NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles

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 Questions (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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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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