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CBSE Class 10 Maths - Chapter 6 - Triangles

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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?

Solution.

Given, AD = 3 cm, DB = 8 cm, AC = 22 cm, DE || BC

∴ We know that
=> 3/11 = AE/22
=> 22 x 3/11 = AE
=> AE = 11 x 3
=> AE = 33

So, the length of AE is 33 cm.
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 = ?

Solution.

Given, DE = 16.5, DH = 5, HF = 6, GH || EF

∴ We know that
=> DG/GE = DH/HF
=> 16.5/GE = 5/6
=> GE = 16.5 X 6/5
=> GE = 3.3 X 6
=> GE = 19.8

So, the length of GE is 19.8 units.
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

Solution.

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.