Rd Sharma 2021 for Class 9 Maths Chapter 12 - Congruent Triangles

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 .

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.

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 .

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 .

Answer:

In it is given that

, and

We have to find.

Since

Then (isosceles triangles)

Now 

(As given)

Thus 

(Property of triangle)

Hence .

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.

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 .

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.

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)

Answer:

It is given that

We have to find the ratio.

Since

And

So we have,

So

Hence .

Answer:

It is given that

Is right angled triangle

And 

We have to find and

Since

(Isosceles triangle)

Now 

(Property of triangle)

()

So

Hence 

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

$$\begin{gathered} \angle TQR+\angle RTQ+\angle QRT={180}^0 \\ \angle TQR+\angle TQR+{150}^0={180}^0 \\ 2\angle TQR+{150}^0={180}^0 \\ 2\angle TQR={180}^0-{150}^0 \\ 2\angle TQR={30}^0 \\ \angle TQR=\frac{{30}^0}{2}={15}^0 \end{gathered}$$

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.

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 = $\frac{180°}{2}$ = 90º

Thus, ∠ABC is a right angle.

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 = $\frac{90°}{2}$ = 45º

∴ ∠C = ∠B = 45º   

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

∴ ∠CAD = ∠BAD$=\frac{\angle \mathrm{A}}{2}=\frac{90°}{2}$ = 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.

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

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.

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.

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.

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.

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.

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 

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.

Answer:

It is given that 

We have to prove that triangle ΔABC 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.

Answer:

It is given that 

We have to find and.

Since so,

Now (property of triangle)

(Since )

Here

Then

Hence 

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.

Answer:

It is given that

In,

We have to prove that

Now 

(Given)

Thus

   ........(1)

In, we have

So, {from equation (1)}

Hence Proved.

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.

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.

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)

$$\begin{gathered} \Rightarrow \frac{\angle \mathrm{ABC}}{2}+\frac{\angle \mathrm{C}}{2}+\angle \mathrm{BOC}=180° \left[\mathrm{Using} \left(2\right) \mathrm{and} \left(3\right)\right] \\ \Rightarrow \frac{\angle \mathrm{ABC}+\angle \mathrm{C}}{2}+\angle \mathrm{BOC}=180° \\ \Rightarrow \frac{2\angle \mathrm{ABC}}{2}+\angle \mathrm{BOC}=180° \left[\mathrm{Using} \left(1\right)\right] \\ \Rightarrow \angle \mathrm{ABC}+\angle \mathrm{BOC}=180° .....\left(4\right) \end{gathered}$$

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.

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.

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.

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.

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.

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.

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.

Answer:

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

Answer:

(1)

(2)

(3)

(4)

(5)

(6)

(7)

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.

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.

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)

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

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

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.

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)

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.

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.

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.

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.

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.

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.

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

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.

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.

Answer:

(1)

(2)

(3)

(4)

(5)

(6)

Answer:

(1)

(2)

(3)

(4)

(5)

(6)

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.

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.

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.

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

Answer:

It is given that

As triangles are congruent, same sides will be equal.

So

Hence (c).

Answer:

It is given that and is isosceles

Since triangles are congruent so as same side are equal

Hence (a).

Answer:

If and is not congruent to

Since and compare corresponding sides you will see

(As )

Hence (a) , is not true.

Answer:

In and

It is given that

Since two sides and an angle are equal so it obeys

Hence (a).

Answer:

In and

It is given that

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

Hence (b).

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).

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)

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).

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).

Answer:

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

Answer:

We have to find

Since triangles are congruent 

So

Now in

(By property of triangle)

Hence (b) .

Answer:

It is given thatand

So and

Now, in triangle ABC,

Therefore,

Hence the correct option is (c).

Answer:

It is given that

And

(Given)

(Given)

So (from above)

Hence 

From (d).

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).

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) .

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) .

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) 

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) .

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 ΔAXB

So 

Hence (c) .

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) .

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__.

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__.

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__.

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___.

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°__.

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 = $\frac{100°}{2}$ = 50°

In ∆ABC, AB = BC and ∠B = 80°. Then ∠A = __50°__.

Answer:

In ∆PQR,

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

$\therefore 70°+\angle \mathrm{Q}+30°=180°$

$\Rightarrow \angle \mathrm{Q}+100°=180°$

$\Rightarrow \angle \mathrm{Q}=180°-100°=80°$



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___.

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___.

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___.

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 _.
 

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___.

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____.

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___.

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.

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__.

Answer:

It is given that

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

Answer:

It is given that

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

Answer:

It is given that

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

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.

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.

Answer:

In equilateral triangle we know that each angle is equal

So

Now (by triangle property)

Hence.

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.

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.

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.