Rd Sharma 2021 Solutions for Class 9 Maths Chapter 12 Congruent Triangles are provided here with simple step-by-step explanations. These solutions for Congruent Triangles are extremely popular among Class 9 students for Maths Congruent Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2021 Book of Class 9 Maths Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2021 Solutions. All Rd Sharma 2021 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 12.15:

#### Question 1:

In the given figure, the sides *BA* and *CA* have been produced such that *BA* = *AD* and *CA* = *AE*. Prove that segment *DE* || *BC*.

#### Answer:

It is given that

We have to prove that

Now considering the two triangles we have

In

(Given)

(Given)

We need to show to prove.

Now

(Vertically opposite angle)

So by congruence criterion we have

So and

Then

, and

Hence from above conditions .

#### Page No 12.15:

#### Question 2:

In a Δ*PQR*, if *PQ* = *QR* and *L*, *M* and *N* are the mid-points of the sides *PQ*, *OR*, and *RP* respectively. Prove that *LN* = *MN*.

#### Answer:

It is given that

And is the mid point of

So

And is the mid point of

So

And is the mid point of

So

We have to prove that

In we have

(Equilateral triangle)

Then

, and

, and

Similarly comparing and we have

, and

And (Since *N* is the mid point of )

So by congruence criterion, we have

Hence.

#### Page No 12.15:

#### Question 3:

Prove that the medians of an equilateral triangle are equal.

#### Answer:

We have to prove that the median of an equilateral triangle are equal.

Let be an equilateral triangle with as its medians.

Let

In we have

(Since similarly)

(In equilateral triangle, each angle)

And (common side)

So by congruence criterion we have

This implies that,

Similarly we have

Hence .

#### Page No 12.15:

#### Question 4:

In a Δ *ABC*, if ∠*A* = 120° and *AB* = *AC*. Find ∠*B* and ∠*C*.

#### Answer:

In, it is given that

, and

We have to find, and

Since and

Then (as AB = AC)

Now

(By property of triangle)

Thus,

, as (given)

So,

Since,, so

Hence .

#### Page No 12.15:

#### Question 5:

In a Δ*ABC*, if *AB* = *AC* and ∠*B* = 70°, find ∠*A*.

#### Answer:

In it is given that

, and

We have to find.

Since

Then (isosceles triangles)

Now

(As given)

Thus

(Property of triangle)

Hence .

#### Page No 12.15:

#### Question 6:

The vertical angle of an isosceles triangle is 100°. Find its base angles.

#### Answer:

Suppose in the isosceles triangle ΔABC it is given that

We have to find the base angle.

Now vertical angle (given)

And

Since then

Now

(By property of triangle)

So

Hence the base angle is.

#### Page No 12.15:

#### Question 7:

In the given figure, *AB* = *AC* and ∠*ACD* = 105°, find ∠*BAC*.

#### Answer:

It is given that

We have to find.

(Isosceles triangle)

Now

Since exterior angle of isosceles triangle is the sum of two internal base angles

Now

So, (By property of triangle)

Hence .

#### Page No 12.15:

#### Question 8:

Find the measure of each exterior angle of an equilateral triangle.

#### Answer:

We have to find the measure of each exterior angle of an equilateral triangle.

It is given that the triangle is equilateral

So, and

Since triangle is equilateral

So,

Now we have to find the exterior angle.

As we know that exterior angle of the triangle is sum of two interior angles

Thus

Hence each exterior angle is.

#### Page No 12.15:

#### Question 9:

If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other.

#### Answer:

It is given that the base of an isosceles triangle is produced on both sides.

We have to prove that the exterior angles so formed are equal to each other.

That is we need to show that

Let the is isosceles having base and equal sides *AB* and *AC*

Then, and

(Isosceles triangles)

Now

.........(1)

And,

.......(2)

Thus

........(3)

Now from equation (2)

.........(4)

Since

Hence from equation (3) and (4)

#### Page No 12.15:

#### Question 10:

In the given figure, *AB* = *AC* and *DB *= *DC*, find the ratio ∠*ABD* : ∠*ACD*.

#### Answer:

It is given that

We have to find the ratio.

Since

And

So we have,

So

Hence .

#### Page No 12.16:

#### Question 11:

Determine the measure of each of the equal angles of a right-angled isosceles triangle.

*ABC*is a right-angled triangle in which ∠

*A*= 90° and

*AB*=

*AC*. Find ∠

*B*and ∠

*C*.

#### Answer:

It is given that

Is right angled triangle

And

We have to find and

Since

(Isosceles triangle)

Now

(Property of triangle)

()

So

Hence

#### Page No 12.16:

#### Question 12:

In the given figure, *PQRS* is a square and *SRT* is an equilateral triangle. Prove that

(i) *PT* = *QT*

(ii) ∠*TQR* = 15°

#### Answer:

It is given that

Δis a square and Δ is an equilateral triangle.

We have to prove that

(1) and (2)

(1)

Since,

(Angle of square)

(Angle of equilateral triangle)

Now, adding both

Similarly, we have

Thus in and we have

(Side of square)

And (equilateral triangle side)

So by congruence criterion we have

Hence.

(2)

Since

QR = RS ( Sides of Square)

RS = RT (Sides of Equilateral triangle)

We get

QR = RT

Thus, we get

$\angle TQR=\angle RTQ$ (Angles opposite to equal sides are equal)

Now, in the triangle TQR, we have

$\angle TQR+\angle RTQ+\angle QRT={180}^{0}\phantom{\rule{0ex}{0ex}}\angle TQR+\angle TQR+{150}^{0}={180}^{0}\phantom{\rule{0ex}{0ex}}2\angle TQR+{150}^{0}={180}^{0}\phantom{\rule{0ex}{0ex}}2\angle TQR={180}^{0}-{150}^{0}\phantom{\rule{0ex}{0ex}}2\angle TQR={30}^{0}\phantom{\rule{0ex}{0ex}}\angle TQR=\frac{{30}^{0}}{2}={15}^{0}\phantom{\rule{0ex}{0ex}}$

#### Page No 12.16:

#### Question 13:

In the given figure, *AB *is a line segment. *P* and *Q* are points on opposite sides of *AB* such that each of them is equidistant from the points *A* and *B*. Show that the line* PQ* is perpendicular bisector of *AB*.

#### Answer:

It is given that

*P* and *Q *are equidistant from *A* and *B* that is

, and

We are asked to show that line *PO* is perpendicular bisector of line *AB.*

First of all we will show that ΔAQP and ΔQBP are congruent to each other and ultimately we get the result.

Consider the triangles AQP and QBP in which

*AP=BP*, *AQ=BQ*, *PQ=PQ*

So by *SSS* property we have

Implies that

Now consider the triangles ΔAPC and ΔPCB in which

And

So by *SAS *criterion we find that,

So this implies that *AC=BC* and

But

Hence *PQ* is perpendicular bisector of *AB*.

#### Page No 12.16:

#### Question 14:

In a ∆*ABC*, *D *is the mid-point of *AC *such that *BD *= $\frac{1}{2}$ *AC*. Show that ∠*ABC *is a right angle.

#### Answer:

In a ∆ABC, D is the mid-point of AC such that BD = $\frac{1}{2}$AC.

D is the mid-point of AC.

