## CONTENT LIST

## CBSE Class 10 Maths-Chapter 6-Triangles

Triangle is a closed two-dimensional three-sided polygon. All sides are made up of straight lines and joint at the vertex. A triangle has three vertices and each vertex forms an angle.

- The sum of all the three interior angles of any triangle is 180 degrees.
- The base angles of an isosceles triangle are same.
- The measure of the exterior angle of any triangle is equal to the sum of the corresponding interior angles.
- The sum of the exterior angles of any triangle is equal to 360 degrees.
- The sum of consecutive interior and exterior angle is supplementary.
- The sum of the lengths of any two sides of any triangle is greater than the length of the third side.
- The shortest side is always opposite the smallest interior angle.

### Similarity of Triangles

**Two triangles are said to similar if they have the same ratio of corresponding sides and equal pair of corresponding angles.**

**Triangle similarity criteria**

- AA : Two pairs of corresponding angles are equal.

- SSS : Three pairs of corresponding sides are proportional.

- SAS : Two pairs of corresponding sides are proportional and the corresponding angles between them are equal.

### Thales'Theorem

**If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.**

Prove:- AD/DB = AE/EC

Proof:-

Let, ABC be triangle in which a line DE parallel to side BC intersect AB and AC at D and E respectively.

Join BE and CD and then draw DM ⊥ AC and EN ⊥ AB, we get

Now, ar(ADE) = 1/2 x base x height = 1/2 x AD x EN

=> ar(BDE) = 1/2 x DB × EN

=> ar(ADE) = 1/2 x AE × DM

=> ar(DEC) = 1/2 x EC × DM

So, by taking the ratio of ar(ADE) and ar(BDE), we get

=> ar(ADE)/ar(BDE) = (1/2 x AE × EN) / (1/2 x DB × EN) = AD/DB ---(i)

Similarly, by taking the ratio of ar(ADE) and ar(DEC), we get

=> ar(ADE)/ar(DEC) = (1/2 x AE × DM) / (1/2 x EC × DM) = AE/EC ---(ii)

So, ar(BDE) = ar(DEC) ---(iii)

From (i), (ii) and (iii), we get

=> AD/DB = AE/EC

### Converse of Thales 'Theorem

**If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.**

Let △ ABC , a line DE divide AB and AC in such a way AD/DB=AE/EC.

We have to prove that, DE ∥ BC

Given, AD/DB=AE/EC or DB/AD=EC/AE

Add 1 in LHS and RHS we get,

(DB/AD)+1=(EC/AE)+1

(DB+AD)/AD=(EC+AE)/AE

AB/AD=AC/AE

AD/AB=AE/EC

Consider △ ADE and △ ABC,

AD/AB=AE/EC

∠A=∠A

∴ △ ADE ~ △ ABC

∠ADC=∠ABC (pair of corresponding angles)

∴DE ∥ BC

### Mid point theorem

**A line segment joining the mid points of two sides of triangle is parallel to third side.**

Let, △ ABC and point D and E are mid points of side AB and AC respectively.

We have to prove that, DE ∥ BC

Given, AD = AB/2 => AD/AB = 1/2 ----(i)

Similarly, AE/AC = 1/2 ----(ii)

From equation (i) and (ii), we get

=> AD/AB = AE/AC

Now, consider △ ADE and △ ABC

∠A = ∠A

AD/AB = AE/AC

By SAS similarity criteria, we get

△ ADE ~ △ ABC

So, ∠ADE = ∠ABC (Corresponding angle)

And, AB is the transversal of cutting two lines DE and BC

∴ DE ∥ BC [Hence Proved]

### Angle bisector theorem

**The internal bisector of any angle of triangle, divides the opposite side internally in the ratio of the sides containing the angle.**

Let, △ ABC, AF is the angle bisector of ∠A.

Then we have to prove that, AB/AC = BF/FC

Construction, Draw a parallel line from point B such that BE ∥ AF

Now, extend CA which meets BE at point E

∵ BE ∥ AF

∴ ∠CAF = ∠AEB (Corresponding angle)

and, ∠FAB = ∠ABE (Alternate angle)

∠FAB = ∠CAF (∵ AF is an angle bisector of ∠CAB)

∴ ∠AEB = ∠ABE

So, AB = AE -----(i)

Consider △ CEB, we get

AF ∥ EB

∴ AC/AE = CF/BF

=> AC/AB = FC/BF (From equation i)

=> AB/AC = BF/FC [Hence Proved]

### PROBLEMS BASED ON TRIANGLE-CLASS 10

## Question 1.

A 15 mtr high tower cast a shadow of 24 mtr long at certain time, and at the same time a electric pole cast a shadow of 16 mts length. Find the height of electric pole?

