NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles 2021

Q: Two triangles are congruent; are they necessarily similar? The inverse of this is true or not?

If two triangles are congruent, they will have the same interior angles and the same sides. Therefore, they must be similar, and however, if two triangles are similar, they need not be congruent.

Q: Should I learn all the congruencies and their proofs from the chapter of Triangles?

It is highly recommended to understand and practice all the congruencies mentioned in the chapter along with attempting practice questions on the same. The NCERT exemplar Class 9 Maths solutions chapter 7 will help you understand the problems in a detailed manner.

NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles

Edited By Safeer PP | Updated on Aug 31, 2022 12:22 PM IST

NCERT exemplar Class 9 Maths solutions chapter 7 discusses a triangle and its properties which are essential in mathematics. The NCERT exemplar Class 9 Maths chapter 7 solutions are curated to get a detailed overview of steps required to solve the problems. These NCERT exemplar Class 9 Maths chapter 7 solutions are prepared by highly qualified and seasoned subject matter experts to provide supporting material while attempting NCERT Class 9 Maths Questions. The NCERT exemplar Class 9 Maths solutions chapter 7 develops a progressive grasp of triangles’ concepts and follows the CBSE Class 9 Syllabus

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Question:1

Which of the following is not a criterion for congruence of triangles?
(A) SAS
(B) ASA
(C) SSA
(D) SSS

Answer:

[C] SSA is correct option
Solution. Consider two triangle as shown:

In ABC and PQR, side AB = PQ and side BC = QR and angle ÐC = ÐP
But we can observe that on placing one triangle upon the other triangle, they do not cover each other completely.
So, ABC and PQR have Side-Side-Angle criterion but does not follow congruence relation PQR and ABC are not congruent.
Hence option (C) is correct.

Question:2

If AB = QR, BC = PR and CA = PQ, then
$\\(A) \triangle ABC \cong \triangle PQR \\ (B) \triangle CBA \cong \triangle PRQ\\ (C) \triangle BAC \cong \triangle RPQ\\ (D) \triangle PQR \cong \triangle BCA\\$

Answer:

[B]
Solution. AB = QR, BC =PR, CA = PQ (Given)

Triangles follow the SSS criterion for congruence
In option $(A) \triangle ABC \cong \triangle PQR$ , from this we conclude that side AB = PQ
But it is given AB = QR and AB = PQ may or may not be possible
In option $(C) \triangle BCA \cong \triangle RPQ$ , by same relation, we say that
side BA = RP may or may not be possible because it is given BA = QP
In option $(D) \triangle PQR \cong \triangle BCA$
side PQ = BC may or may not be possible
In option $(B) \triangle CBA \cong \triangle PQR$ ,
CB = PR, BA = RQ, AC = QP (all are given.)
Hence option (B) is correct.

Question:3

In $\triangle ABC$ , AB = AC and $\angle B = 50^{\circ}$ . Then $\angle C$ is equal to
$(A) 40^{\circ}\\ (B) 50^{\circ}\\ (C) 80^{\circ}\\ (D) 130^{\circ}\\$

Answer:

[B] Solution.
We know that angles opposite to equal sides of a triangle are equal
AB = AC

Since, $\angle$ B is opposite to side AC and $\angle$ C is opposite to side AB
$\therefore$ $\angle$ B = $\angle$ C = 50° ( $\because$ $\angle$ B = 50° given)
Hence option (B) is correct.

Question:4

In $\triangle$ ABC, BC = AB and $\angle$ B = $80^{\circ}$ . Then $\angle$ A is equal to
$\\(A) 80^{\circ}\\ (B) 40^{\circ}\\ (C) 50^{\circ}\\ (D) 100^{\circ}\\$

Answer:

[C]
Solution. We know, angles opposite to equal sides of a triangle are equal

Here, AB = BC and $\angle$ C is opposite to side AB and $\angle$ A is opposite to side BC
$\therefore \angle$ C = $\angle$ A = x (Assume)
$\angle$ B = 80° (Given)
$\because$ $\angle$ A + $\angle$ B + $\angle$ C = 180°
$\Rightarrow$ x + 80° + x = 180°
$\Rightarrow$ 2x = 100°
$\Rightarrow$ x = 50°
Hence option (C) is correct.

Question:5

In $\triangle$ PQR, $\angle$ R = $\angle$ P and QR = 4 cm and PR = 5 cm. Then the length of PQ is
(A) 4 cm
(B) 5 cm
(C) 2 cm
(D) 2.5 cm

Answer:

[A]
Solution. We know that sides opposite to equal angle of a triangle are equal.

Here $\angle$ R = $\angle$ P (Given)
Side QR is opposite to $\angle$ P and side PQ is opposite to $\angle$ R
$\therefore$ PQ = QR = 4 cm
$\Rightarrow$ PQ = 4 cm
Hence option (A) is correct.

Question:6

D is a point on the side BC of a $\triangle$ ABC such that AD bisects $\angle$ BAC. Then
(A) BD = CD
(B) BA > BD
(C) BD > BA
(D) CD > CA

Answer:

[A]
Solution.
BC is the side of $\triangle$ ABC, D is the point on side BC.

In $\triangle$ ABC,
$\angle$ BAD = $\angle$ CAD (Given that AD bisects ÐBAC)
In $\triangle$ ADC,
$\angle$ BDA > $\angle$ BAD ( $\because$ Exterior angle of triangle is greater than interior opposite angle)
$\angle$ BAD = $\angle$ CAD
Then BA > BD ( $\because$ Side opposite to greater angle is longer)
Hence option (A) is correct.

