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NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles

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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

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, \angleB is opposite to side AC and \angleC is opposite to side AB
\therefore \angleB = \angleC = 50° (\because \angleB = 50° given)
Hence option (B) is correct.

Question:4

In \triangleABC, BC = AB and \angleB = 80^{\circ}. Then \angleA 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 \angleC is opposite to side AB and \angleA is opposite to side BC
\therefore \angleC = \angleA = x (Assume)
\angleB = 80° (Given)
\because \angleA + \angleB + \angleC = 180°
\Rightarrow x + 80° + x = 180°
\Rightarrow 2x = 100°
\Rightarrow x = 50°
Hence option (C) is correct.

Question:5

In \trianglePQR, \angleR = \angleP 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 \angleR = \angleP (Given)
Side QR is opposite to \angleP and side PQ is opposite to \angleR
\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 \triangleABC such that AD bisects \angleBAC. Then
(A) BD = CD
(B) BA > BD
(C) BD > BA
(D) CD > CA

Answer:

[A]
Solution.
BC is the side of \triangleABC, D is the point on side BC.

In \triangleABC,
\angleBAD = \angleCAD (Given that AD bisects ÐBAC)
In \triangleADC,
\angleBDA > \angleBAD (\because Exterior angle of triangle is greater than interior opposite angle)
\angleBAD = \angleCAD
Then BA > BD (\because Side opposite to greater angle is longer)
Hence option (A) is correct.

Question:7

It is given that \triangleABC \cong \triangleFDE and AB = 5 cm, \angleB = 40° and \angleA = 80°. Then which of the following is true?
(A) DF = 5 cm, \angleF = 60°
(B) DF = 5 cm, \angleE = 60°
(C) DE = 5 cm, \angleE = 60°
(D) DE = 5 cm, \angleD = 40°

Answer:

[B]
Solution. In \triangleABC, \angleA = 80°, \angleB = 40° and \angleC = x ( assume)

\because \angleA + \angleB + \anglex = 180° (angle sum property of a triangle)
\Rightarrow 80° + 40° + x = 180°
\Rightarrow x = 180° – 120°
\Rightarrowx = 60°
\Rightarrow ÐC = 60°
\because \triangleABC \cong \triangleFDE
By ASA congruence relation
AB = FD, \angleA = \angleF, \angleB = \angleD
\Rightarrow DF = 5 cm, \angleF = 80°, \angleB = 40° = \angleD
\because Also, \angleC = \angleE
\Rightarrow\angleE = 60°
Therefore, DF = 5 cm, \angleE = 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 \trianglePQR, if \angleR > \angleQ, 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 \angleR > \angleQ
Therefore, PQ > PR
Hence option (B) is correct.

Question:10

In triangles ABC and PQR, AB = AC, \angleC = \angleP and \angleB =\angleQ. 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 \angleB = \angleC
We also know that, sides opposite to equal angles are also equal
Here,\angleQ =\angleP
\Rightarrow PR = QR
Hence, two sides of both the triangles are respectively equal.
So \triangleABC and \trianglePQR are isosceles triangle.
But may or may not be congruent.
Because sides of \triangleABC may not be equal to side of \trianglePQR.
Hence option (A) is correct.

Question:1

In triangles ABC and PQR, \angleA = \angleQ and ÐB = ÐR. Which side of \trianglePQR should be equal to side AB of \triangleABC so that the two triangles are congruent? Give reason for your answer.

Answer:

Given : In triangles ABC and PQR, \angleA = \angleQ and\angleB =\angleR

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

Question:2

In triangles ABC and PQR, \angleA =\angleQ and \angleB = \angleR. Which side of \triangle PQR should be equal to side BC of \triangleABC 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
\angleA = \angleQ and \angleB = \angleR.

By congruency, corresponding angles and sides must be equal to each other in both triangles.
By the AAS criterion of congruence BC = PR.
Then \triangleABC \cong \triangleQRP
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
\angleC = \angleR
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 \triangleABC and \trianglePQR

AC = PR
\angleB = \angleQ, \angleC = \angleR
\Rightarrow \triangleABC \cong \trianglePQR
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 \trianglePQR\cong \triangleEDF, then is it true to say that PR = EF? Give reason for your answer.

Answer:

True
Solution.
Let us consider \trianglePQR and \triangleEDF
\because \trianglePQR \cong \triangleEDF (Given)

By congruency, corresponding angles and sides must be equal
\angleP = \angleE, \angleQ = \angleD, \angleR = \angleF
and
PQ = ED, QR = DF, PR = EF
Therefore the given statement is true.

Question:8

In \trianglePQR, \angleP = 70° and \angleR = 30°. Which side of this triangle is the longest? Give reason for your answer.

