NCERT solutions for Class 9 Maths Chapter 7 Triangles is one of the most important topic for the students to practice. For solving questions, it is crucial to have good knowledge of the topic. Students can study various concepts of maths by visiting our site BYJU’S. To ease the fear of maths for the students, we at BYJU’S provide NCERT Solution for Class 9 Maths Chapter 7 Triangles. Student can download the NCERT Solution for class 9 Maths Chapter 7 pdf or can view it online by following the link. These solution are provided in detailed manner, where one can find step-by-step solution to all the questions of **NCERT Solutions for class 9 maths chapter 7**.

NCERT solutions for class 9 maths chapter 7 pdf are prepared by our subject experts under the guidelines of NCERT to assists students in their examination.

### NCERT Solutions For Class 9 Maths Chapter 7 Exercises

- NCERT Solutions For Class 9 Maths Chapter 7 Triangles Exercise 7.1
- NCERT Solutions For Class 9 Maths Chapter 7 Triangles Exercise 7.2
- NCERT Solutions For Class 9 Maths Chapter 7 Triangles Exercise 7.3
- NCERT Solutions For Class 9 Maths Chapter 7 Triangles Exercise 7.4
- NCERT Solutions For Class 9 Maths Chapter 7 Triangles Exercise 7.5

__Exercise 7.1__

**Question 1:**

*In quadrilateral PQRS, PR = PQ and PQ bisects ∠P (look at the fig.). Show that the △PQR = △PQS. What can you say about QR and QS?*

**Solution:**

In

PR = PS

PQ = PQ (common)

Hence Proved.

Therefore, QR = QS (CPCT)

*Question 2:*

*ABCD is a quadrilateral in which AD = BC and ∠DAB= ∠CBA (see Fig.). Prove that*

*(i) △ABD=△BAC*

*(ii) BD = AC*

*(iii) ∠ABD=∠BAC*

**Solution:**

In the given figure, ABCD is a quadrilateral in which AD = BC and

In

AD = BC [Given]

AB = AB [Common]

BD = AC [CPCT]

and

Hence Proved.

**Question 3:**

*AD and BC are equal perpendiculars to a line segment AB (see Fig.). Show that CD bisects AB.*

**Solution:**

In

AD = BC [Given]

Also, AO = BO [CPCT]

Hence, CD bisects AB Proved.

**Question 4:**

*L and m are two para;;e; lines intersected by another pair of parallel lines p and q (see fig.). Show that △ABC+△CDA .*

**Solution:**

In the given figure, ABCD is a parallelogram in which AC is a diagonal, i.e., AB || DC and BC || AD.

In

AC = AC (Common)

Hence Proved.

**Question 5:**

*Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendicular from B to the arms of ∠A (See fig.). Show that:*

(i)

(ii) BP = BQ or B is equidistant from the arms of

**Solution:**

In

AB = AB

Also, BP = BQ (By CPCT)

i.e, B is equidistant from the arms of

Hence proved.

**Question 6:**

*In the fig, AC = AE, AB = AD and △BAD=△EAC . Show that BC = DE.*

**Solution:**

Now, in

AB = AD

AC = AE

BC = DE (CPCT)

Hence Proved.

**Question 7:**

*AB is a line segment and p is its mid-point. D and E are points on the same side of AB such that ∠BAD=∠ABE and ∠EPA=∠DPB (see fig.). Show that*

* (i) △DAP=△EBP*

* (ii) AD = BE*

*Solution:*

In

AP = BP (P is the midpoint of the line segment AB)

AD = BE (ASA)

AD = BE

Hence Proved.

**Question 8:**

*In right triangle ABC, right angles at C, M is the mid-point of hypotenuse AB, C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see fig.). Show that :*

* (i) △AMC=△BMD*

* (II) ∠DBC is a right angle.*

* (iii) △DBC=△ACB*

* (iv) CM=12AB*

**Solution:**

In

(i) DM = CM (given)

BM = AM (M is the midpoint of AB)

Hence Proved.

(ii) AC || BD

Hence proved.

(iii) In

DB = AC (CPCT)

BC = BC (Common)

Hence proved.