∴ AD = CD = $\frac{1}{2}$AC

⇒ AD = CD = BD (BD = $\frac{1}{2}$AC)

In ∆ABD,

AD = BD

∴ ∠ABD = ∠A .....(1) (In a triangle, equal sides have equal angles opposite to them)

In ∆CBD,

CD = BD

∴ ∠CBD = ∠C .....(2) (In a triangle, equal sides have equal angles opposite to them)

Adding (1) and (2), we get

∠ABD + ∠CBD = ∠A + ∠C

⇒ ∠B = ∠A + ∠C .....(3)

In ∆ABC,

∠A + ∠B + ∠C = 180º (Angle sum property of triangle)

⇒ ∠B + ∠B = 180º [Using (3)]

⇒ 2∠B = 180º

⇒ ∠B = $\frac{180\xb0}{2}$ = 90º

Thus, ∠ABC is a right angle.

#### Page No 12.16:

#### Question 15:

∆*ABC *is a right triangle with *AB *= *AC*. Bisector of ∠*A* meets *BC *at *D*. Prove that *BC *= 2*AD*.

#### Answer:

In ∆ABC,

AB = AC (Given)

∴ ∠C = ∠B .....(1) (In a triangle, angles opposite to equal sides are equal)

Also, ∠A = 90º (Given)

Now,

∠A + ∠B + ∠C = 180º (Angle sum property of triangle)

⇒ 90º + 2∠B = 180º [Using (1)]

⇒ 2∠B = 180º − 90º = 90º

⇒ ∠B = $\frac{90\xb0}{2}$ = 45º

∴ ∠C = ∠B = 45º

It is given that, AD is the bisector of ∠A.

∴ ∠CAD = ∠BAD$=\frac{\angle \mathrm{A}}{2}=\frac{90\xb0}{2}$ = 45º

In ∆ABD,

∠B = ∠BAD (Each measure 45º)

∴ AD = BD .....(2) (In a triangle, sides opposite to equal angles are equal)

In ∆ACD,

∠C = ∠CAD (Each measure 45º)

∴ AD = CD .....(3) (In a triangle, sides opposite to equal angles are equal)

Adding (2) and (3), we have

AD + AD = BD + CD

⇒ 2AD = BC

Or BC = 2AD

Hence proved.

#### Page No 12.16:

#### Question 16:

∆*ABC *is a right triangle right angled at *B* such that ∠*BCA *= 2∠*BAC*. Show that *AC *= 2*BC*.

#### Answer:

∆ABC is a right triangle right angled at B such that ∠BCA = 2∠BAC.

Produce CB to D such that BC = BD.

In ∆ABD and ∆ABC,

BD = BC (Construction)

∠ABD = ∠ABC (90º each)

AB = AB (Common)

∴∆ABD ≅ ∆ABC (SAS congruence axiom)

So, AD = AC .....(1) (CPCT)

∠BAD = ∠BAC .....(2) (CPCT)

Now,

∠BCA = 2∠BAC

⇒ ∠BCA = ∠BAC + ∠BAC

⇒ ∠BCA = ∠BAC + ∠BAD [Using (2)]

⇒ ∠BCA = ∠CAD

In ∆ACD,

∠DCA = ∠CAD (Proved above)

⇒ AD = CD (Sides opposite to equal angles in a triangle are equal)

⇒ AC = BC + BD [Using (1)]

⇒ AC = BC + BC (BC = BD)

⇒ AC = 2BC

#### Page No 12.25:

#### Question 1:

BD and CE are bisectors of ∠B and ∠C of an isosceles ΔABC with AB = AC. Prove that BD** =** CE.

#### Answer:

It is given that

Is bisector of and is bisector of.

And is isosceles with

We have to prove that

If will be sufficient to prove to show that

Now in these two triangles

Since, so

Now as BD and CE are bisector of the respectively**, so**

**, **and

*BC=BC*

So by congruence criterion we have

Hence Proved.

#### Page No 12.25:

#### Question 2:

In the given figure, it is given that *RT* =* TS*, ∠1 = 2∠2 and ∠4 = 2∠3. Prove that Δ*RBT* **≅**Δ*SAT**.*

#### Answer:

It is given that

We have to prove that

Now

In we have

(Isosceles triangle) .......(1)

Now we have

(Vertically opposite angles)

(Since, given)

.......(2)

Subtracting equation (2) from equation (1) we have

Now in and we have

(Given)

So all the criterion for the two triangles and are satisfied to be congruent

Hence by congruence criterion we have proved.

#### Page No 12.25:

#### Question 3:

Two lines *AB* and *CD* intersect at *O *such that *BC* is equal and parallel to *AD*. Prove that the lines *AB *and *CD* bisect at *O*.

#### Answer:

It is given that

We have to prove that the lines and bisect at.

If we prove that, then

We can prove and bisects at.

Now in and

(Given)

(Since and is transversal)

And (since and is transversal)

So by congruence criterion we have,

, so

Hence and bisect each other at.

#### Page No 12.47:

#### Question 1:

In two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.

#### Answer:

It is given that

We are asked to show that

Let us assume

, and are right angled triangle.

Thus in and, we have

And (given)

Hence by AAs congruence criterion we haveProved.

#### Page No 12.47:

#### Question 2:

If the bisector of the exterior vertical angle of a triangle be parallel to the base. Show that the triangle is isosceles.

#### Answer:

We have to prove that is isosceles.

Let Δ be such that the bisector of is parallel to

The base, we have

(Corresponding angles)

(Alternate angle)

(Since)

Hence is isosceles.

#### Page No 12.47:

#### Question 3:

In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.

#### Answer:

In the triangle *ABC* it* *is given that the vertex angle is twice of base angle.

We have to calculate the angles of triangle.

Now, let be an isosceles triangle such that

Then

(Given)

()

Now (property of triangle)

Hence

#### Page No 12.47:

#### Question 4:

Prove that each angle of an equilateral triangle is 60°

#### Answer:

We have to prove each angle of an equilateral triangle is.

Here

(Side of equilateral triangle)

...........(1)

And

(Side of equilateral triangle)

..........(2)

From equation (1) and (2) we have

Hence

Now

That is (since)

Hence Proved.

#### Page No 12.47:

#### Question 5:

Angles *A*, *B*, *C* of a triangle *ABC* are equal to each other. Prove that Δ*ABC* is equilateral.

#### Answer:

It is given that

We have to prove that triangle Δ*ABC* is equilateral.

Since (Given)

So, ..........(1)

And (given)

So ........(2)

From equation (1) and (2) we have

Now from above equation if we have

Given condition satisfy the criteria of equilateral triangle.

Hence the given triangle is equilateral.

#### Page No 12.47:

#### Question 6:

*ABC* is a right angled triangle in which ∠*A* = 90° and *AB* = *AC*. Find ∠*B* and ∠*C*.

#### Answer:

It is given that

We have to find and.

Since so,

Now (property of triangle)

(Since )

Here

Then

Hence

#### Page No 12.47:

#### Question 7:

*PQR* is a triangle in which *PQ* = *PR* and *S* is any point on the side *PQ*. Through *S*, a line is drawn parallel to *QR* and intersecting *PR* at *T*. Prove that *PS* = *PT*.