Answer:

[PR]
Solution.
In \trianglePQR

\therefore \angleP + \angleQ +\angleR = 180° (angle sum property)
\angleP = 70° and \angleR = 30° (Given)
70° + \angleQ + 30° = 180°
\angleQ = 80°
\therefore Greatest angle is \angleQ 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 \triangleABD
AB + BD > AD …(i)
In \triangleADC
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 \angleBAC. Is it true to say that perimeter of the triangle is greater than 2AM ? Give reason for your answer.

Answer:

True
Solution.

Perimeter of \triangleABC = AB + BC + AC
In \triangleAMB and \triangleAMC
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
\thereforeAB + BC + AC > 2AM
=> Perimeter of \triangleABC > 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 \triangleABC \cong \triangleDEF.

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 \anglePQR is exterior angle of \trianglePSQ
\therefore \anglePQR > \anglePSQ
\because PQ = PR (Given)
\anglePQR = \anglePRQ (Angles opposite to equal sides are equal)
Then, \anglePRQ = \anglePRS
\therefore \anglePRS > 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 \trianglePQR. Show that:PQ + QR + RP > 2 PS.

Answer:

Given, \trianglePQR, 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 \trianglePQS,
PQ + QS > PS
and in \trianglePSR,
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 \triangleABC 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 \triangleABC
AB = AC (Given)
\angleABC = \angleACB (angles opposite to equal sides are equal)
In \triangleABC and \angleDBC
\angleABC > \angleDBC
\Rightarrow \angleACB > \angleDBC
\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 \triangleACM and \triangleDMB
\angleACM = \angleBDM (Alternate interior angles are equal)
\angleAMC = \angleDMB (vertically opposite angles are equal)
AM = MB (given)
then by AAS criterion of congruence
\triangleACM \cong \triangleBDM
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 \angleMOC = \angleABC.

Answer:

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

To prove : \angleMOC = \angleABC
Proof :
AB = AC
\angleABC = \angleACB (Angles opposite to two equal sides of triangle are equal)
In \triangleOBC,
\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 \angleABC is equal to \angleBOC

Answer:

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

To prove : External angle adjacent to \angleABC is equal to \angleBOC
Proof : Produce line CB to D in \triangleABC
AB = AC (Given)
\angleACB = \angleABC (angles opposite to equal sides are equal)
\angleOCB = \angleOBC (Bisector of angle B and C, respectively)
In \triangleBOC,
\angleOBC + \angleOCB + \angleBOC = 180°
\Rightarrow 2\angleOBC + \angleBOC = 180° (From above)
\Rightarrow \angleABC + \angleBOC = 180°
(\because \angleABO + \angleOBC = \angleABC)
\Rightarrow \angleABC+ \angleOBA = 180°
\Rightarrow \angleOBA = \angleBOC.
Hence proved.

Question:11

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

Answer:

Given: In \triangleABC, AD is bisector of \angleBAC.
To prove: AB > BD
Proof: We know that
Exterior angle of triangle is greater than each of opposite interior angles then
In \triangleADC
\angleADB > \angleCAD (From above result)
Now, \angleBAD = \angleDAC (\becauseAD is bisector of \angleBAC)
\Rightarrow \angleADB > \angleBAD
\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
\angleLCA = 90 – i, \angleMCD = 90 – r
\because \angleACN = \angleDCN = i = r (Given)
\Rightarrow \angleLCA = \angleMCD = 90 – i = 90 – r (Since, i = r )
\Rightarrow \angleBCL = 90 – r (vertically opposite angles)
\because \angleBCL = \angleLCA
\angleL 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 \angleBAD = \angleCAD, a student proceeded as follows:
In \triangleABD and \triangleACD,
AB = AC (Given)
\angleB = \angleC (because AB = AC)
and \angleADB = \angleADC
Therefore, \triangleABD \cong \triangleACD (AAS)
So, \angleBAD =\angleCAD (CPCT)
What is the defect in the above arguments?

Answer:

[\angleABD = ACD is defect.]
Solution.
In \triangleABC, AB = AC (Given)

\angleACB = \angleABC (angles opposite to equal side are equal)
In \triangleABD and \triangleACD
\angleABD = \angleACD (given)
\angleABD = \angleACD (from above)
\angleADB = \angleADC (each 90°)
\therefore \triangleABD \cong \triangleACD
\angleBAD = \angleCAD
So, the defect in above given argument, firstly prove
\angleABD = \angleACD

Question:4

P is a point on the bisector of \angleABC. 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
\angle1 = \angle2 (\because BP is bisector of \angleABC)
\angle1 = \angle3 (\because PQ is parallel to BA)
\therefore\angle2 = \angle3
\Rightarrow PQ = BQ (If two angles are equal their opposite sides also equal)
\Rightarrow \trianglePBQ 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 :- \angleADB =\angleBDC
and \angleABD = \angleDBC
Proof : In \triangleABD and \triangleCBD
AB = CD (Given)
BD = BD (common)
AB = BC (given)
\triangleABD \cong \triangleCBD (SSS congruency)
\angleABD = \angleCBD (by CPCT)
And \angleADB = \angleBDC (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 \angleA meets BC at D. Prove that BC = 2 AD.

