# CBSE Class 10 Maths - Chapter 6 - Triangles

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.
∠A= ∠B  ,  ∠B= ∠D
•  SSS : Three pairs of corresponding sides are proportional.
AB/ED = BC/DF=AC/EF
•  SAS : Two pairs of corresponding sides are proportional and the corresponding angles between them are equal.
AB/ED =  BC/DF,  ∠B= ∠D
AB/ED =  AC/EF , ∠A= ∠E
CA/FE =  CB/FD,  ∠C= ∠F

### 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.

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

### 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

Add 1 in LHS and RHS we get,

Consider  △ ADE and  △ ABC,

∠A=∠A

∴ △ ADE ~  △ ABC

∴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

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

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

Now, consider △ ADE and △ ABC

∠A = ∠A

By SAS similarity criteria, we get

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.

Given, height of tower (AB) = 15 m, shadow cast by the tower (BC) = 24 m, shadow cast by the electric pole (DF) = 16 m
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 =-------?

Solution.

Given, In △ABC, DE || AB, CD = 3 cm, EC = 4 cm, BE = 6 cm
Let, length of DA be x.

∴ According to the question
=> CD / DA = CE / EB
=> 3/x = 4/6
=> 3/x = 2/3
=> 2x = 3 x 3
=> x = 9/2
=> x = 4.5 cm

So, the length of DA is 4.5 cm.
Question 3.
In a rhombus if d1 = 16 cm, d2 = 12 cm, then find the length of the side of the rhombus?

Solution.

Given, AC = 16 cm, BD = 12 cm.

=> Length of BO = 12/2 = 6 cm
=> Length of AO = 16/2 = 8 cm

Using Pyathogorus theorem, we get
=> (AB)² = (BO)² + (AO)²
=> (AB)² = (6)² + (8)²
=> (AB)² = 36 + 64
=> (AB)² = 100
=> AB = √(100)
=> AB = 10 cm

So, length of each side is 10 cm.
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) .

Solution.

Given, AD = 2 cm, BD = 3 cm, BC = 7.5 cm.
Let, the length of DE be x cm.

∴ According to the question
=> 2/5 = x/7.5
=> x = 7.5 x 2/5
=> x = 1.5 x 2
=> x = 3 cm

So, the length of DE is 3 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
=> AE/CE = 5/4

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?

Solution.

Given, PQ = 6 cm, PR = 8 cm, QS = 3 cm, PS is the bisector of angle QPR.
Let the length of SR be x.

∴ According to the question
=> PQ/PR = QS/SR [by angle bisector theorem
=> 6/8 = 3/x
=> x = 3 x 8/6
=> x = 4 cm

So, the length of SR is 4 cm.
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.

Given, PQ || RS, PQ = 3(RS)

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 14.

In the above figure, PQ||BC, then the values of AQ and QC are

Solution.

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.

Given, ∆ABC ~ ∆DEF, AB = 4 cm, DE = 6 cm, EF = 9 cm, FD = 12 cm

∴ 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.

Given, height of boy (CE) = 90 cm = 0.9 m, speed of boy = 1.2 m/s, height of lamp (AB) = 3.6 m, time taken = 4 sec.

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

∵ △CDE and △ADB are similar
=> 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
Question 20.
State whether the above quadrilaterals are similar or not:

Solution.
Since, the ratio of corresponding sides are same but corresponding angles are not same. So, the given figure are not similar.