(iv) AB = CD

Hence,

__Exercise 7.2:__

**Question 1:**

*In an isosceles triangle XYZ with XY=XZ ,the bisector of ∠B and ∠C*

*Intersect each other at O .join A at O .Show that:*

*(1)OY=OZ (2)XO bisects ∠A*

**Solution:**

[OY and OZ are bisector of

OBY=OZ [ sides opposite to equal angles are equal]

Again,

In

XY=XZ [Given]

OY=OZ [proved XYove]

XO bisects **proved **

**Question 2:**

In

** Solution:**

In

YO=ZO [ XO bisects YZ]

XO=XO [common]

Hence, **Proved**

*Question 3:*

*ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and*

*AB respectively (see figure).Show that these altitudes are equal.*

**Solution:**

In

AB=AC [Given]

Now in right triangles BFC and CEB ,

BC=BC

Hence, BE=CF [CPCT] **proved**

** **

**Question 4:**

*ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see.fig) show that*

(1)

(2) AB=AC i,e ABC is an isosceles triangle.

Solution(1) in

BE=CF [Given]

So,

(2) also, AB=AC [ CPCT]

i,e., ABC is an isosceles triangle **Proved.**

**Question 5:**

*ABC and DBC are two isosceles triangle on the same base BC .show that*

Solution. In isosceles

AB= AC

[Angles opposite to equal sides are equal]

Now , in isosceles

BD=CD

[Angle opposite to equal sides are equal]

Adding (i) and (ii) , we have

**proved**

**Question 6:**

**Solution:**

AB=AC

[Angles opposite to equal sides are equal]

AB=AD [Given]

AD=AC [AB=AC]

[Angles opposite to equal sides are equal]

Adding (i) and (ii)

Now in

**Question 7:**

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

Solution. In

AB =AC

We know that angles opposite to equal sides of an isosceles triangle are equal

So,

Since

Hence

**Question 8:**

**Show that the angles of an equilateral triangle are 60 each.**

**Solution:**

As

So, AB=BC=AC

Now, AB=AC

Again BC=AC

Now in

Also from (i) and (ii)

Hence each angle of an equilateral triangle is 60° **Proved**

__Exercise 7.3__

**Question 1:**

*On the same base BC **⧍**XYZ and **⧍**DYZ are two isosceles triangles and vertices X and D are on the same side of YZ. If XD is extended to intersect YZ at P, show that*

*(i) **⧍**XYD **⩬**⧍**XZD*

*(ii) **⧍**XYP **⩬**⧍**XZP*

*(iii) XP bisects **∠**X as well as **∠**D.*

*(iv) XP is the perpendicular and bisector of YZ.*

**Solution:**

(i) In ⧍XYD and ⧍XZD, we have

XY = XZ [Given]

YD = ZD [Given]

XD = XD [Common]

⧍XYD ⩬ ⧍XZD [SSS congruence].

(ii) In ⧍XYP and ⧍XZP, we have

XY = XZ

∠YXP = ∠ZXP

XP = XP

⧍XYP ⩬ ⧍XZP [SAS congruence].

(iii)⧍XYD ⩬ ⧍XDY

∠XDY = ∠XDZ

=> 180° — ∠XDY = 180° — ∠XDZ

Also, from part (ii), ∠YXPD = ∠ZXP

XP bisects DX as well as ∠D.

(iv) Now, YP = ZP and ∠YPX = ∠ZPX

But ∠YPX + ∠ZPX = 180°

So, 2∠BPA = 180°

Or, ∠YPX = 90°

Since YP = ZP, therefore XP is perpendicular bisector of YZ.

*Question 2:*

*Isosceles triangle XYZ has an altitude XD, in which XY = XZ. Show that*

*(i) XD bisects YZ*

*(ii) XD bisects **∠**X*

**Solution:**

(i) In ⧍ABD and ⧍ACD, we have ∠ADB = ∠ADC

XY = XZ

XD = XD

⧍XYD ⩬ ⧍XZD.

YD = ZD

Hence, XD bisects YC.

(ii) Also, ∠YXD = ∠ZXD

Hence XD bisects ∠X.

**Question 3:**

*Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of **⧍**PQR. Show that:*

*(i) **⧍**ABM **⩬** **APQN (ii) **⧍**ABC **⩬**⧍**PQR*

**Solution:**

(i) In ⧍ABM and ⧍PQN,

we have

BM = QN

AB = PQ

AM = PN

⧍ABM = ⧍PQN [ SSS Congruence ]

∠ABM = ∠PQN.

(ii) Now, in ⧍ABC and ⧍PQR, we have

AB = PQ

∠ABC = ∠PQR

BC = QR

⧍ABC ⩬ ⧍PQR [SAS congruence]

**Question 4:**

*In triangle ABC, BE and CF are two equal altitudes. By Using RHS congruence rule, prove that the triangle ABC is isosceles.*

**Solution:**

BE and CF are altitudes of a ⧍ABC.