Question:7

It is given that $\triangle$ ABC $\cong$ $\triangle$ FDE and AB = 5 cm, $\angle$ B = 40° and $\angle$ A = 80°. Then which of the following is true?
(A) DF = 5 cm, $\angle$ F = 60°
(B) DF = 5 cm, $\angle$ E = 60°
(C) DE = 5 cm, $\angle$ E = 60°
(D) DE = 5 cm, $\angle$ D = 40°

Answer:

[B]
Solution. In $\triangle$ ABC, $\angle$ A = 80°, $\angle$ B = 40° and $\angle$ C = x ( assume)

$\because$ $\angle$ A + $\angle$ B + $\angle$ x = 180° (angle sum property of a triangle)
$\Rightarrow$ 80° + 40° + x = 180°
$\Rightarrow$ x = 180° – 120°
$\Rightarrow$ x = 60°
$\Rightarrow$ ÐC = 60°
$\because$ $\triangle$ ABC $\cong$ $\triangle$ FDE
By ASA congruence relation
AB = FD, $\angle$ A = $\angle$ F, $\angle$ B = $\angle$ D
$\Rightarrow$ DF = 5 cm, $\angle$ F = 80°, $\angle$ B = 40° = $\angle$ D
$\because$ Also, $\angle$ C = $\angle$ E
$\Rightarrow$ $\angle$ E = 60°
Therefore, DF = 5 cm, $\angle$ E = 60°
Hence option (B) is correct.

Question:8

Two sides of a triangle are of lengths 5 cm and 1.5 cm. The length of the third side of the triangle cannot be
(A) 3.6 cm
(B) 4.1 cm
(C) 3.8 cm
(D) 3.4 cm

Answer:

[D]
Solution. We know that, sum of two sides of a triangle is greater than the third side.

(A) BC = 1.5 cm, AC = 3.6 cm (Assume)
Then, BC + AC = 1.5 + 3.6 cm = 5.1 cm
Sum of two side BC and AC is more than third side AB = 5 cm
Therefore, AC = 3.6 cm is possible.
(B) BC = 1.5 cm, AC = 4.1 cm (Assume)
Then, BC + AC = 1.5 + 4.1 cm = 5.6 cm
Sum of two side BC and AC is more than third side AB = 5 cm
Therefore, AC = 4.1 cm is possible.
(C) BC = 1.5 cm, AC = 3.8 cm (Assume)
Then, BC + AC = 1.5 + 3.8 cm = 5.3 cm
Sum of two side BC and AC is more than third side AB = 5 cm
Therefore, AC = 3.8 cm is possible.
(D) BC = 1.5 cm, AC = 3.4 cm (Assume)
Then, BC + AC = 1.5 + 3.4 cm = 4.9 cm
Sum of two side BC and AC is less than third side AB = 5 cm
Therefore, AC = 3.4 cm cannot be possible.
Hence option (D) is correct.

Question:5

In $\triangle$ PQR, if $\angle$ R > $\angle$ Q, then
(A) QR > PR
(B) PQ > PR
(C) PQ < PR
(D) QR < PR

Answer:

[B]
Solution.
We know that side opposite to the greater angle is longer.

Here $\angle$ R > $\angle$ Q
Therefore, PQ > PR
Hence option (B) is correct.

Question:10

In triangles ABC and PQR, AB = AC, $\angle$ C = $\angle$ P and $\angle$ B = $\angle$ Q. The two triangles are
(A) isosceles but not congruent
(B) isosceles and congruent
(C) congruent but not isosceles
(D) neither congruent nor isosceles

Answer:

[A]
Solution. We know that, angles opposite to equal sides are equal

$\because$ AB = AC
$\Rightarrow$ $\angle$ B = $\angle$ C
We also know that, sides opposite to equal angles are also equal
Here, $\angle$ Q = $\angle$ P
$\Rightarrow$ PR = QR
Hence, two sides of both the triangles are respectively equal.
So $\triangle$ ABC and $\triangle$ PQR are isosceles triangle.
But may or may not be congruent.
Because sides of $\triangle$ ABC may not be equal to side of $\triangle$ PQR.
Hence option (A) is correct.

Question:1

In triangles ABC and PQR, $\angle$ A = $\angle$ Q and ÐB = ÐR. Which side of $\triangle$ PQR should be equal to side AB of $\triangle$ ABC so that the two triangles are congruent? Give reason for your answer.

Answer:

Given : In triangles ABC and PQR, $\angle$ A = $\angle$ Q and $\angle$ B = $\angle$ R

By the ASA criterion of congruence, side AB = QR for a triangle to be congruent.
If, AB = QR then $\triangle$ ABC $\cong$ $\triangle$ QRP.

Question:2

In triangles ABC and PQR, $\angle$ A = $\angle$ Q and $\angle$ B = $\angle$ R. Which side of $\triangle$ PQR should be equal to side BC of $\triangle$ ABC so that the two triangles are congruent? Give reason for your answer.

Answer:

Side BC should be equal to side PR.
Solution:
In triangles ABC and PQR
$\angle$ A = $\angle$ Q and $\angle$ B = $\angle$ R.

By congruency, corresponding angles and sides must be equal to each other in both triangles.
By the AAS criterion of congruence BC = PR.
Then $\triangle$ ABC $\cong$ $\triangle$ QRP
Hence, side BC should be equal to side PR.

Question:3

“If two sides and an angle of one triangle are equal to two sides and an angle of another triangle, then the two triangles must be congruent.” Is the statement true? Why?

Answer:

If two sides and the included angle of one triangle are equal to the two side and the included angle of the other triangle then only the triangles can be congruent by SAS criterion of congruence otherwise not.

AB = PQ
BC = QR
$\angle$ C = $\angle$ R
But the triangles are not congruent
Therefore the given statement is false.

Question:4

“If two angles and a side of one triangle are equal to two angles and a side of another triangle, then the two triangles must be congruent.” Is the statement true? Why?

Answer:

[True]
Solution. If two angles and a side of one triangle are equal to two angles and a side of another triangle, then the two triangles must be congruent.
Yes, the given statement is true because by the AAS criterion of congruence, triangles are congruent.
Let us consider $\triangle$ ABC and $\triangle$ PQR

AC = PR
$\angle$ B = $\angle$ Q, $\angle$ C = $\angle$ R
$\Rightarrow$ $\triangle$ ABC $\cong$ $\triangle$ PQR
Therefore the given statement is true.

Question:5

Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm? Give reason for your answer.