**Solution.**

Let, the height of the electric pole be x (ED).

∴ According to the question

=> AB/ED = BC/DF

=> 15/x = 24/16

=> 15/x = 3/2

=> x = 15 x 2/3

=> x = 10 m

So, the height of the electric pole is 10 m.

## Question 2.

In △ABC, DE || AB. If CD = 3 cm, EC = 4 cm, BE = 6 cm, then Find DA =-------?

## Question 3.

In a rhombus if d1 = 16 cm, d2 = 12 cm, then find the length of the side of the rhombus?

## Question 4.

D and E are respectively the points on the sides AB and AC of a triangle ABC such that AD = 2 cm, BD = 3 cm, BC = 7.5 cm and DE || BC. Then, Determine length of DE (in cm) .

## Question 5.

In given figure, AD = 3 cm, AE = 5 cm, BD = 4 cm, CE = 4 cm, CF = 2 cm, BF = 2.5 cm, then

Choose correct answer from below 4options.

(a) DE || BC

(b) DF || AC

(c) EF || AB

(d) none of these

**Solution.**

Given, AD = 3 cm, AE = 5 cm, BD = 4 cm, CE = 4 cm, CF = 2 cm, BF = 2.5 cm.

Using the theorem, if a line is drawn parallel to one side of a triangle to intersect the other two side in distinct points, the other two sides are divided in the same ratio.

In case 1, DE || BC, we get

=> AD/BD = AE/CE

=> AD/BD = 3/4

=> AE/CE = 5/4

So, AD/BD ≠ AE/EC

In case 2, DF || AC, we get

=> BD/DA = BF/CF

=> BD/DA = 4/3

=> BF/CF = 2.5/2

So, BD/DA ≠ BF/CF

In case 3, EF || AB, we get

=> CE/AE = CF/FB

=> CE/AE = 4/5

=> CF/FB = 2/2.5

So, CE/AE = CF/FB

Option (c) EF || AB, is the correct answer

## Question 6.

In triangle PQR, if PQ = 6 cm, PR = 8 cm, QS = 3 cm, and PS is the bisector of angle QPR, what is the length of SR?

## Question 7.

Diagonals of a trapezium PQRS intersect each other at the point O, PQ || RS and PQ = 3(RS), Then find the ratio of areas of triangles POQ and ROS?

**Solution.**

From △ROS and △POQ, we get

=> ∠ROS = ∠POQ [Opposite angle]

=> ∠OPQ = ∠ORS [Alternate angles]

By AA similarity criteria, we get

=> △ROS ~ △POQ

∴ According to the question

=> (Area of ROS)/(Area of POQ) = (RS)²/(PQ)² = (RS/PQ)² = (RS/3RS)² = (1/3)² = 1/9

=> (Area of ROS)/(Area of POQ) = (RS/PQ)² = (RS/3RS)² = (1/3)² = 1/9

=> (Area of ROS)/(Area of POQ) = (RS/3RS)² = (1/3)² = 1/9

=> (Area of ROS)/(Area of POQ) = (1/3)² = 1/9

So, the ratio of areas of triangles POQ and ROS are 1:9.

## Question 8.

ABCD is a trapezium in which AB || DC and P, Q are points on AD and BC respectively such that PQ || DC. If PD = 18 cm, BQ = 35 cm and QC = 15 cm, find AD.

**Solution.**

Given, AB|| DC, PD = 18 cm, BQ = 35 cm, QC = 15 cm.

Draw a line which is joining point A and C.

So, we get

=> AE/AC = AP/PD [PE || DC]

=> AE/AC = BQ/QC [QE || AB]

=> AP/PD = BQ/QC

=> x/18 = 35/15

=> x = 18 x 7/3

=> x = 42 cm

So, the length of AD is 42 + 18 = 60 cm.

## Question 9.

Corresponding sides of two similar triangles are in the ratio of 2 : 3. If the area of the smaller triangle is 48 cm², then the area of the larger triangle is:

**Solution.**

Given, ratio of sides of two similar triangles = 2 : 3, area of smaller triangle is 48 cm².