#### Answer:

It is given that

We have to prove

In we have

(Given)

So,

Now (Given)

Since corresponding angle are equal, so

That is,

Henceproved.

#### Page No 12.47:

#### Question 8:

In a Δ*ABC*, it is given that *AB* = *AC *and the bisectors of ∠*B* and ∠*C* intersect at *O*. If *M* is a point on *BO* produced prove that ∠*MOC* = ∠*ABC*.

#### Answer:

It is given that

In,

We have to prove that

Now

(Given)

Thus

........(1)

In, we have

So, {from equation (1)}

Hence Proved.

#### Page No 12.47:

#### Question 9:

P is a point on the bisector of an angle ∠ABC. If the line through P parallel to AB meets BC at Q, prove that triangle BPQ is isosceles.

#### Answer:

In the following figure it is given that sides *AB* and *PQ* are parallel and BP is bisector of

We have to prove that is an isosceles triangle.

(Since BP is the bisector of) ........(1)

(Since and are parallel) .......(2)

Now from equation (1) and (2) we have

So

Now since and is a side of.

And since two sides and are equal, so

Hence is an isosceles triangle.

#### Page No 12.47:

#### Question 10:

*ABC* is a triangle in which ∠*B* = 2∠*C*. *D* is a point on *BC* such that *AD* bisects ∠*BAC* and *AB* = *CD*. Prove that ∠*BAC* = 72°.

#### Answer:

It is given that in

And bisects

We have to prove that

Now let

(Given)

Sinceis a bisector of so let

Let be the bisector of

If we join we have

In

So

In triangle and we have

(Given)

(Proved above)

So by congruence criterion, we have

And

, and (since)

In we have

Since,

And,

So,

In we have

Here,

HenceProved.

#### Page No 12.47:

#### Question 11:

Bisectors of angles *B* and *C* of an isosceles triangle *ABC *with *AB *= *AC *intersect each other at *O*. Show that external angle adjacent to ∠*ABC *is equal to ∠*BOC*.

#### Answer:

∠ABD is the external angle adjacent to ∠ABC.

∆ABC is an isosceles triangle.

AB = AC (Given)

∴ ∠C = ∠ABC .....(1) (In a triangle, equal sides have equal angles opposite to them)

Also, OB and OC are the bisectors of ∠B and ∠C, respectively.

$\therefore \angle \mathrm{OBC}=\frac{\angle \mathrm{ABC}}{2}.....\left(2\right)$

Similarly, $\angle \mathrm{OCB}=\frac{\angle \mathrm{C}}{2}.....\left(3\right)$

In ∆BOC,

∠OBC + ∠OCB + ∠BOC = 180º (Angle sum property of triangle)

$\Rightarrow \frac{\angle \mathrm{ABC}}{2}+\frac{\angle \mathrm{C}}{2}+\angle \mathrm{BOC}=180\xb0\left[\mathrm{Using}\left(2\right)\mathrm{and}\left(3\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\angle \mathrm{ABC}+\angle \mathrm{C}}{2}+\angle \mathrm{BOC}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2\angle \mathrm{ABC}}{2}+\angle \mathrm{BOC}=180\xb0\left[\mathrm{Using}\left(1\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{ABC}+\angle \mathrm{BOC}=180\xb0.....\left(4\right)$

Now,

∠ABD + ∠ABC = 180º .....(5) (Linear pair)

From (4) and (5), we have

∠ABD + ∠ABC = ∠ABC + ∠BOC

⇒ ∠ABD = ∠BOC

Thus, the external angle adjacent to ∠ABC is equal to ∠BOC.

#### Page No 12.57:

#### Question 1:

In the given figure, it is given that *AB* = *CD* and *AD* = *BC*. Prove that Δ*ADC* **≅**Δ*CBA*.

#### Answer:

It is given that

We have to prove that.

Now in triangles and we have

(Given)

(Given)

So (common)

Each side of is equal to .

Hence, by congruence criterion we have Proved.

#### Page No 12.58:

#### Question 2:

In a Δ*PQR*, if *PQ* = *QR* and *L*, *M* and *N* are the mid-points of the sides *PQ*, *QR* and *RP* respectively. Prove that *LN* = *MN*.

#### Answer:

It is given that

and *L*, *M*, *N* are the mid points of sides, , and respectively.

We have to prove that

Now using the mid point theorem, we have

And

Similarly we have

In triangle and we have

(Proved above)

(Proved above)

And (common)

So, by congruence criterion, we have

And

Then

HenceProved.

#### Page No 12.61:

#### Question 1:

*ABC* is a triangle and *D* is the mid-point of *BC*. The perpendiculars from *D* to *AB* and *AC* are equal. Prove that the triangle is-isosceles.

#### Answer:

We have to prove thatis isosceles.

Let and be perpendicular from D on *AB* and *AC* respectively.

In order to prove that

We will prove that

Now in and we have

(Since *D* is mid point of *BC*)

(Given)

So by congruence criterion we have

And

Hence is isosceles.

#### Page No 12.61:

#### Question 2:

*ABC* is a triangle in which *BE* and *CF* are, respectively, the perpendiculars to the sides *AC* and *AB*. If *BE* = *CF*, prove that Δ*ABC* is isosceles.

#### Answer:

It is given that

, and

And.

We have to prove is isosceles.

To prove is isosceles we will prove

For this we have to prove

Now comparing and we have

(Given)

(Common side)

So, by right hand side congruence criterion we have

So (since sides opposite to equal angle are equal)

Hence is isosceles.

#### Page No 12.61:

#### Question 3:

If perpendiculars from any point within an angle on its arms are congruent, prove that it lies on the bisector of that angle.

#### Answer:

Let *P* be a point within such that

We have to prove that *P* lies on the bisector of

In and we have

(We have)

(Common)

So by right hand side congruence criterion, we have

So,

Hence *P* lies on the bisector of proved.

#### Page No 12.61:

#### Question 4:

In the given figure, *AD* ⊥ *CD* and *CB* ⊥ *CD*. If *AQ* = *BP* and *DP* = *CQ*, prove that

∠*DAQ* = ∠*CBP*.

#### Answer:

It is given that

, and

If and

We have to prove that

In triangles and we have

(Since given)

So

And (given)

So by right hand side congruence criterion we have

So

HenceProved.

#### Page No 12.62:

#### Question 5:

Which of the following statements are true (T) and which are false (F):

(i) Sides opposite to equal angles of a triangle may be unequal.

(ii) Angles opposite to equal sides of a triangle are equal.

(iii) The measure of each angle of an equilateral triangle is 60°.

(iv) If the altitude from one vertex of a triangle bisects the opposite side, then the triangle may be isosceles.

(v) The bisectors of two equal angles of a triangle are equal.

(vi) If the bisector of the vertical angle of a triangle bisects the base, then the triangle may be isosceles.

(vii) The two altitudes corresponding to two equal sides of a triangle need not be equal.

(viii) If any two sides of a right triangle are respectively equal to two sides of other right triangle, then the two triangles are congruent.

(ix) Two right triangles are congruent if hypotenuse and a side of one triangle are respectively equal equal to the hypotenuse and a side of the other triangle.

#### Answer:

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

#### Page No 12.62:

#### Question 6:

Fill in the blanks in the following so that each of the following statements is true.