Answer:

Not possible.
Solution.
We know that the sum of two sides of a triangle must be greater than third side.
Let ABC is any triangle, AB = 4 cm, BC = 3 cm and CA = 7 cm
AB + BC = 4 + 3 = 7 cm = CA
=> AB + BC = CA
It is not greater than the third side.
Therefore the given statement is false.

Question:7

If $\triangle$ PQR $\cong$ $\triangle$ EDF, then is it true to say that PR = EF? Give reason for your answer.

Answer:

True
Solution.
Let us consider $\triangle$ PQR and $\triangle$ EDF
$\because$ $\triangle$ PQR $\cong$ $\triangle$ EDF (Given)

By congruency, corresponding angles and sides must be equal
$\angle$ P = $\angle$ E, $\angle$ Q = $\angle$ D, $\angle$ R = $\angle$ F
and
PQ = ED, QR = DF, PR = EF
Therefore the given statement is true.

Question:8

In $\triangle$ PQR, $\angle$ P = 70° and $\angle$ R = 30°. Which side of this triangle is the longest? Give reason for your answer.

Answer:

[PR]
Solution.
In $\triangle$ PQR

$\therefore$ $\angle$ P + $\angle$ Q + $\angle$ R = 180° (angle sum property)
$\angle$ P = 70° and $\angle$ R = 30° (Given)
70° + $\angle$ Q + 30° = 180°
$\angle$ Q = 80°
$\therefore$ Greatest angle is $\angle$ Q and we know that side opposite to greatest angle will be the longest.
Therefore, PR is the longest side

Question:9

AD is a median of the triangle ABC. Is it true that AB + BC + CA > 2AD?
Give reason for your answer.

Answer:

True
Solution.
Let triangle be AB with median AD

AD is the line bisecting BC
$\Rightarrow$ BD = CD
$\because$ sum of two sides of a triangle is greater that third side.
In $\triangle$ ABD
AB + BD > AD …(i)
In $\triangle$ ADC
AC + DC > AD …(ii)
Adding (i) and (ii)
AB + BD + DC + AC > 2AD
( $\because$ BD + DC = BC)
AB + BC + AC > 2AD
Hence proved

Question:10

M is a point on side BC of a triangle ABC such that AM is the bisector of $\angle$ BAC. Is it true to say that perimeter of the triangle is greater than 2AM ? Give reason for your answer.

Answer:

True
Solution.

Perimeter of $\triangle$ ABC = AB + BC + AC
In $\triangle$ AMB and $\triangle$ AMC
We know that sum of two sides of a triangle is greater than third side.
AB + BM > AM ......(1)
and AC + MC > AM ......(2)
By adding (1) and (2) equation
AB + BM + MC + AC > 2AM
Since, BC = BM + MC
$\therefore$ AB + BC + AC > 2AM
=> Perimeter of $\triangle$ ABC > 2AM
Hence the given statement is true.

Question:11

Is it possible to construct a triangle with lengths of its sides as 9 cm, 7 cm and 17 cm? Give reason for your answer.

Answer:

[Not possible]
Solution.
We know that the sum of two sides of triangle must be greater than the third side.
Let AB = 9 cm, BC = 7 cm, AC = 17 cm
AB + BC = 9 + 7 cm = 16 cm < 17 cm = AC
AB + BC < AC
So, the triangle is not possible.

Question:12

Is it possible to construct a triangle with lengths of its sides as 8 cm, 7 cm and
4 cm? Give reason for your answer.

Answer:

Yes
Solution.
Sum of two sides of triangle must be greater than third side. Here in DABC.
Let AB = 8 cm, BC = 7 cm, AC = 4 cm,
AB + BC = 8 + 7 cm = 15 cm
AB + BC > AC
and AB + AC = 8 + 4 cm = 12 cm > BC
and BC + AC = 7 + 4 cm = 11 cm > AB
Hence the answer is yes, it is possible to construct such a triangle.

Question:4

In Figure, BA $\perp$ AC, DE $\perp$ DF such that BA = DE and BF = EC. Show that $\triangle$ ABC $\cong$ $\triangle$ DEF.

Answer:

Given, BA $\perp$ AC
DE $\perp$ DF
BA = DE and BF = EC
To show :- $\triangle$ ABC $\cong$ $\triangle$ DEF
Proof :-
BF = EC (given)
BF + FC = EC + FC (adding FC to LHS and RHS)
BC = FE
Also,
$\angle$ CAB = $\angle$ FDE = 90o (BA $\perp$ AC, DE $\perp$ DF given)
AB = DE (given)
By RHS criterion of congruence.
$\triangle$ ABC $\cong$ $\triangle$ DEF
Hence proved.

Question:5

Q is a point on the side SR of a DPSR such that PQ = PR. Prove that PS > PQ.

Answer:

Given: Q is point on side SR in DPSR and PQ = PR

To Prove = PS > PQ
Proof: We know that exterior angle of any triangle is greater than each of the opposite interior angles
$\therefore$ $\angle$ PQR is exterior angle of $\triangle$ PSQ
$\therefore$ $\angle$ PQR > $\angle$ PSQ
$\because$ PQ = PR (Given)
$\angle$ PQR = $\angle$ PRQ (Angles opposite to equal sides are equal)
Then, $\angle$ PRQ = $\angle$ PRS
$\therefore$ $\angle$ PRS > PSR
Now, side opposite to greater angle is longer in a triangle
Then PS > PR
PS > PQ ( $\because$ PQ = PR)
Hence Proved.

Question:6

S is any point on side QR of a $\triangle$ PQR. Show that:PQ + QR + RP > 2 PS.

Answer:

Given, $\triangle$ PQR, S is any point on QR.
To prove: PQ + QR + RP > 2PS

Proof:- we know that sum of two sides of a triangle is greater than the third side.
In $\triangle$ PQS,
PQ + QS > PS
and in $\triangle$ PSR,
PR + SR > PS
On adding above both,
PQ + QS + SR + PR > PS + PS
$\because$ QS + SR = QR
$\Rightarrow$ PQ + QR + PR > 2PS
Hence proved.

Question:7

D is any point on side AC of a $\triangle$ ABC with AB = AC. Show that CD < BD.