Let, the area of larger triangle be x cm².

∴ According to the question

=> 48/x = (2/3)²

=> 48/x = 4/9

=> 48 x 9/4 = x

=> x = 108 cm²

So, the area of larger triangle is 108 cm².

## Question 10.

ΔABC ~ ΔPQR, ∠B = 50° and ∠C = 70° then Find ∠P.

**Solution.**

Given, ΔABC ~ ΔPQR, ∠B = 50° and ∠C = 70°

∵ ΔABC ~ ΔPQR, so ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R

∴ We know that

=> ∠P + ∠Q + ∠R = 180°

=> ∠P + ∠B + ∠C = 180°

=> ∠P + 50° + 70° = 180°

=> ∠P + 120° = 180°

=> ∠P = 180° - 120° = 60°

So, the value of ∠P is 60°.

## Question 11.

In triangle ABC, DE || BC, AD = 3 cm, DB = 8 cm AC = 22 cm. At what distance from A does the line DE cut AC?

## Question 12.

In triangle DEF, GH is a line parallel to EF cutting DE in G and DF in H. If DE = 16.5, DH = 5, HF = 6 then GE = ?

## Question 13.

In the given figure, DE ‖ BC, AE = 15 cm, EC = 9 cm, NC = 6 cm and BN = 24 cm. Find lengths of ME and DM.

**Solution.**

Given, DE ‖ BC, AE = 15 cm, EC = 9 cm, NC = 6 cm, BN = 24 cm

In △AMC, we have

=> AE/AC = ME/NC

=> 15/24 = ME/6

=> ME = 6 x 15/24

=> ME = 15/4

=> ME = 3.75

=> AE/AC = AM/AN

=> 15/24 = AM/AN

In △ABN, we have

=> DM ‖ BN

=> AM/AN = DM/BN

=> DM/24 = 15/24

=> DM = 15/24 x 24

=> DM = 15

So, the length of ME and DM are 3.75 cm and 15 cm respectively.

## Question 15.

Find the value of x for which DE || AB in above figure

**Solution.**

Given, DE || AB

∴ We know that

=> CD/DA = CE/EB

=> (x + 3)/(3x + 19) = x/(3x + 4)

=> (x + 3)(3x + 4) = x(3x + 19)

=> 3x² + 13x + 12 = 3x² + 19x

=> -6x + 12 = 0

=> -6x = -12

=> x = (-12)/(-6)

=> x = 2

So, the value of x is 2.

## Question 16.

If ∆ABC = 4 + 6 + 8 = 18 cm ~ ∆DEF, AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm, find the perimeter of ∆ABC.

**Solution.**

∴ We know that

=> AB/DE = BC/EF and AB/DE = AC/FD

=> 4/6 = BC/9 and 4/6 = AC/12

=> 2/3 = BC/9 and 2/3 = AC/12

=> 9 x 2/3 = BC and 12 x 2/3 = AC

=> 3 x 2 = BC and 4 x 2 = AC

=> BC = 6 and AC = 8

So, the perimeter of ∆ABC = 4 + 6 + 8 = 18 cm

## Question 17.

O is a point on side PQ of a △PQR such that PO = QO = RO, then choose the correct from 4options given below?

(a) RS² = PR × QR

(b) PR² + QR² = PQ²

(c) QR² = QO² + RO²

(d) PO² + RO² = PR²

## Question 18.

In above figer find the value of x?

**Solution.**

Given, PB = 3.2 cm, BR = 4.8 cm, PA = 2.4 cm, AQ = 3.6 cm, AB = 2 cm, QR = x.

So, the length of PR = PB + BR = 3.2 + 4.8 = 8 cm

∴ We know that

=> PB/PR = AB/QR

=> 3.2/8 = 2/x

=> x = 2 x 8/3.2

=> x = 1/0.2

=> x = 5

So, the length of QR is 5 cm.

## Question 19.

A boy of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of his shadow after 4 seconds.

**Solution.**

So, distance covered by the boy after 4 sec (AC) = 1.2 x 4 = 4.8 m

∵ △CDE and △ADB are similar

=> CD/AD = CE/AB

=> x/(x + 4.8) = 0.9/3.6

=> x/(x + 4.8) = 1/4

=> 4x = x + 4.8

=> 3x = 4.8

=> x = 4.8/3

=> x = 1.6

So, the length of shadow casted by boy after 4 sec is 1.6 m