(i) Sides opposite to equal angles of a triangle are .......

(ii) Angle opposite to equal sides of a triangle are .......

(iii) In an equilateral triangle all angles are ........

(iv) In a *ABC* if ∠*A* = ∠*C*, then *AB* = ............

(v) If altitudes *CE* and *BF* of a triangle *ABC* are equal, then *AB* = ........

(vi) In an isosceles triangle *ABC* with *AB* = *AC*, if *BD* and *CE* are its altitudes, then *BD* is ...... *CE*.

(vii) In right triangles *ABC* and *DEF*, if hypotenuse *AB* = *EF* and side *AC* = *DE*, then Δ*ABC* **≅** Δ .........

#### Answer:

(1)

(2)

(3)

(4)

(5)

(6)

(7)

#### Page No 12.62:

#### Question 7:

*ABCD* is a square, *X* and *Y* are points on sides *AD* and *BC* respectively such that *AY*= *BX*. Prove that *BY* = *AX* and ∠*BAY* = ∠*ABX*.

#### Answer:

It is given *ABCD* is a square and

We have to prove that and

In right angled trianglesand Δ we have

And, and

So by right hand side congruence criterion we have

So (since triangle is congruent)

HenceProved.

#### Page No 12.62:

#### Question 8:

*ABCD *is a quadrilateral such that *AB *= *AD *and *CB *= *CD*. Prove that *AC *is the perpendicular bisector of *BD*.

#### Answer:

In quadrilateral ABCD, AB = AD and BC = CD. Let AC and BD intersect at O.

In ∆ABC and ∆ADC,

AB = AD (Given)

AC = AC (Common)

BC = CD (Given)

∴ ∆ABC ≅ ∆ADC (SSS congruence criterion)

⇒ ∠BAC = ∠DAC (CPCT)

Now, in ∆ABO and ∆ADO,

AB = AD (Given)

∠BAO = ∠DAO (Proved above)

AO = AO (Common)

∴ ∆ABO ≅ ∆ADO (SAS congruence axiom)

⇒ OB = OD .....(1) (CPCT)

∠AOB = ∠AOD .....(2) (CPCT)

Now,

∠AOB + ∠AOD = 180º (Linear pair)

⇒ 2∠AOB = 180º [Using (2)]

⇒ ∠AOB = 90º .....(3)

From (1) and (3), we conclude that AC is the perpendicular bisector of BD.

#### Page No 12.62:

#### Question 9:

∆*ABC *is a right triangle such that *AB *= *AC *and bisector of angle *C* intersects the side *AB *at *D*. Prove that *AC *+ *AD *= *BC*.

#### Answer:

∆ABC is a right triangle such that AB = AC. CD is the bisector of ∠C which intersects AB at D.

Let AB = AC = *x* units

In right ∆ABC,

${\mathrm{BC}}^{2}={\mathrm{AB}}^{2}+{\mathrm{AC}}^{2}$ (Pythagoras theorem)

$\Rightarrow {\mathrm{BC}}^{2}={x}^{2}+{x}^{2}=2{x}^{2}$

$\Rightarrow \mathrm{BC}=\sqrt{2}x$

In ∆ABC, CD is the bisector of ∠C.

$\therefore \frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AC}}{\mathrm{BC}}$ (The bisector of an angle of a triangle divides the opposite side in the ratio of sides containing the angle)

$\Rightarrow \frac{\mathrm{AD}}{x-\mathrm{AD}}=\frac{x}{\sqrt{2}x}$

$\Rightarrow \sqrt{2}\mathrm{AD}=x-\mathrm{AD}$

$\Rightarrow \mathrm{AD}\left(\sqrt{2}+1\right)=x$

$\Rightarrow \mathrm{AD}=\frac{x}{\sqrt{2}+1}=x\left(\sqrt{2}-1\right)$

Now,

$\mathrm{AC}+\mathrm{AD}=x+\sqrt{2}\left(x-1\right)=\sqrt{2}x=\mathrm{BC}$

Hence proved.

#### Page No 12.62:

#### Question 10:

*O* is a point in the interior of a square *ABCD *such that *OAB *is an equilateral triangle. Show that ∆*OCD *is an isosceles triangle.

#### Answer:

O is a point in the interior of square ABCD. ∆OAB is an equilateral triangle.

Now,

∠DAB = ∠CBA .....(1) (Measure of each angle of a square is 90º)

∠OAB = ∠OBA .....(2) (Measure of each angle of an equilateral triangle is 60º)

Subtracting (2) from (1), we get

∠DAB − ∠OAB = ∠CBA − ∠OBA

⇒ ∠OAD = ∠OBC

In ∆OAD and ∆OBC,

OA = OB (Sides of an equilateral triangle are equal)

∠OAD = ∠OBC (Proved above)

AD = BC (Sides of a square are equal)

∴ ∆OAD ≅ ∆OBC (SAS congruence axiom)

⇒ OD = OC (CPCT)

In ∆OCD,

OC = OD

∴ ∆OCD is an isosceles triangle. (A triangle whose two sides are equal is an isosceles triangle)

#### Page No 12.62:

#### Question 11:

*ABCD *is a quadrilateral in which *AB *= *BC *and *AD *= *CD*. Show that *BD *bisects both the angles ∠*ABC *and ∠*ADC*.

#### Answer:

In quadrilateral ABCD, AB = BC and AD = CD.

In ∆ABD and ∆CBD,

AB = CB (Given)

BD = BD (Common)

AD = CD (Given)

∴ ∆ABD ≅ ∆CBD (SSS congruence criterion)

So, ∠ABD = ∠CBD .....(1) (CPCT)

∠ADB = ∠CDB .....(2) (CPCT)

From (1) and (2), we conclude that

BD bisects both ∠ABC and ∠ADC.

#### Page No 12.62:

#### Question 12:

Line segment joining the mid-points *M *and *N *of parallel sides *AB *and *DC*, respectively of a trapezium *ABCD *is perpendicular to both the sides *AB *and *DC*. Prove that *AD *= *BC*.

#### Answer:

ABCD is a trapezium with AB || CD. M and N are the mid-points of sides AB and AC, respectively.

Join AN and BN.

In ∆AMN and ∆BMN,

AM = BM (M is the mid-point of AB)

∠AMN = ∠BMN (MN ⊥ AB)

MN = MN (Common)

∴ ∆AMN ≅ ∆BMN (SAS congruence axiom)

So, AN = BN .....(1) (CPCT)

∠ANM = ∠BNM (CPCT)

Now,

∠DNM = ∠CNM (90º each)

∴ ∠DNM − ∠ANM = ∠CNM − ∠BNM

⇒ ∠AND = ∠BNC .....(2)

In ∆AND and ∆BNC,

DN = CN (N is the mid-point of CD)

∠AND = ∠BNC [From (2)]

AN = BN [From (1)]

∴ ∆AND ≅ ∆BNC (SAS congruence axiom)

So, AD = BC (CPCT)

Hence proved.

#### Page No 12.81:

#### Question 1:

In Δ*ABC*, if ∠*A* = 40° and ∠*B* = 60°. Determine the longest and shortest sides of the triangle.

#### Answer:

In the triangle *ABC* it is given that

We have to find the longest and shortest side.

