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