Answer:

Given: ABC is a triangle and D is any point on AC.
AB = AC

To prove: CD < BD.
Proof: In $\triangle$ ABC
AB = AC (Given)
$\angle$ ABC = $\angle$ ACB (angles opposite to equal sides are equal)
In $\triangle$ ABC and $\angle$ DBC
$\angle$ ABC > $\angle$ DBC
$\Rightarrow$ $\angle$ ACB > $\angle$ DBC
$\Rightarrow$ BD > CD (Side opposite to greater angle is larger)
$\therefore$ CD < BD
Hence proved.

Question:8

In Figure, l $\parallel$ m and M is the mid-point of a line segment AB. Show that M is also the mid-point of any line segment CD, having its end points on l and m, respectively.

Answer:

Given : $l \parallel m$
M is mid-point of AB then AM = MB
To show :- M is mid-point of CD.
Proof :- $l \parallel m$
Then, In $\triangle$ ACM and $\triangle$ DMB
$\angle$ ACM = $\angle$ BDM (Alternate interior angles are equal)
$\angle$ AMC = $\angle$ DMB (vertically opposite angles are equal)
AM = MB (given)
then by AAS criterion of congruence
$\triangle$ ACM $\cong$ $\triangle$ BDM
Then DM = CM (CPCT)
and CD = DM + MC = 2 DM
DM = CM = CD
Then M is also mid-point of CD.
Hence proved.

Question:9

Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. BO is produced to a point M.
Prove that $\angle$ MOC = $\angle$ ABC.

Answer:

Given: ABC is a isosceles triangle AB = AC
BO is produced to point M.

To prove : $\angle$ MOC = $\angle$ ABC
Proof :
AB = AC
$\angle$ ABC = $\angle$ ACB (Angles opposite to two equal sides of triangle are equal)
In $\triangle$ OBC,
$\angle BOC +\frac{\angle B}{2}+\frac{\angle C}{2} = 180^{\circ}$ ( $\because$ OB and OC are bisector of ÐB, ÐC)
$\Rightarrow \angle BOC + 2 \times \left ( \frac{\angle B}{2} \right )=180^{\circ}$ $\left ( \because \angle B = \angle C \right )$
$\Rightarrow \angle BOC + \angle B = 180^{\circ}\\ \Rightarrow \angle BOC = 180^{\circ}- \angle B \cdots \cdots (i)\\ So\; \angle MOC + \angle BOC = 180^{\circ} \; \; \; (Linear\; pair)\\ \Rightarrow \angle MOC = 180 ^{\circ}-\angle BOC\\ \Rightarrow \angle MOC = 180^{\circ} - (180^{\circ} -\angle B) \; \; (From i)\\ \Rightarrow \angle MOC = 180^{\circ} - 180^{\circ} + \angle B\\ \Rightarrow \angle MOC = \angle B\\ \Rightarrow \angle MOC=\angle ABC$
Hence proved

Question:10

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

Answer:

Given, ABC is an isosceles triangle
AB = AC (Given))

To prove : External angle adjacent to $\angle$ ABC is equal to $\angle$ BOC
Proof : Produce line CB to D in $\triangle$ ABC
AB = AC (Given)
$\angle$ ACB = $\angle$ ABC (angles opposite to equal sides are equal)
$\angle$ OCB = $\angle$ OBC (Bisector of angle B and C, respectively)
In $\triangle$ BOC,
$\angle$ OBC + $\angle$ OCB + $\angle$ BOC = 180°
$\Rightarrow$ 2 $\angle$ OBC + $\angle$ BOC = 180° (From above)
$\Rightarrow$ $\angle$ ABC + $\angle$ BOC = 180°
( $\because$ $\angle$ ABO + $\angle$ OBC = $\angle$ ABC)
$\Rightarrow$ $\angle$ ABC+ $\angle$ OBA = 180°
$\Rightarrow$ $\angle$ OBA = $\angle$ BOC.
Hence proved.

Question:11

In given figure, AD is the bisector of $\angle$ BAC. Prove that AB > BD.

Answer:

Given: In $\triangle$ ABC, AD is bisector of $\angle$ BAC.
To prove: AB > BD
Proof: We know that
Exterior angle of triangle is greater than each of opposite interior angles then
In $\triangle$ ADC
$\angle$ ADB > $\angle$ CAD (From above result)
Now, $\angle$ BAD = $\angle$ DAC ( $\because$ AD is bisector of $\angle$ BAC)
$\Rightarrow$ $\angle$ ADB > $\angle$ BAD
$\therefore$ AB > BD (Since, side opposite to greater angle is longer)
Hence proved.

Question:1

Find all the angles of an equilateral triangle.

Answer:

$\left [60^{\circ} \right ]$
Solution. In equilateral triangle all sides and angles are equal.

Let $\angle PQR = \angle PRQ = \angle RPQ = x^{\circ}$
Then Ð $\angle PQR = \angle PRQ = \angle RPQ = 180^{\circ}$ (angle sum property)
$x^{\circ} +x^{\circ} +x^{\circ} = 180^{\circ}\\ 3x^{\circ} =180^{\circ}\\ x^{\circ} =60^{\circ}\\$
Thus, each angle of an equilateral triangle is 60°.

Question:2

The image of an object placed at a point A before a plane mirror LM is seen at the point B by an observer at D as shown in Figure. Prove that the image is as far behind the mirror as the object is in front of the mirror.

Answer:

Given : ABC is an triangle and let O is a point on AB.
To prove : OA = OB

Proof : Since i and r is angle of incidence and reflection respectively.
CN is normal then
$\angle$ LCA = 90 – i, $\angle$ MCD = 90 – r
$\because$ $\angle$ ACN = $\angle$ DCN = i = r (Given)
$\Rightarrow$ $\angle$ LCA = $\angle$ MCD = 90 – i = 90 – r (Since, i = r )
$\Rightarrow$ $\angle$ BCL = 90 – r (vertically opposite angles)
$\because$ $\angle$ BCL = $\angle$ LCA
$\angle$ L bisects BA
Hence, OB = OA,
Hence, proved.