Here

Now is the largest angle of the triangle.

So the side in front of the largest angle will be the longest side.

Hence will be the longest

Since is the shortest angle so that side in front of it will be the shortest.

And is shortest side

Hence Is longest and is shortest.

#### Page No 12.81:

#### Question 2:

In a Δ*ABC*, if ∠*B* = ∠*C* = 45°, which is the longest side?

#### Answer:

In the triangle *ABC* it is given that

We have to find the longest side.

Here

(Since)

Now is the largest angle of the triangle.

So the side in front of the largest angle will be the longest side.

Hence *BC* will be the longest side.

#### Page No 12.81:

#### Question 3:

In Δ *ABC*, side *AB* is produced to *D* so that *BD* = *BC.* If ∠*B* = 60° and ∠*A* = 70°, prove that :

(i) *AD* > *CD* (ii) *AD*>*AC*

#### Answer:

It is given that

, and

We have to prove that

(1)

(2)

(1)

Now

And since *BD=BC*, so, and

That is,

Now

And, so

Hence (1) (Side in front of greater angle will be longer)

And (2) Proved.

#### Page No 12.81:

#### Question 4:

Is a possible to draw a triangle with sides of length 2 cm, 3 cm and 7 cm?

#### Answer:

As we know that a triangle can only be formed if

The sum of two sides is greater than the third side.

Here we have 2 cm, 3 cm and 7 cm as sides.

If we add

(Since 5 is less than 7)

Hence the sum of two sides is less than the third sides

So, the triangle will not exist.

#### Page No 12.81:

#### Question 5:

In Δ *ABC*, ∠B = 35°, ∠*C* = 65° and the bisector of ∠*BAC* meets *BC* in *P*. Arrange *AP*, *BP* and *CP* in descending order.

#### Answer:

It is given that

*AP* is the bisector of

We have to arrange, and in descending order.

In we have

(As *AP *is the bisector of)

So (Sides in front or greater angle will be greater) ........(1)

In we have

(As *AP *is the bisector of)

Since,

So ..........(2)

Hence

From (1) & (2) we have

#### Page No 12.81:

#### Question 6:

Prove that the perimeter of a triangle is greater than the sum of its altitudes.

#### Answer:

We have to prove that the perimeter of a triangle is greater than the sum of its altitude.

In

, ,

We have to prove

Since

So and

By adding, we have

........(1)

Now consider then

, and

Now by adding .......(2)

Again consider

, and

By adding ...........(3)

Adding (1), (2) and (3), we get

Hence the perimeter of a triangle is greater than the sum of all its altitude.

#### Page No 12.81:

#### Question 7:

In the given figure, prove that:

(i) *CD* + *DA* + *AB* + *BC* > 2*AC*

(ii) *CD* + *DA* + *AB* > *BC*

#### Answer:

(1) We have to prove that

In we have

(As sum of two sides of triangle is greater than third one) ........(1)

In we have

(As sum of two sides of triangle is greater than third one) .........(2)

Hence

Adding (1) & (2) we get

Proved.

(2) We have to prove that

In we have

(As sum of two sides of triangle is greater than third one)

(Adding both sides)

Proved.

#### Page No 12.81:

#### Question 8:

Which of the following statements are true (T) and which are false (F)?

(i) Sum of the three sides of a triangle is less than the sum of its three altitudes.

(ii) Sum of any two sides of a triangle is greater than twice the median drawn to the third side.

(iii) Sum of any two sides of a triangle is greater than the third side.

(iv) Difference of any two sides of a triangle is equal to the third side.

(v) If two angles of a triangle are unequal, then the greater angle has the larger side opposite to it.

(vi) Of all the line segments that can be drawn from a point to a line not containing it, the perpendicular line segment is the shortest one.

#### Answer:

(1)

(2)

(3)

(4)

(5)

(6)

#### Page No 12.82:

#### Question 9:

Fill in the blanks to make the following statements true.

(i) In a right triangle the hypotenuse is the ...... side.

(ii) The sum of three altitudes of a triangle is ....... than its perimeter.

(iii) The sum of any two sides of a triangle is ..... than the third side.

(iv) If two angles of a triangle are unequal, then the smaller angle has the ..... side opposite to it.

(v) Difference of any two sides of a triangle is ...... than the third side.

(vi) If two sides of a triangle are unequal, then the larger side has ...... angle opposite to it.

#### Answer:

(1)

(2)

(3)

(4)

(5)

(6)

#### Page No 12.82:

#### Question 10:

*O* is any point in the interior of Δ *ABC.* Prove that

(i) *AB* + *AC* > *OB* + *OC*

(ii) *AB* + *BC* + *CA* > *OA* + *OB* + *OC*

(iii) *OA* + *OB* + *OC* > $\frac{1}{2}$(*AB* + *BC* + *CA*)

#### Answer:

It is given that, is any point in the interior of

We have to prove that

(1) Produced to meet at.

In we have

.........(1)

And in we have

.........(2)

Adding (1) & (2) we get

HenceProved.

(2) We have to prove that

From the first result we have

..........(3)

And

.........(4)

Adding above (4) equation

HenceProved.

(3) We have to prove that

In triangles, and we have

Adding these three results

HenceProved.

#### Page No 12.82:

#### Question 11:

Prove that in a quadrilateral the sum of all the sides is greater than the sum of its diagonals.

#### Answer:

We have to prove that the sum of four sides of quadrilateral is greater than sum of diagonal.

Since the sum of two sides of triangle is greater than third side.

In we have

..........(1)

In we have

..........(2)

In we have

.........(3)

In we have

.........(4)

Adding (1) & (2) & (3) and (4) we get

Hence Proved.

#### Page No 12.82:

#### Question 12:

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

#### Answer:

Let AC be the longest side in the ∆ABC.

Now,

AC > AB

⇒ ∠B > ∠C .....(1) (In a triangle, greater side has greater angle opposite to it)

Also,

AC > BC

⇒ ∠B > ∠A .....(2) (In a triangle, greater side has greater angle opposite to it)

From (1) and (2), we have

∠B + ∠B > ∠A + ∠C

⇒ ∠B + ∠B + ∠B > ∠A + ∠B + ∠C

⇒ 3∠B > 180º (Using angle sum property of a triangle)

⇒ ∠B > $\frac{1}{3}$ × 180º

Or ∠B > $\frac{2}{3}$ × 90º

Thus, the angle opposite to the longest side is greater than $\frac{2}{3}$of a right angle.

Hence proved.

#### Page No 12.82:

#### Question 13:

*D *is any point on side *AC *of a ∆*ABC *with *AB *= *AC*. Show that *CD *< *BD*.

#### Answer:

It is given that, D is any point on side AC of a ∆ABC with AB = AC.

In ∆ABC,

AB = AC (Given)

∴ ∠ACB = ∠ABC (In a triangle, equal sides have equal angles opposite to them)

Now, ∠ABC > ∠DBC

⇒ ∠ACB > ∠DBC (∠ACB = ∠ABC)

In ∆BCD,

∠DCB > ∠DBC

⇒ BD > CD (In a triangle, greater angle has greater side opposite to it)

Or CD < BD

#### Page No 12.84:

#### Question 1:

Mark the correct alternative in each of the following:

If *ABC *$\cong $ Δ*LKM*, then side of Δ*LKM* equal to side AC of Δ*ABC* is

(a) *LK*

(b) *KM*

(c) *LM*

(d) None of these

#### Answer:

It is given that

As triangles are congruent, same sides will be equal.

