cong $$\u25b3$BAD.

There option (iii) is true.

#### Page No 16.9:

#### Question 8:

In Fig. 19, ∆ *ABC* is isosceles with *AB* = *AC*, *D* is the mid-point of base *BC*.

(i) Is ∆ *ADB* ≅ ∆ *ADC*?

(ii) State the three pairs of matching parts you use to arrive at your answer.

#### Answer:

We have AB = AC.

Also since D is the midpoint of BC, BD = DC.

And AD = DA.

Therefore by SSS condition, $\u25b3ABD\cong \u25b3ADC$.

We have used AB, AC : BD, DC and AD, DA.

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#### Question 9:

In Fig. 20, ∆ *ABC* is isosceles with *AB* = *AC*. State if ∆ *ABC* ≅ ∆ *ACB*. If yes, state three relations that you use to arrive at your answer.

#### Answer:

Yes $\u25b3ABC\cong \u25b3ACB$ by SSS condition.

Since ABC is an isosceles triangle, AB = AC, BC = CB and AC = AB.

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