Question:3

ABC is an isosceles triangle with AB = AC and D is a point on BC such that AD $\perp$ BC (Figure). To prove that $\angle$ BAD = $\angle$ CAD, a student proceeded as follows:
In $\triangle$ ABD and $\triangle$ ACD,
AB = AC (Given)
$\angle$ B = $\angle$ C (because AB = AC)
and $\angle$ ADB = $\angle$ ADC
Therefore, $\triangle$ ABD $\cong$ $\triangle$ ACD (AAS)
So, $\angle$ BAD = $\angle$ CAD (CPCT)
What is the defect in the above arguments?

Answer:

[ $\angle$ ABD = ACD is defect.]
Solution.
In $\triangle$ ABC, AB = AC (Given)

$\angle$ ACB = $\angle$ ABC (angles opposite to equal side are equal)
In $\triangle$ ABD and $\triangle$ ACD
$\angle$ ABD = $\angle$ ACD (given)
$\angle$ ABD = $\angle$ ACD (from above)
$\angle$ ADB = $\angle$ ADC (each 90°)
$\therefore$ $\triangle$ ABD $\cong$ $\triangle$ ACD
$\angle$ BAD = $\angle$ CAD
So, the defect in above given argument, firstly prove
$\angle$ ABD = $\angle$ ACD

Question:4

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

Answer:

Given : P is a point on the bisector at ÐABC.

To prove:- BPQ is on an isosceles triangle
$\angle$ 1 = $\angle$ 2 ( $\because$ BP is bisector of $\angle$ ABC)
$\angle$ 1 = $\angle$ 3 ( $\because$ PQ is parallel to BA)
$\therefore$ $\angle$ 2 = $\angle$ 3
$\Rightarrow$ PQ = BQ (If two angles are equal their opposite sides also equal)
$\Rightarrow$ $\triangle$ PBQ is on the isosceles triangle.
Hence proved.

Question:5

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

Answer:

Given: ABCD is a quadrilateral in which AB = BC and AD = CD

To prove :- $\angle$ ADB = $\angle$ BDC
and $\angle$ ABD = $\angle$ DBC
Proof : In $\triangle$ ABD and $\triangle$ CBD
AB = CD (Given)
BD = BD (common)
AB = BC (given)
$\triangle$ ABD $\cong$ $\triangle$ CBD (SSS congruency)
$\angle$ ABD = $\angle$ CBD (by CPCT)
And $\angle$ ADB = $\angle$ BDC (by CPCT)
Therefore, BD bisects both the angle ABC and ADC.
Hence proved

Question:6

ABC is a right triangle with AB = AC. Bisector of $\angle$ A meets BC at D. Prove that BC = 2 AD.

Answer:

Given : ABC is a right triangle AB = AC
Bisector of $\angle$ A meets BC at D.

To prove : BC = 2AD
Proof : In $\triangle$ ABC, AB = AC (given)
Then $\angle$ B = $\angle$ C (If sides are equal in a triangle, then opposite angle are also equal)
Now,
$\angle$ BAC + $\angle$ ABC + $\angle$ BCA = 180° (angle sum property)
$\Rightarrow$ 90° + 2 $\angle$ ABC = 180° (Q $\angle$ ABC = $\angle$ ACB)
$\Rightarrow$ $\angle$ B = 45° = $\angle$ ACB
$\because$ $\angle$ BAD = $\angle$ CAD (AD is bisector ÐA)
BD = AD, CD = AD (Sides opposite to equal angles are equal).
Adding both,
BD + CD = AD + AD
BC = AD + AD
BC = 2AD ( $\because$ BD + DC = BC)
Hence Proved.

Question:7

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

Answer:

ABCD is square. O is any interior point of the square and OAB is an equilateral triangle.

To prove: $\triangle$ OCD is an isosceles triangle.
Proof : $\angle$ AOB = $\angle$ OAB = $\angle$ OBA = 60°
And, DA = AB = OB ( $\because$ $\triangle$ OAB is equilateral)
Now, $\angle$ DCB = $\angle$ BAD = $\angle$ ABD = $\angle$ CDA = 90°
DA = AB = CB = CD ( $\because$ ABCD is square)
Let, $\angle$ OAB = $\angle$ OBA = 60° ……(1)
$\angle$ BAD = $\angle$ ABD = 90° …….(2)
Then subtract eqn. (1) from (2), we get
$\angle$ BAD – $\angle$ OAB = $\angle$ ABD – $\angle$ OBA = 90° – 60° = 30°
$\angle$ DAO = $\angle$ CBO = 30° …….(3)
In $\triangle$ AOD and $\triangle$ BOC
AO = BO (OAB is an equilateral triangle)
$\angle$ DAO = $\angle$ CBO (from 3)
AD = BC (ABCD is square)
$\therefore$ $\triangle$ AOD $\cong$ $\triangle$ BOC (by SAS criterion of congruency)
$\Rightarrow$ DO = OC (by CPCT)
$\therefore$ $\triangle$ OCD is an isosceles triangle.
Hence proved

Question:8

ABC and DBC are two triangles on the same base BC such that A and D lie on the opposite sides of BC, AB = AC and DB = DC. Show that AD is the perpendicular bisector of BC.

Answer:

Given : AD intersects BC at O
ABC, DBC are triangles on same base BC and
AB = AC, DB = DC
To prove : $\angle$ AOB = $\angle$ AOC = 90°
$\angle$ BOD = $\angle$ COD = 90°
Proof : In $\triangle$ ABD and $\triangle$ ACD
AB = AC (given)
AD = AD (common)
BD = CD (given)
$\therefore$ $\triangle$ ABD $\cong$ $\triangle$ ACD (By SSS congruence)
$\angle$ BAO = $\angle$ CAO (by CPCT)
Now; In $\triangle$ AOB and $\triangle$ AOC
AB = AC (given)
AO = OA (common)
$\angle$ BAO = $\angle$ CAO (from above)
$\therefore$ $\triangle$ AOB $\cong$ $\triangle$ AOC (By SAS congruence)
$\Rightarrow$ BO = OC
And, $\angle$ AOB = $\angle$ AOC
Then $\angle$ AOB + $\angle$ AOC = 180°
$\angle$ AOB + $\angle$ AOB = 180° (from above)
$\angle$ AOB = 90°, $\angle$ AOC = 90°
and $\angle$ AOB = $\angle$ COD, $\angle$ AOC = $\angle$ BOD (Vertically opposite angles)
$\therefore$ AO $\perp$ ^r bisector of BC.
Hence proved.

