Class 10 Maths Solutions of Important Questions of 6 th chapter -Triangle - Future Study Point

Class 10 Maths Solutions of Important Questions of 6 th chapter -Triangle

chapter 6 triangles class 10 imp questions

Class 10 CBSE Maths Solutions of Important Questions of 6 th chapter -Triangle

chapter 6 triangles class 10 imp questions

Class 10 Maths Solutions of Important Questions of 6 th chapter -Triangle will help you to get the idea of solving the questions of the chapter 6 Triangle on your Maths question paper of Class 10 CBSE board exam .Study of Solutions of important questions of the class 10 chapter 6 contains selected questions of the chapter 6 triangle which has been asked in the previous year’s question papers of CBSE board exam . All questions are solved by an expert of maths who has 25 years of experience in teaching maths from 9 to 12 classes.

 See the video-Solutions for class 10 maths exercise 6.1 Triangle

Class 10 Maths Solutions of Important Questions of 6 th chapter -Triangle

Q1. In the given figure ,if LM ∥ CB  and LN ∥ CD.Prove that

Imp questions triangle ch.1

Ans.

GIVEN: In the figure LM ∥ CB  and LN ∥ CD

TO PROVE:

PROOF : In ΔABC ,we have

LM ∥ CB (given)

In ΔADC ,we have

LN ∥ CD (given)

From equation (i) and equation (ii)

Adding 1 both sides,we have

Hence,proved


Q2. In the given figure DE OQ and DF OR. Show that EF QR.

Imp questions triangle Q2 class 10

 

GIVEN:DE ∥ OQ and DF ∥ OR

TO PROVE EF ∥ QR.

PROOF: In ΔPOQ

DE ∥ OQ (given)

In ΔPOR

DF ∥ OR (given)

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

EF ∥ QR(converse of BPT theorem),Hence proved

See the video-Class 10 Maths NCERT Solurions Exercise 6.2 chapter 6 Triangles

Q3. Using BPT ,prove that a line drawn through the mid point of one side of a triangle parallel to another side bisects the third side .(Recall that you have proved it in class IX).

Ans.

Q3 imp question class 10 triangle

 

GIVEN: ΔABC in which DE ∥ BC

TO PROVE: DE bisects the sides AC

PROOF: In ΔABC

DE ∥ BC (given)

Since D is the mid point of  AB

∴ AD = BD

Putting the value of AD in above equation

AE = CE

Therefore DE bisects AC, Hence proved

Q4.ABCD is a trapizium in which AB DC and its diagonals intersects each other at the point O. Show that 

Ans.

Q4. Imp question triangle class 10

 

GIVEN:  Trapezium ABCD in which diagonal AC and BD bisect each other at O in which AB ∥ DC

TO PROVE:

CONSTRUCTION: Drawing OE ∥ AB ∥ DC

PROOF: In ΔABD

OE ∥ AB (constructed)

In ΔADC

OE ∥ DC (constructed)

From equation (i) and equation (ii)

Hence proved

Q5.The diagonal of a quadrilateral ABCD intersect each other at  the point O such that 

Show that ABCD is a trapizium

Ans.

Q5 imp questions triangles class 10

 

GIVEN: The  quadrilateral ABCD  in which diagonal AC and BD intersect each other at  the point O such that

TO PROVE: ABCD is a trapizium

CONSTRUCTION: Drawing EF ∥ DC

PROOF: In ΔADC

EF ∥ DC (constructed)

∴OE ∥ DC

From equation (i) and equation (ii)

∴OE ∥ AB (converse of BPT)

We also have OE ∥ DC

∴AB ∥ DC

Hence ABCD is a trapizium

Q6.Diagonals AC and BD of a trapezium ABCD with AB DC intersect each other at the point O. Using a similarity criterion for two triangles, show that 

Ans.

Q6 imp questions triangles class 10

GIVEN: Trapezium ABCD with AB ∥ DC in which diagonal AC and BD bisect each other at O

TO PROVE:

PROOF : In ΔAOB and ΔCOD

AB ∥ DC (given) and AC and BD  are  transversal

∴ ∠BAO = ∠OCD (alternate angle)

∴∠ABO = ∠ODC (alternate angle)

ΔAOB ∼ ΔCOD (AA similarity)

Applying the rule of similar triangles

Hence Proved

Q7.In the given figure       and ∠1 = ∠2, show that ΔPQR ∼ ΔTQR.

 

Q7 ex 6 triangle imp question class10

 

Ans.

GIVEN: ∠1 = ∠2

TO PROVE:ΔPQR ∼ ΔTQR

PROOF: In ΔPQR and ΔTQR

∠1 = ∠2 (given)

PQ = PR

Substituting the value of PR =PQ in the given equation

∠1 = ∠1 (common)

ΔPQR ∼ ΔTQR (SAS criteria of similar triangle)

Q8.D is a point on the side BC of a triangle ABC, such that ∠ADC = ∠BAC. Show that CA² = CB.CD.

Ans.

TRIANGLE imp questions class 10 Q8

GIVEN: ΔABC and D is a point on AC

Such that ∠ADC = ∠BAC

TO PROVE:  CA² = CB.CD

PROOF: In ΔABC and ΔADC

∠DAC = ∠ABC

∠BAC = ∠ADC

ΔABC∼ ΔDAC (AA criteria of the similar triangle)

According to the rule of similar triangle

CA² = CB.CD, Hence proved

Q9.Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆PQR (see in given figure). Show that ∆ABC ~ ∆PQR.

Triangle imp questions class 10 Q9

Ans.

GIVEN: ΔABC and ΔPQR in which AD and PM are the medians of the triangles

TO PROVE:∆ABC ~ ∆PQR

PROOF: In ΔABC and ΔPQR

BD = CD = BC/2 (AD is the median of the triangle ABC)

QM = MR = QR/2 (PM is the median of the triangle PQR)

Rewriting the given relationship of the sides

ΔABD ∼ ΔPQM (SSS rule)

∠B = ∠Q ( corresponding angles of similar triangles)

∴∆ABC ~ ∆PQR ( SAS rule criteria of similar triangles)

Q10.Let ∆ABC ~ ∆DEF and their areas be, respectively, 64 cm² and 121 cm². If EF = 15.4 cm, find BC.

Ans. We are given that

∆ABC ~ ∆DEF and ar ∆ABC = 64 cm² , ar ∆DEF =121 cm²

Since ∆ABC ~ ∆DEF, therefore

It is given to us that EF = 15.4 cm

BC = √(125.44)= 11.2

Hence the value of BC is 11.2 cm

Q11.A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Ans. We are given the height of the vertical pole is = 6 m , Let height of the tower is h, the length of its shadow given to us is 28 m

Q11 imp questions triangles

In ΔABC and ΔPQR

∠B = ∠Q ( each of 90°)

∠C = ∠R (son’s altitude is same)

ΔABC∼ ΔPQR (AA rule )

According to the rule of similar triangle

4h = 168

h = 42

Hence the height of the tower is 42 m

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