So

Hence (c).

#### Page No 12.84:

#### Question 2:

If Δ*ABC* $\cong $ Δ*ABC* is isosceles with

(a) *AB* = *AC*

(b) *AB* = *BC*

(c) *AC* = *BC*

(d) None of these

#### Answer:

It is given that and is isosceles

Since triangles are congruent so as same side are equal

Hence (a).

#### Page No 12.84:

#### Question 3:

If Δ*ABC** *$\cong $ Δ*PQR* and Δ*ABC* is not congruent to Δ*RPQ*, then which of the following is not true:

(a) *BC* = *PQ*

(b) *AC* = *PR*

(c) *AB* = *PQ*

(d) *QR* = *BC*

#### Answer:

If and is not congruent to

Since and compare corresponding sides you will see

(As )

Hence (a) , is not true.

#### Page No 12.84:

#### Question 4:

In triangles *ABC* and *PQR* three equality relations between some parts are as follows:

*AB* = *QP,* ∠*B* = ∠*P* and *BC* = *PR*

State which of the congruence conditions applies:

(a) SAS

(b) ASA

(c) SSS

(d) RHS

#### Answer:

In and

It is given that

Since two sides and an angle are equal so it obeys

Hence (a).

#### Page No 12.84:

#### Question 5:

In triangles *ABC* and *PQR*, if ∠*A* = ∠*R*, ∠*B** *= ∠*P** *and *AB* = *RP*, then which one of the following congruence conditions applies:

(a) SAS

(b) ASA

(c) SSS

(d) RHS

#### Answer:

In and

It is given that

Since given two sides and an angle are equal so it obeys

Hence (b).

#### Page No 12.84:

#### Question 6:

In Δ*PQR* $\cong $ Δ*EFD* then *ED* =

(a) *PQ*

(b) *QR*

(c)* PR*

(d) None of these

#### Answer:

If

We have to find

Since, as in congruent triangles equal sides are decided on the basis of “how they are named”.

Hence (c).

#### Page No 12.85:

#### Question 7:

If Δ*PQR* $\cong $ Δ*EFD*, then ∠*E* =

(a) ∠*P*

(b) ∠*Q*

(c) ∠*R*

(d) None of these

#### Answer:

If

Then we have to find

From the given congruence, as equal angles or equal sides are decided by the location of the letters in naming the triangles.

Hence (a)

#### Page No 12.85:

#### Question 8:

In a Δ*ABC*, if *AB* = *AC* and *BC* is produced to *D* such that ∠*ACD* = 100°, then ∠*A* =

(a) 20°

(b) 40°

(c) 60°

(d) 80°

#### Answer:

In the triangle *ABC* it is given that

We have to find

Now (linear pair)

Since

So, (by isosceles triangle)

This implies that

Now,

(Property of triangle)

Hence (a).

#### Page No 12.85:

#### Question 9:

In an isosceles triangle, if the vertex angle is twice the sum of the base angles, then the measure of vertex angle of the triangle is

(a) 100°

(b) 120°

(c) 110°

(d) 130°

#### Answer:

Let be isosceles triangle

Then

Now it is given that vertex angle is 2 times the sum of base angles

(As)

Now

(Property of triangle)

(Since, and )

Hence (b).

#### Page No 12.85:

#### Question 10:

Which of the following is not a criterion for congruence of triangles?

(a) SAS

(b) SSA

(c) ASA

(d) SSS

#### Answer:

(b) ,as it does not follow the congruence criteria.

#### Page No 12.85:

#### Question 11:

In the given figure, the measure of ∠*B'A'C**'* is

(a) 50°

(b) 60°

(c) 70°

(d) 80°

#### Answer:

We have to find

Since triangles are congruent

So

Now in

(By property of triangle)

Hence (b) .

#### Page No 12.85:

#### Question 12:

If *ABC* and *DEF* are two triangles such that Δ*ABC* $\cong $ Δ*FDE* and *AB* = 5cm, ∠*B* = 40°

(a) *DF* = 5cm, ∠*F* = 60°

(b) *DE* = 5cm, ∠*E* = 60°

(c) *DF* = 5cm, ∠*E* = 60°

(d) *DE* = 5cm, ∠*D* = 40°

#### Answer:

It is given thatand

So and

Now, in triangle ABC,

Therefore,

Hence the correct option is (c).

#### Page No 12.85:

#### Question 13:

In the given figure, *AB* ⊥ *BE* and *FE* ⊥ *BE*. If *BC* = *DE* and *AB* = *EF*, then Δ*ABD* is congruent to

(a) Δ*EFC*

(b) Δ*ECF*

(c) Δ*CEF*

(d) Δ*FEC*

#### Answer:

It is given that

And

(Given)

(Given)

So (from above)

Hence

From (d).

#### Page No 12.86:

#### Question 14:

In the given figure, if *AE* || *DC* and *AB* = *AC*, the value of ∠*ABD* is

(a) 70°

(b) 110°

(c) 120°

(d) 130°

#### Answer:

We have to find the value of in the following figure.

It is given that

(Vertically apposite angle)

Now (linear pair) …… (1)

Similarly (linear pair) …… (2)

From equation (1) we have

Now (same exterior angle)

(Interior angle)

Now

So

Since

Hence (b).

#### Page No 12.86:

#### Question 15:

In the given figure, *ABC* is an isosceles triangle whose side *AC* is produced to *E*. Through *C*, *CD* is drawn parallel to *BA*. The value of *x *is

(a) 52°

(b) 76°

(c) 156°

(d) 104°

#### Answer:

We are given that;

, is isosceles

And

We are asked to find angle *x*

From the figure we have

Therefore,

Since, so

Now

Hence (d) .

#### Page No 12.86:

#### Question 16:

In the given figure, if *AC* is bisector of ∠*BAD* such that *AB* = 3 cm and *AC* = 5 cm, then *CD* =

(a) 2 cm

(b) 3 cm

(c) 4 cm

(d) 5 cm

#### Answer:

It is given that

, is bisector of

We are to find the side *CD*

Analyze the figure and conclude that

(As in the two triangles are congruent)

In

So

Hence (c) .

#### Page No 12.86:

#### Question 17:

*D, E, F* are the mid-point of the sides *BC*, *CA* and *AB* respectively of Δ*ABC*. Then Δ*DEF* is congruent to triangle

(a) *ABC*

(b) *AEF*

(c) *BFD*, *CDE*

(d) *AFE*, *BFD*, *CDE*

#### Answer:

It is given that, and are the mid points of the sides, andrespectively of

(By mid point theorem)

(As it is mid point)

Now in and

(Common)

(Mid point)

(Mid point)

Hence (d)

#### Page No 12.86:

#### Question 18:

*ABC* is an isosceles triangle such that *AB* = *AC* and *AD* is the median to base *BC*. Then, ∠*BAD* =

(a) 55°

(b) 70°

(c) 35°

(d) 110°

#### Answer:

It is given that, AB=AC and Ad is the median of BC

We know that in isosceles triangle the median from he vertex to the unequal side divides it into two equal part at right angle.

