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NCERT New Syllabus Mathematics
Class: 10
Exercise 6.2
Topic: Triangles
Understanding Triangles: A Key Concept in Class 10 Mathematics
Triangles form one of the foundational blocks in geometry, and in the Class 10 NCERT syllabus, this topic holds significant importance. From learning the basic properties of triangles to diving deeper into congruence, similarity, and the Pythagoras theorem, the chapter on triangles equips students with essential tools to solve geometric problems.
In this blog, we will systematically explore each concept, providing clear explanations and practical solutions to the exercises. By the end of this discussion, you'll have a solid understanding of triangles, enabling you to confidently approach simple and complex problems.
Let's begin our journey into the world of triangles and uncover the logic that shapes this fascinating topic!
EXERCISE 6.2
Q1. In the following fig, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
Solution:
1) In ∆ ABC, DE ‖ BC, find EC in fig (i)
a) In ∆ ABC and ∆ ADE,
< ABC = < ADE (corresponding angles)
< ACB = < AED (corresponding angles)
< BAC = < DAE (common angle)
So, ∆ ABC ~ ∆ ADE
b) Using basic proportionality theorem,
(AD)/(DB) = (AE)/(EC)
(1.5)/3 = 1/(EC)
1/2 = 1/(EC)
(EC) = 2
c) So, here EC = 2 cm.
2) In ∆ ABC, DE ‖ BC, find AD in fig (ii)
a) In ∆ ABC and ∆ ADE,
< ABC = < ADE (corresponding angles)
< ACB = < AED (corresponding angles)
< BAC = < DAE (common angle)
So, ∆ ABC ~ ∆ ADE
b) Using basic proportionality theorem,
(AD)/(DB) = (AE)/(EC)
(AD)/(7.2) = (1.8)/(5.4)
(AD) = [(1.8)(7.2)]/(5.4)
(AD) = (1.8)[(72)/(54)]
(AD) = (1.8)[(8)/(6)]
(AD) = (18/10)[(8)/(6)]
(AD) = (3/10)[(8)]
(AD) = [(3)(8)]/10
(AD) = 24/10
(AD) = 2.4
c) So, here AD = 2.4 cm.
3) Therefore, EC = 2 cm, and AD = 2.4 cm.
Q2. E and F are points on the sides PQ and PR respectively of a ∆ PQR. For each of the following cases, state whether EF || QR:
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
Solution:
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
1) Using basic proportionality theorem,
a) If EF ‖ QR, then we must have,
(PE)/(EQ) = (PF)/(FR) ----------- equation 1
b) LHS = (PE)/(EQ)
LHS = (3.9)/(3)LHS = 1.3 ----------- equation 2
c) RHS = (PF)/(FR)
RHS = (3.6)/(2.4)
RHS = (36)/(24)
RHS = 3/2
RHS = 1.5 ----------- equation 3
2) So, from equations 2 and 3, we can say that (PE)/(EQ) ≠ (PF)/(FR).
3) Therefore, EF is not paraller to QR.
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
1) Using basic proportionality theorem,
a) If EF ‖ QR, then we must have,
(PE)/(EQ) = (PF)/(FR) ----------- equation 1
b) LHS = (PE)/(EQ)
LHS = (4)/(4.5)
LHS = (40)/(45)
LHS = 8/9 ----------- equation 2
c) RHS = (PF)/(FR)
RHS = (8)/(9)
RHS = 8/9 ----------- equation 3
2) So, from equations 2 and 3, we can say that (PE)/(EQ) = (PF)/(FR).
3) Therefore, EF is paraller to QR.