∠BEC = ∠CFB = 90°

Now, in right triangles BEB and CFB, we have

Hyp. BC = Hyp. BC

Side BE= Side CF

⧍BEC ⩬ ⧍CFB

∠BCE = ∠CBF

Now, in ⧍ABC, ∠B = ∠C

AB = AC

Hence, ⧍ABC is an isosceles triangle.

**Question 5:**

AB = AC in an isosceles triangle ABC. Draw AP perpendicular BC to show that ∠B= ∠C.

**Solution:**

Draw AP perpendicular BC. In ⧍ABP and ⧍ACP, we have

AB = AC

∠APB = ∠APC

AB = AP [Common]

⧍ABP = ⧍ACP [By RHS congruence rule]

Also, ∠B = ∠C Proved. [CPCT]

**Exercise 7.4**

**Question 1:**

*Show that in a right-angled triangle, the hypotenuse is the longest side.*

**Solution:**

** ** ABC is a right angle triangle, right angled at B.

Now

Angles A and C are less than 90^{o}

Now,

AC > BC ….(i)

(Side opposite to greater angle is longer)

Again,

AC > AB …(ii)

(Side opposite to greater angle are longer)

Hence, from (i) and (ii), we can say that AC (hypotenuse) is the longest side.

Hence proved.

**Question 2:**

*In the figure, sides AB and AC of △ABC are extended to points P and Q respectively. Also, ∠PBC<∠QCB . Show that AC > AB.*

**Solution:**

Similarly,

It is given that

Or

Hence proved.

**Question 3:**

*In the figure, ∠B<∠A and ∠C<∠D. Show that AD < BC.*

**Solution:**

BO > AO …(i)

(Side opposite to greater angle is longer)

CO > DO …(ii)

(Same reason)

Adding (i) and (ii)

BO + CO > AO + DO

BC > AD

AD < BC

Hence Proved.

**Question 4:**

*AB and CD are respectively the smallest and longest side of a quadrilateral ABCD (see fig.) Show that ∠A>∠C and∠B>∠D .*

* *

**Solution:**

Join AC.

Mark the angles as shown in the figure..

In

BC > AB (AB is the shortest side)

[Angle opposite to longer side is greater]

In

CD > AD (CD is the longest side)

[Angle opposite to longer side is greater]

Adding (i) and (ii), we have

Similarly, by joining BD, we can prove that

**Question 5:**

*In the figure, PR > PQ and PS bisects ∠QPR . Prove that ∠PSR>∠PSQ*

**Solution:**

PR > PQ

[Angle opposite to longer side is greater]

In

Similarly in

From (i), we know that

So from (iii) and (iv),

Hence proved.

**Question 6:**

*Show that of all the segments drawn from a given point not on it, the perpendicular line segment is the shortest. *

**Solution:**

We have a line I and O is the point not on I

We have to prove that OP < OQ, OP < OR and OP < OS.

OP < OS

In

Therefore,

Hence, OP < OQ (Side opposite to greater angle is longer)

Similarly, we can prove that OP is shorter than OR, OS, etc.

Hence proved

__Exercise 7.5__

** ***Question 1:*

*Find a point in the interior of △DEF which is at an equal distance or equidistant from all the vertices of △DEF.*

**Solution: **

Draw perpendicular bisectors of sides DE,

EF and FD, which meets at O.

Hence, O is the required point.

** **

*Question 2:*

*Find a point in the interior of a triangle such that it is at equal distances from all the sides of the triangle.*

** Solution:**

** **

*Question 3:*

* People are concentrated at three points in a park namely A,B and C. (see Fig.).*

** A: is where swings and slides for children are present**

** B: is where a lake is present**

** C: is where there is a large parking lot and exit**

**Where do you think an ice – cream parlor should be set up such that the maximum number of people can access it? Draw bisectors ∠A,∠B and ∠C of △ABC. Let these angle bisectors meet at O. O is the required point.**

**Solution: **Join AB, BC and CA to get a triangle ABC. Draw the perpendicular bisector of AB and BC. Let them meet at O. Then O is equidistant from A, B and C. Hence, the parlor should be set up at O so that all the other points are equidistant from it.

** **

*Question 4:*

*Fill the star shaped and hexagonal rangolies [see fig.(i) and (ii)] by filling them with as many equilateral triangles as you can of side 1 cm. What is the number of triangles in both the cases? Which one has the most number of triangles?*

** **

** **

**Solution:**