Answer:

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

To prove : BC = 2AD
Proof : In \triangleABC, AB = AC (given)
Then \angleB = \angleC (If sides are equal in a triangle, then opposite angle are also equal)
Now,
\angleBAC + \angleABC + \angleBCA = 180° (angle sum property)
\Rightarrow 90° + 2\angleABC = 180° (Q \angleABC = \angleACB)
\Rightarrow \angleB = 45° = \angleACB
\because \angleBAD = \angleCAD (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 \triangleOCD is an isosceles triangle.

Answer:

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

To prove: \triangleOCD is an isosceles triangle.
Proof : \angleAOB = \angleOAB = \angleOBA = 60°
And, DA = AB = OB (\because \triangleOAB is equilateral)
Now, \angleDCB = \angleBAD = \angleABD = \angleCDA = 90°
DA = AB = CB = CD (\because ABCD is square)
Let, \angleOAB = \angleOBA = 60° ……(1)
\angleBAD = \angleABD = 90° …….(2)
Then subtract eqn. (1) from (2), we get
\angleBAD – \angleOAB =\angleABD – \angleOBA = 90° – 60° = 30°
\angleDAO = \angleCBO = 30° …….(3)
In \triangleAOD and \triangleBOC
AO = BO (OAB is an equilateral triangle)
\angleDAO = \angleCBO (from 3)
AD = BC (ABCD is square)
\therefore\triangleAOD \cong \triangleBOC (by SAS criterion of congruency)
\Rightarrow DO = OC (by CPCT)
\therefore\triangleOCD 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 : \angleAOB = \angleAOC = 90°
\angleBOD = \angleCOD = 90°
Proof : In \triangleABD and \triangleACD
AB = AC (given)
AD = AD (common)
BD = CD (given)
\therefore \triangleABD \cong \triangleACD (By SSS congruence)
\angleBAO = \angleCAO (by CPCT)
Now; In \triangleAOB and \triangleAOC
AB = AC (given)
AO = OA (common)
\angleBAO = \angleCAO (from above)
\therefore \triangleAOB \cong\triangleAOC (By SAS congruence)
\Rightarrow BO = OC
And, \angleAOB = \angleAOC
Then \angleAOB + \angleAOC = 180°
\angleAOB + \angleAOB = 180° (from above)
\angleAOB = 90°, \angleAOC = 90°
and \angleAOB = \angleCOD, \angleAOC = \angleBOD (Vertically opposite angles)
\therefore AO \perpr 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: \triangleABC is isosceles triangle AB = AC
AD and BE are altitudes at BC and AC

To Prove: AE = BD
Proof: In \triangleABD & \triangleACD
AD = AD (common)
\angleADB = \angleADC (AD is altitude at BC)
AB = AC (Given)
\therefore \triangleABD \cong \triangleACD (by SAS congruence)
\angleBAD = \angleDAC (by CPCT)
\angleBAD = \angleDAE (ÐDAC = ÐDAE) (i)
Now, in \triangle ABD and \triangle ABE
AB = BA (common)
\angleADB = \angleAEB = 90° (AD and BE are altitudes)
\angleBAD = \angleDAE (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 \triangleABD and \triangleECD
CD = DB (D is mid point)
\angleBDA = \angleCDE (Vertically opposite angle)
ED = AD (by construction)
\therefore \triangle ABD \cong \triangleECD (by SAS congruence)
Þ EC = AB (by CPCT)
In \triangleACE, 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 \triangleAOB, AB < OA + OB
In \triangleBOC, BC < OB + OC
In \triangleCOD, CD < OC + OD
In \triangleAOD, 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 \triangleABC, AB + BC > AC
In \triangleBCD, BC + CD > BD
In \triangleCAD, AD + CD > AC
In \triangleBAD, 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 : \angleABC = 90°