Question:9

ABC is an isosceles triangle in which AC = BC. AD and BE are respectively two altitudes to sides BC and AC. Prove that AE = BD.

Answer:

Given: $\triangle$ ABC is isosceles triangle AB = AC
AD and BE are altitudes at BC and AC

To Prove: AE = BD
Proof: In $\triangle$ ABD & $\triangle$ ACD
AD = AD (common)
$\angle$ ADB = $\angle$ ADC (AD is altitude at BC)
AB = AC (Given)
$\therefore$ $\triangle$ ABD $\cong$ $\triangle$ ACD (by SAS congruence)
$\angle$ BAD = $\angle$ DAC (by CPCT)
$\angle$ BAD = $\angle$ DAE (ÐDAC = ÐDAE) (i)
Now, in $\triangle$ ABD and $\triangle$ ABE
AB = BA (common)
$\angle$ ADB = $\angle$ AEB = 90° (AD and BE are altitudes)
$\angle$ BAD = $\angle$ DAE (From i)
$\therefore$ $\triangle$ ABD $\cong$ $\triangle$ ABE (by AAS congruency)
$\therefore$ AE = BD (by CPCT)
Hence proved.

Question:10

Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side.

Answer:

Given: ABC is a triangle
AD is median at BC drawn from A.
Produce D to E and DE = AD, Join DE
To Prove: AB + AC > AD
Proof: Let $\triangle$ ABD and $\triangle$ ECD
CD = DB (D is mid point)
$\angle$ BDA = $\angle$ CDE (Vertically opposite angle)
ED = AD (by construction)
$\therefore$ $\triangle$ ABD $\cong$ $\triangle$ ECD (by SAS congruence)
Þ EC = AB (by CPCT)
In $\triangle$ ACE, AC + EC > AE (Sum of two sides is greater than the third side)
Then AC + AB > AD + DE (QAB = EC and AE = AD + DE as D is the mid-point)
AC + AB > 2AD
AC + AB > AD
Hence proved.

Question:11

Show that in a quadrilateral ABCD, AB + BC + CD + DA < 2 (BD + AC)

Answer:

Given : ABCD is a quadrilateral

To prove : AB + BC + CD + DA + < 2 (BD + AC)
Proof : Since, we know that sum of lengths of any two sides in a triangle should be greater than the third side.
$\therefore$ In $\triangle$ AOB, AB < OA + OB
In $\triangle$ BOC, BC < OB + OC
In $\triangle$ COD, CD < OC + OD
In $\triangle$ AOD, DA < OD + OA
Adding all of the above,
$\Rightarrow$ AB + BC + CD + AD < 2OA + 2OB + 2OC + 2OD
$\Rightarrow$ AB + BC + CD + AD < 2((AO + OC) + (DO + OB))
$\Rightarrow$ AB + BC + CD + AD < 2(AC + BD)
( $\because$ AC = AO + OC, BD = DO + OB)
Hence proved.

Question:12

Show that in a quadrilateral ABCD, AB + BC + CD + DA > AC + BD

Answer:

Given, ABCD is a quadrilateral

To Prove : AB + BC + CD + DA > AC + BD
Proof - Since, we know that sum of the two sides of a triangle is greater than the third side.
In $\triangle$ ABC, AB + BC > AC
In $\triangle$ BCD, BC + CD > BD
In $\triangle$ CAD, AD + CD > AC
In $\triangle$ BAD, BA + AD > BD
Adding all the above equations,
2(AB + BC + CA + AD) > 2(AC + BD)
$\Rightarrow$ 2(AB + BC + CA + AD) > 2(AC + BD)
$\Rightarrow$ AB + BC + CA + AD > AC + BD
Hence, proved.

Question:13

In a triangle ABC, D is the mid-point of side AC such that BD equal to AC. Show that angle ABC is a right angle.

Answer:

ABC is triangle and D is mid-point at AC
BD = AC

To prove : $\angle$ ABC = 90°

Proof : AD = CD = ½ AC ( $\because$ D is mid-point)
BD = AC (given)
So, AD = BD = CD
Let AD = BD
$\angle$ BAD = $\angle$ ABD (angles opposite to equal sides are equal)
Now, CD = BD
$\angle$ BCD = $\angle$ CBD (angles opposite to equal sides are equal)
In $\triangle$ ABC,
$\angle$ ABC + $\angle$ BAC + $\angle$ BCA = 180° (angle sum property)
$\angle$ ABC + $\angle$ BAD + $\angle$ BCD = 180°
Now, $\angle$ ABC + $\angle$ ABD + $\angle$ CBD = 180° ( $\because$ $\angle$ BAD = $\angle$ ABD, $\angle$ BCD = $\angle$ CBD)
Then $\angle$ ABC + $\angle$ ABC = 180° ( $\because$ $\angle$ ABD + $\angle$ CBD = $\angle$ ABC)
$\Rightarrow$ $\angle$ ABC = 90°
Hence, proved.

Question:14

In a right triangle, prove that the line-segment joining the mid-point of the hypotenuse to the opposite vertex is half the hypotenuse.

Answer:

Given : ABC is right angle triangle
D is mid point of AC
Construction: Construct EC parallel to AB, AE parallel to BC. Join DE.