Therefore,

(Property of triangle)

Hence (a) .

#### Page No 12.87:

#### Question 19:

In the given figure, *X* is a point in the interior of square *ABCD*. *AXYZ* is also a square. If *DY* = 3 cm and *AZ* = 2 cm, then *BY* =

(a) 5 cm

(b) 6 cm

(c) 7 cm

(d) 8 cm

#### Answer:

In the following figure we are given

Where *ABCD* is a square and *AXYZ* is also a square

We are asked to find *BY*

From the above figure we have *XY=YZ=AZ=AX*

Now in the given figure

So,

Now inn triangle Δ*AXB*

So

Hence (c) .

#### Page No 12.87:

#### Question 20:

In the given figure, *ABC* is a triangle in which ∠*B* = 2∠*C*. D is a point on side *BC* such that *AD* bisects ∠*BAC* and *AB* = *CD*. *BE* is the bisector of ∠*B*. The measure of ∠BAC is

(a) 72°

(b) 73°

(c) 74°

(d) 95°

#### Answer:

It is given that

,

*AB = CD*

We have to find

Now AB = CD

AB = BD

Now the triangle is isosceles

Let

So

Now

Since

Hence (a) .

#### Page No 12.87:

#### Question 1:

If *AB *= *QR*, *BC *= *PR *and *CA *= *PQ*, then triangle _________ ≅ triangle _________.

#### Answer:

It is given that, AB = QR, BC = PR and CA = PQ.

So, A ↔ Q, B ↔ R and C ↔ P

∴ ∆ABC ≅ ∆QRP

If AB = QR, BC = PR and CA = PQ, then triangle ____ABC____ ≅ triangle ____QRP____.

#### Page No 12.87:

#### Question 2:

In triangles *ABC *and *DEF*, *AB *= *FD *and ∠*A *= ∠*D*. The two triangles will be congruent by *SAS *axiom, if __________.

#### Answer:

SAS congruence axiom states that two triangles are congruent if two sides and the included angle of one are equal to the corresponding sides and the included angle of the other triangle.

In ∆ABC, ∠A is included between the sides AB and AC.

In ∆DEF, ∠D is included between the sides DF and DE.

∴ ∆ABC ≅ ∆DEF by SAS axiom if AC = DE.

In triangles ABC and DEF, AB = FD and ∠A = ∠D. The two triangles will be congruent by SAS axiom, if ____AC = DE____.

#### Page No 12.87:

#### Question 3:

In ∆*ABC *and *PQR*, *AB *= *AC*, ∠*C *= ∠*P *and ∠*B *= ∠*Q*. The two triangles are __________ but not __________.

#### Answer:

If two sides of a triangle are equal, then it is an isosceles triangle.

In ∆ABC, AB = AC

∴ ∆ABC is an isosceles triangle.

In ∆ABC,

AB = AC (Given)

∴ ∠C = ∠B (In a triangle, equal sides have equal angles opposite to them)

It is given that, ∠C = ∠P and ∠B = ∠Q.

∴ ∠P = ∠Q

In ∆PQR,

∠P = ∠Q (Proved)

∴ QR = PR (In a triangle, equal sides have equal angles opposite to them)

So, ∆PQR is an isosceles triangle.

However, it **cannot** be proved that the corresponding sides of ∆ABC are congruent to the corresponding sides of ∆PQR. Hence, the triangles are not congruent.

In ∆ABC and ∆PQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles are ____isosceles____ but not ____congruent____.

#### Page No 12.87:

#### Question 4:

In ∆*PQR*, ∠*P* = ∠*R*, *QR *= 4 cm and *PR *= 5 cm. Then *PQ *= ________.

#### Answer:

In ∆PQR,

∠P = ∠R (Given)

∴ QR = PQ (Sides opposite to the equal angles of a triangle are equal)

⇒ PQ = 4 cm (QR = 4 cm)

In ∆PQR, ∠P = ∠R, QR = 4 cm and PR = 5 cm. Then PQ = _____4 cm_____.

#### Page No 12.87:

#### Question 5:

In ∆*ABC*, *AB *= *AC *and ∠*B *= 50°. Then, ∠*C* = __________.

#### Answer:

In ∆ABC,

AB = AC (Given)

∴ ∠C = ∠B (Angles opposite to the equal sides of a triangle are equal)

⇒ ∠C = 50° (∠B = 50°)

In ∆ABC, AB = AC and ∠B = 50°. Then, ∠C = ____50°____.

#### Page No 12.87:

#### Question 6:

In ∆*ABC*, *AB *= *BC *and ∠*B* = 80°. Then ∠*A* = __________.

#### Answer:

In ∆ABC,

AB = BC (Given)

∴ ∠C = ∠A ..... (1) (Angles opposite to the equal sides of a triangle are equal)

Now,

∠A + ∠B + ∠C = 180° (Angle sum property of triangle)

∴ ∠A + 80° + ∠A = 180°

⇒ 2∠A = 180° − 80° = 100°

⇒ ∠A = $\frac{100\xb0}{2}$ = 50°

In ∆ABC, AB = BC and ∠B = 80°. Then ∠A = ____50°____.

#### Page No 12.87:

#### Question 7:

If in ∆*PQR*, ∠*P* = 70° and ∠*R* = 30°, then the longest side of ∆*PQR *is ___________.

#### Answer:

In ∆PQR,

*$\angle \mathrm{P}+\angle \mathrm{Q}+\angle \mathrm{R}=180\xb0$ *(Angle sum property of a triangle)

*$\therefore 70\xb0+\angle \mathrm{Q}+30\xb0=180\xb0$
$\Rightarrow \angle \mathrm{Q}+100\xb0=180\xb0$
$\Rightarrow \angle \mathrm{Q}=180\xb0-100\xb0=80\xb0$*

So, ∠Q is the greatest angle in the ∆PQR.

We know that, in a triangle the greater angle has the longer side opposite to it.

∴ PR is the longest side of ∆PQR.

If in ∆PQR, ∠P = 70° and ∠R = 30°, then the longest side of ∆PQR is

_____PR_____.

#### Page No 12.87:

#### Question 8:

In ∆*PQR*, if ∠*R* > ∠*Q*, then __________.

#### Answer:

In ∆PQR,

∠R > ∠Q (Given)

∴ PQ > PR (In a triangle, the greater angle has the longer side opposite to it)

In ∆PQR, if ∠R > ∠Q, then _____PQ > PR_____.

#### Page No 12.87:

#### Question 9:

*D* is a point on side *BC *of a ∆*ABC *such that *AD *bisects ∠*BAC*. Then ________.

#### Answer:

In ∆ABC, AD is the bisector of ∠A.

∴ ∠BAD = ∠CAD .....(1)

We know that exterior angle of a triangle is greater than each of interior opposite angle.

In ∆ABD,

∠ADC > ∠BAD

⇒ ∠ADC > ∠CAD [Using (1)]

In ∆ADC,

∠ADC > ∠CAD

∴ AC > CD (In a triangle, the greater angle has the longer side opposite to it)

Similarly, AB > BD

D is a point on side BC of a ∆ABC such that AD bisects ∠BAC. Then _____AC > CD and AB > BD_____.