Proof : AD = CD = ½ AC (\because D is mid-point)
BD = AC (given)
So, AD = BD = CD
Let AD = BD
\angleBAD = \angleABD (angles opposite to equal sides are equal)
Now, CD = BD
\angleBCD = \angleCBD (angles opposite to equal sides are equal)
In \triangleABC,
\angleABC + \angleBAC + \angleBCA = 180° (angle sum property)
\angleABC + \angleBAD + \angleBCD = 180°
Now, \angleABC + \angleABD + \angleCBD = 180° (\because\angleBAD = \angleABD, \angleBCD = \angleCBD)
Then \angleABC + \angleABC = 180° (\because \angleABD + \angleCBD = \angleABC)
\Rightarrow \angleABC = 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 \triangleADB and \triangleEDC
AD = CD (D is midpoint)
BD = DE (D is midpoint)
\angleADB = \angleEDC (Vertically opposite angles)
\angleADB \cong \triangleEDC (by SAS congruence)
\therefore AB = EC (by CPCT)
Now, \angleABC + \angleBCE = 180°
Þ 90° + \angleBCE = 180° (\angleABC = 90°, Given)
Þ\angleBCE = 90°
And EC || AB (\because \angleBAD and \angleDCE are alternate angle)
In \triangleABC and \angleEBC
BC = BC (common)
AB = EC (From above)
\angleABC = \angleBCE (From above)
\therefore \triangleABC \cong \triangleECB (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: \angleQOP = \angleROP
Proof: In \trianglePOR and \trianglePOQ
PQ = PR (Given)
\anglePQO = \anglePRO = 90° (P is equidistant from l and m)
So PQ and PR should be perpendicular to lines l and m respectively
\therefore \trianglePOR \cong\trianglePOQ (by RHS congruence)
\angleROP = \angleQOP (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: \triangleANM and \triangleBNM
AM = BM (M is mid-point)
\angleAMN = \angleBMN = 90° (given)
MN = MN (common)
\triangleANM \cong \triangleBNM (congruency of SAS)
AN = BN (by CPCT) …(i)
We know that
\angleANM = \angleBNM (by CPCT)
Then 90° – \angleANM = 90° – \angleBNM
\Rightarrow \angleAND = \angleBNC …(ii)
In \triangleAND and \triangleBNC,
AN = BN (From i)
\angleAND = \angleBNC (From ii)
DN = CN (N is mid-point)
\triangleAND \cong \triangleBNC (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 \angleA and \angleC

To prove: AB = AD, CB = CD
In \triangleABC and \triangleADC
\angleBAC = \angleDAC (\because AC bisects \angleA)
AC = AC (common)
\angleBAC = \angleDCA (\becauseAC bisects \angleC)
\therefore \triangle ABC \cong \triangleADC (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: \becauseAB = 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 \angleB and \angleD decide which is greater

Answer:

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

Join BD
In \triangleBCD
DC < BC (\because CD is Greatest side)
\angleCBD > \angleBDC (Angle opposite to greater side is greater) …(i)
In \triangleABD
AD > AB (AB is the smallest side)
\angleABD > \angleADB (Angle opposite to greater side is greater) …(ii)
Adding (i) and (ii)
\angleCBD + \angleABD > \angleBDC + \angleADB
\angleABC >\angleADC
Hence \angleB 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 > \angleC (angle opposite to longest side is greaten)
\angleB > \angleA
By adding both
\angleB + \angleB > \angleC + \angleA
2\angleB > \angleC +\angleA
2\angleB + \angleB > \angleC + \angleA + \angleB (adding \angleB to LHS and RHS)
3\angleB >\angleC + \angleA + \angleB
3\angleB > 180° (Sum of all interior angles in triangle is 180°)
\angleB > 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 \triangleABC and \triangleADC
AB = AD (Given)
BC = CD (Given)
AC = AC (Common)
\triangleABC \cong \triangleADC (by SSS congruency)
Then by CPCT,
ÐBAC = ÐDAC …(i)
Now in \triangleABO and \triangleADO
AB = AD (Given)
AO = AC (common)
\angleBAO = \angleDAO (from i)
\triangleABO \cong \triangleDAO (by SAS congruency)
Then by CPCT,
\therefore \angleAOB = \angleDOA
\because\angleBOD = 180°
\angleAOB + \angleAOD = 180° (\because\angleBOD = \angleAOB + \angleAOD)
\angleAOB + \angleAOB = 180° (\because\triangleAOB \cong \triangleAOD)
\angleAOB = 90° (\angleAOB = \angleAOD)
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:

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

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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Get answers from students and experts

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

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

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

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 .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

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.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

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

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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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Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
Photographer

Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

2 Jobs Available
Producer

An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story. 

They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.

2 Jobs Available
Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

5 Jobs Available
Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article. 

3 Jobs Available
Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
Reporter

Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.  

2 Jobs Available
Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available
Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
QA Manager
4 Jobs Available
Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
Production Manager
3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

2 Jobs Available
Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software. 

2 Jobs Available
Process Development Engineer

The Process Development Engineers design, implement, manufacture, mine, and other production systems using technical knowledge and expertise in the industry. They use computer modeling software to test technologies and machinery. An individual who is opting career as Process Development Engineer is responsible for developing cost-effective and efficient processes. They also monitor the production process and ensure it functions smoothly and efficiently.

2 Jobs Available
QA Manager
4 Jobs Available
AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

4 Jobs Available
Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

4 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
ITSM Manager
3 Jobs Available
Automation Test Engineer

An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process. 

2 Jobs Available
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