To prove = $BD = \frac{1}{2} AC$
Proof:
In $\triangle$ ADB and $\triangle$ EDC
AD = CD (D is midpoint)
BD = DE (D is midpoint)
$\angle$ ADB = $\angle$ EDC (Vertically opposite angles)
$\angle$ ADB $\cong$ $\triangle$ EDC (by SAS congruence)
$\therefore$ AB = EC (by CPCT)
Now, $\angle$ ABC + $\angle$ BCE = 180°
Þ 90° + $\angle$ BCE = 180° ( $\angle$ ABC = 90°, Given)
Þ $\angle$ BCE = 90°
And EC || AB ( $\because$ $\angle$ BAD and $\angle$ DCE are alternate angle)
In $\triangle$ ABC and $\angle$ EBC
BC = BC (common)
AB = EC (From above)
$\angle$ ABC = $\angle$ BCE (From above)
$\therefore$ $\triangle$ ABC $\cong$ $\triangle$ ECB (by ASA congruence)
AC = EB (by CPCT)
$\frac{1}{2}AC = \frac{1}{2}EB\\ \frac{1}{2}AC = BD$
Hence proved

Question:15

Two lines l and m intersect at the point O and P is a point on a line n passing through the point O such that P is equidistant from l and m. Prove that n is the bisector of the angle formed by l and m.

Answer:

Given: l and m intersect at point O and P is a point on a line n passing through O and
PQ = PR

To prove: $\angle$ QOP = $\angle$ ROP
Proof: In $\triangle$ POR and $\triangle$ POQ
PQ = PR (Given)
$\angle$ PQO = $\angle$ PRO = 90° (P is equidistant from l and m)
So PQ and PR should be perpendicular to lines l and m respectively
$\therefore$ $\triangle$ POR $\cong$ $\triangle$ POQ (by RHS congruence)
$\angle$ ROP = $\angle$ QOP (by CPCT)
Hence Proved.

Question:16

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:

Given : ABCD is trapezium
M, N are mid-point of AB and DC respectively and perpendicular to both AB and DC

To prove : AD = BC
Proof: $\triangle$ ANM and $\triangle$ BNM
AM = BM (M is mid-point)
$\angle$ AMN = $\angle$ BMN = 90° (given)
MN = MN (common)
$\triangle$ ANM $\cong$ $\triangle$ BNM (congruency of SAS)
AN = BN (by CPCT) …(i)
We know that
$\angle$ ANM = $\angle$ BNM (by CPCT)
Then 90° – $\angle$ ANM = 90° – $\angle$ BNM
$\Rightarrow$ $\angle$ AND = $\angle$ BNC …(ii)
In $\triangle$ AND and $\triangle$ BNC,
AN = BN (From i)
$\angle$ AND = $\angle$ BNC (From ii)
DN = CN (N is mid-point)
$\triangle$ AND $\cong$ $\triangle$ BNC (by SAS congruency)
AD = BC (by CPCT)
Hence, proved.

Question:17

ABCD is a quadrilateral such that diagonal AC bisects the angles A and C. Prove that AB = AD and CB = CD.

Answer:

Given: ABCD is quadrilateral
AC bisects $\angle$ A and $\angle$ C

To prove: AB = AD, CB = CD
In $\triangle$ ABC and $\triangle$ ADC
$\angle$ BAC = $\angle$ DAC ( $\because$ AC bisects $\angle$ A)
AC = AC (common)
$\angle$ BAC = $\angle$ DCA ( $\because$ AC bisects $\angle$ C)
$\therefore$ $\triangle$ ABC $\cong$ $\triangle$ ADC (by ASA criterion)
AB = AD (by CPCT)
BC = CD. (by CPCT)
Hence proved.

Question:18

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:

Given, ABC is right angle triangle
AB = AC, CD is bisector at AB
To prove: AC + AD = BC

Proof: $\because$ AB = AC (Given)
By Pythagoras theorem
BC² =AB²+ AC² = AC² + AC² = AC²+ AC² ( $\because$ AB = AC)
$BC =\sqrt{2}AC$
By angle bisector theorem:
$\frac{AD}{BD}=\frac{AC}{BC}=\frac{AC}{\sqrt{2}AC}=\frac{1}{\sqrt{2}}\left ( From \;above \right )$
Let AB = a, AD = b
$\frac{b}{a-b}=\frac{1}{\sqrt{2}}\\ \Rightarrow b=\frac{a}{1+\sqrt{2}}\\ \Rightarrow b=a\left ( \sqrt{2}-1 \right )\\ \Rightarrow a+b=a\sqrt{2}$
AC + AD = BC ( $\because$ AB = AC)

$\Rightarrow \frac{b}{a-b}-\frac{a}{a\sqrt{2}}=a\left ( \sqrt{2}-1 \right )\\ \Rightarrow a+b=a\sqrt{2}$

We know, AC = a, AD = b, BC = $a\sqrt{2}$
=> AC + AD = BC
Hence proved

Question:19

AB and CD are the smallest and largest sides of a quadrilateral ABCD. Out of $\angle$ B and $\angle$ D decide which is greater

Answer:

ABCD is quadrilateral
AB and CD are smallest and greatest side of ABCD.

Join BD
In $\triangle$ BCD
DC < BC ( $\because$ CD is Greatest side)
$\angle$ CBD > $\angle$ BDC (Angle opposite to greater side is greater) …(i)
In $\triangle$ ABD
AD > AB (AB is the smallest side)
$\angle$ ABD > $\angle$ ADB (Angle opposite to greater side is greater) …(ii)
Adding (i) and (ii)
$\angle$ CBD + $\angle$ ABD > $\angle$ BDC + $\angle$ ADB
$\angle$ ABC > $\angle$ ADC
Hence $\angle$ B is greater.

Question:20

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

Answer:

Given: ABC is triangle and AC is longest side

To Prove: $\angle ABC > \frac{2}{3}\times 90^{\circ}$
Proof: AC is longest side
$\Rightarrow$ $\angle$ B > $\angle$ C (angle opposite to longest side is greaten)
$\angle$ B > $\angle$ A
By adding both
$\angle$ B + $\angle$ B > $\angle$ C + $\angle$ A
2 $\angle$ B > $\angle$ C + $\angle$ A
2 $\angle$ B + $\angle$ B > $\angle$ C + $\angle$ A + $\angle$ B (adding $\angle$ B to LHS and RHS)
3 $\angle$ B > $\angle$ C + $\angle$ A + $\angle$ B
3 $\angle$ B > 180° (Sum of all interior angles in triangle is 180°)
$\angle$ B > 60°
$\Rightarrow \angle B > \frac{2}{3}\times 90^{\circ}$
Hence proved.