#### Page No 12.87:

#### Question 10:

Two sides of a triangle are of lengths 5 cm and 1.5 cm. The length of the third side of the triangle lies between _________ and _________.

#### Answer:

In a triangle, the sum of two sides is greater than the third side.

∴ 5 cm + 1.5 cm > Third side

Or Third side < 6.5 cm

In a triangle, the difference of two sides is less than the third side.

∴ 5 cm − 1.5 cm < Third side

Or Third side > 3.5 cm

So,

3.5 cm < Third side < 6.5 cm

Thus, the length of the third side of the triangle lies between 3.5 cm and 6.5 cm.

Two sides of a triangle are of lengths 5 cm and 1.5 cm. The length of the third side of the triangle lies between ____3.5 cm____ and ___6.5 cm ___.

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#### Question 11:

If *AD *is a median of ∆*ABC*, then the perimeter of ∆*ABC *cannot be less than or equal to ___________.

#### Answer:

AD is the median of the ∆ABC.

In ∆ABD,

AB + BD > AD .....(1) (In a triangle, the sum of any two sides is greater than the third side)

In ∆ACD,

CD + CA > AD .....(2) (In a triangle, the sum of any two sides is greater than the third side)

Adding (1) and (2), we get

AB + BD + CD + CA > AD + AD

⇒ AB + BC + CA > 2AD (BC = BD + CD)

⇒ Perimeter of ∆ABC > 2AD

Or the perimeter of ∆ABC cannot be less than or equal to 2AD

If AD is a median of ∆ABC, then the perimeter of ∆ABC cannot be less than or equal to _____2AD_____.

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#### Question 12:

If ∆*PQR *≅ ∆*EDF*, then *PR *= ___________.

#### Answer:

If ∆PQR ≅ ∆EDF, then

P ↔ E, Q ↔ D and R ↔ F

So, PR = EF, PQ = ED and QR = DF (If two triangles are congruent, then their corresponding sides are congruent)

If ∆PQR ≅ ∆EDF, then PR = ______EF______.

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#### Question 13:

It is given that ∆*ABC *≅ ∆*RPQ*, then *BC *= __________.

#### Answer:

If ∆ABC ≅ ∆RPQ, then

A ↔ R, B ↔ P and C ↔ Q

So, AB = RP, BC = PQ and CA = QR (If two triangles are congruent, then their corresponding sides are congruent)

It is given that ∆ABC ≅ ∆RPQ, then BC = _____PQ_____.

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#### Question 14:

In ∆*ABC *and ∆*PQR*, if ∠*A* = ∠*Q*, ∠*B* = ∠*R* and *PR *= *AC*, then two triangles are congruent by __________ criterion.

#### Answer:

In ∆ABC and ∆PQR,

∠A = ∠Q

∠B = ∠R

AC = PR

∴ ∆ABC ≅ ∆QRP (AAS congruence criterion)

AAS congruence criterion states that if any two angles and a non-included side of one triangle are equal to the corresponding angles and side of another triangle, then the two triangles are congruent.

In ∆ABC and ∆PQR, if ∠A = ∠Q, ∠B = ∠R and PR = AC, then two triangles are congruent by ____AAS congruence____ criterion.

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#### Question 15:

In ∆*ABC *and ∆*PQR*, ∠*A* = ∠*Q*, ∠*B* = ∠*R* and *AB *= *QR*, then these triangles are congruent by _______ criterion.

#### Answer:

In ∆ABC and ∆PQR,

∠A = ∠Q

∠B = ∠R

AB = QR

∴ ∆ABC ≅ ∆QRP (ASA congruence criterion)

ASA congruence criterion states that two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle.

In ∆ABC and ∆PQR, ∠A = ∠Q, ∠B = ∠R and AB = QR, then these triangles are congruent by ____ASA congruence criterion____.

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#### Question 1:

In two congruent triangles *ABC* and *DEF*, if *AB* = *DE* and *BC* = *EF*. Name the pairs of equal angles.

#### Answer:

It is given that

Since, the triangles ABC and DEF are congruent, therefore,

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#### Question 2:

In two triangles *ABC* and *DEF*, it is given that ∠*A* = ∠*D*, ∠*B* = ∠*E* and ∠*C* =∠*F**.* Are the two triangles necessarily congruent?

#### Answer:

It is given that

For necessarily triangle to be congruent, sides should also be equal.

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#### Question 3:

If *ABC* and *DEF* are two triangles such that *AC* = 2.5 cm, *BC* = 5 cm, ∠*C* = 75°, *DE* = 2.5 cm, *DF* = 5cm and ∠*D* = 75°. Are two triangles congruent?

#### Answer:

It is given that

Since, two sides and angle between them are equal, therefore triangle ABC and DEF are congruent.

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#### Question 4:

In two triangles *ABC* and *ADC*, if *AB *= *AD* and *BC* = *CD*. Are they congruent?

#### Answer:

The given information and corresponding figure is given below

From the figure, we have

And,

Hence, triangles ABC and ADC are congruent to each other.

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#### Question 5:

In triangles *ABC* and *CDE*, if *AC* = *CE*, *BC* = *CD*, ∠*A* = 60°, ∠*C* = 30° and ∠*D* = 90°.

Are two triangles congruent?

#### Answer:

For the triangles ABC and ECD, we have the following information and corresponding figure:

In triangles ABC and ECD, we have

The SSA criteria for two triangles to be congruent are being followed. So both the triangles are congruent.

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#### Question 6:

*ABC* is an isosceles triangle in which *AB* = *AC*. *BE* and *CF* are its two medians. Show that *BE* = *CF*.

#### Answer:

In the triangle *ABC* it is given that

, and are medians.

We have to show that

To show we will show that

In triangle *ΔBFC* and *ΔBEC*

As, so

.........(1)

*BC=BC *(common sides) ........(2)

Since,

As *F* and *E* are mid points of sides *AB* and *AC* respectively, so

*BF = CE *..........(3)

From equation (1), (2), and (3)

HenceProved.

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#### Question 7:

Find the measure of each angle of an equilateral triangle.

#### Answer:

In equilateral triangle we know that each angle is equal

So

Now (by triangle property)

Hence.

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#### Question 8:

CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that ΔADE **≅** ΔBCE.

#### Answer:

We have to prove that

Given is a square

So

Now in is equilateral triangle.

So

In and

(Side of triangle)

(Side of equilateral triangle)

And,

So

Hence from congruence Proved.

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#### Question 9:

Prove that the sum of three altitudes of a triangle is less than the sum of its sides.

#### Answer:

We have to prove that the sum of three altitude of the triangle is less than the sum of its sides.

In we have

, and

We have to prove

As we know perpendicular line segment is shortest in length

Since

So ........(1)

And

........(2)

Adding (1) and (2) we get

........(3)

Now, so

.......(4)

And again, this implies that

........(5)

Adding (3) & (4) and (5) we have

HenceProved.

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#### Question 10:

In the given figure, if *AB* = *AC* and ∠*B* = ∠*C**.* Prove that *BQ* = *CP.*

#### Answer:

It is given that

, and

We have to prove that

We basically will prove to show

In and

(Given)

(Given)

And is common in both the triangles

So all the properties of congruence are satisfied

So

Hence Proved.

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