Question:21

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

Answer:

Given: ABCD is quadrilateral
And AB = AD, CB = CD

To prove : AC is $\perp^{r}$ bisector at BD
In $\triangle$ ABC and $\triangle$ ADC
AB = AD (Given)
BC = CD (Given)
AC = AC (Common)
$\triangle$ ABC $\cong$ $\triangle$ ADC (by SSS congruency)
Then by CPCT,
ÐBAC = ÐDAC …(i)
Now in $\triangle$ ABO and $\triangle$ ADO
AB = AD (Given)
AO = AC (common)
$\angle$ BAO = $\angle$ DAO (from i)
$\triangle$ ABO $\cong$ $\triangle$ DAO (by SAS congruency)
Then by CPCT,
$\therefore$ $\angle$ AOB = $\angle$ DOA
$\because$ $\angle$ BOD = 180°
$\angle$ AOB + $\angle$ AOD = 180° ( $\because$ $\angle$ BOD = $\angle$ AOB + $\angle$ AOD)
$\angle$ AOB + $\angle$ AOB = 180° ( $\because$ $\triangle$ AOB $\cong$ $\triangle$ AOD)
$\angle$ AOB = 90° ( $\angle$ AOB = $\angle$ AOD)
Hence, proved.

Important Topics for NCERT Exemplar Solutions Class 9 Maths Chapter 7:

Salient topics covered in NCERT exemplar Class 9 Maths solutions chapter 7 are mentioned below:

◊ Learning of trivial properties and theorems about triangles.

◊ Type of triangles.

◊ Fundamental inequalities of a triangle.

◊ Congruency of triangles and methods to prove it.

◊ NCERT exemplar Cass 9 Maths chapter 7 solutions discuss examples such as “the sum of two sides of a triangle will be more than me third side’s length,” which is very important and useful inequality.

NCERT Class 9 Exemplar Solutions for Other Subjects:

NCERT Class 9 Maths Exemplar Solutions for Other Chapters:

Chapter 1 Number System

Chapter 2 Polynomials

Chapter 3 Coordinate geometry

Chapter 4 Linear equations in Two Variable

Chapter 5 Introduction to Euclid’s Geometry

Chapter 6 Lines and Angles

Chapter 8 Quadrilaterals

Chapter 9 Area of Parallelograms and Triangles

Chapter 10 Circles

Chapter 11 Constructions

Chapter 12 Heron’s Formula

Chapter 13 Surface Areas and Volumes

Chapter 14 Statistics and Probability

Features of NCERT Exemplar Class 9 Maths Solutions Chapter 7:

These Class 9 Maths NCERT exemplar chapter 7 solutions provide a student with basic concepts of Triangles so that the student can solve challenging geometry problems and coordinate geometry. Fundamental inequalities of a triangle will help the student solve many vectors and complex number-related problems later in higher classes. These Class 9 Maths NCERT exemplar solutions chapter 7 Triangles consist of a wide range of practice problems available in the exemplar which provides a student with adequate opportunity to learn and practice the concepts of triangles. The Class 9 Maths NCERT exemplar solutions chapter 7 Triangles builds a good practice regime along with clarity on the concepts such that students can easily attempt other books as RD Sharma Class 9 Maths, NCERT Class 9 Maths, RS Aggarwal Class 9 Maths, et cetera.

A distinct feature of these solutions can be accessed through NCERT exemplar Class 9 Maths solutions chapter 7 pdf download, where students are made available the pdf versions of these solutions which comes in handy while studying NCERT exemplar Class 9 Maths chapter 7 in an offline mode.

Check the Solutions of Questions Given in the Book

Chapter No.	Chapter Name
Chapter 1	Number Systems
Chapter 2	Polynomials
Chapter 3	Coordinate Geometry
Chapter 4	Linear Equations In Two Variables
Chapter 5	Introduction to Euclid's Geometry
Chapter 6	Lines And Angles
Chapter 7	Triangles
Chapter 8	Quadrilaterals
Chapter 9	Areas of Parallelograms and Triangles
Chapter 10	Circles
Chapter 11	Constructions
Chapter 12	Heron’s Formula
Chapter 13	Surface Area and Volumes
Chapter 14	Statistics
Chapter 15	Probability

Also, Read NCERT Solution Subject Wise

Check NCERT Notes Subject Wise

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Question (FAQs)

1. What is the difference between a similar triangle and a congruent triangle?

If two triangles are similar, their corresponding interior angles will be the same; however, the side length will not be the same. If two triangles are congruent, then their corresponding angles and side-lengths will be the same.

2. Can a triangle have side lengths 5 m, 8 m, and 14 m?

No, it is not possible to make a triangle with these three side lengths. Sum of two sides must be more than the third side length.

As, 5 + 8 <14

Therefore, 5,8, and 14 cannot be the sides of a triangle.

3. Two triangles are congruent; are they necessarily similar? The inverse of this is true or not?

If two triangles are congruent, they will have the same interior angles and the same sides. Therefore, they must be similar, and however, if two triangles are similar, they need not be congruent.

4. Should I learn all the congruencies and their proofs from the chapter of Triangles?

It is highly recommended to understand and practice all the congruencies mentioned in the chapter along with attempting practice questions on the same. The NCERT exemplar Class 9 Maths solutions chapter 7 will help you understand the problems in a detailed manner.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles

Question:21

Answer:

Important Topics for NCERT Exemplar Solutions Class 9 Maths Chapter 7:

NCERT Class 9 Exemplar Solutions for Other Subjects:

NCERT Class 9 Maths Exemplar Solutions for Other Chapters:

Features of NCERT Exemplar Class 9 Maths Solutions Chapter 7:

Check the Solutions of Questions Given in the Book

Also, Read NCERT Solution Subject Wise

Check NCERT Notes Subject Wise

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Question (FAQs)

Also Read

Articles

Upcoming School Exams

National Means Cum-Merit Scholarship

National Institute of Open Schooling 12th Examination

National Institute of Open Schooling 10th examination