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
1) First we will find EQ and FR
a) EQ = PQ – PE
EQ = 1.28 – 0.18EQ = 1.10 ----------- equation 1
b) FR = PR – PFFR = 2.56 – 0.36
FR = 2.20 ----------- equation 2
2) Using basic proportionality theorem,
a) If EF ‖ QR, then we must have,
(PE)/(EQ) = (PF)/(FR) ----------- equation 3
b) LHS = (PE)/(EQ)
LHS = (0.18)/(1.10)
LHS = (18)/(110)
LHS = (9)/(55)
LHS = 9/55 ----------- equation 4
c) RHS = (PF)/(FR)
RHS = (0.36)/(2.20)
RHS = (36)/(220)
RHS = (18)/(110)
RHS = (9)/(55)
RHS = 9/55 ----------- equation 5
2) So, from equations 4 and 5, we can say that (PE)/(EQ) = (PF)/(FR).
3) Therefore, EF is paraller to QR.
Q3. In the following fig. , if LM || CB and LN || CD, prove that
(AM)/(AB) = (AN)/(AD).
Solution:
1) In ∆ ABC, LM ‖ CB,
a) In ∆ AML and ∆ ABC,
< AML = < ABC (corresponding angles)
< ALM = < ACB (corresponding angles)
< MAL = < BAC (common angle)
So, ∆ AML ~ ∆ ABC
b) Using basic proportionality theorem,
(AM)/(AB) = (AL)/(AC) ----------- equation 1
2) In ∆ ADC, LN ‖ CD,
a) In ∆ ANL and ∆ ADC,
< ANL = < ADC (corresponding angles)
< ALN = < ACD (corresponding angles)
< NAL = < DAC (common angle)
So, ∆ ANL ~ ∆ ADC
b) Using basic proportionality theorem,
(AN)/(AD) = (AL)/(AC) ----------- equation 2
3) From equations 1 and 2, we have (AM)/(AB) = (AN)/(AD). Hence proved.
Q4. In the following Fig., DE || AC and DF || AE. Prove that BF/FE = BE/EC
Solution:
1) In ∆ ABC, DE ‖ AC,a) Using basic proportionality theorem,
(BD)/(DA) = (BE)/(EC) ----------- equation 1
2) In ∆ BAE, DF ‖ AE, a) Using basic proportionality theorem,
(BD)/(DA) = (BF)/(FE) ----------- equation 2
3) From equations 1 and 2, we have (BE)/(EC) = (BF)/(FE). Hence proved.
Q5. In the following Fig., DE || OQ and DF || OR. Show that EF || QR.
1) In ∆ ABC, DE ‖ AC,
a) Using basic proportionality theorem,
(BD)/(DA) = (BE)/(EC) ----------- equation 1
2) In ∆ BAE, DF ‖ AE,
a) Using basic proportionality theorem,
(BD)/(DA) = (BF)/(FE) ----------- equation 2
3) From equations 1 and 2, we have (BE)/(EC) = (BF)/(FE). Hence proved.
Q5. In the following Fig., DE || OQ and DF || OR. Show that EF || QR.
Solution:
1) In ∆ PQO, DE ‖ OQ,a) Using basic proportionality theorem,
(PD)/(DO) = (PE)/(EQ) ----------- equation 1
2) In ∆ POR, DF ‖ OR, a) Using basic proportionality theorem,
(PD)/(DO) = (PF)/(FR) ----------- equation 2
3) From equations 1 and 2, we have (PE)/(EQ) = (PF)/(FR).4) So, EF ‖ QR. Hence proved.
Q6. In the following Fig., A, B, and C are points on OP, OQ, and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
1) In ∆ PQO, DE ‖ OQ,
a) Using basic proportionality theorem,
(PD)/(DO) = (PE)/(EQ) ----------- equation 1
2) In ∆ POR, DF ‖ OR,
a) Using basic proportionality theorem,
(PD)/(DO) = (PF)/(FR) ----------- equation 2
3) From equations 1 and 2, we have (PE)/(EQ) = (PF)/(FR).
4) So, EF ‖ QR. Hence proved.
Q6. In the following Fig., A, B, and C are points on OP, OQ, and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
Solution:
1) In ∆ OPQ, AB ‖ PQ,a) Using basic proportionality theorem,
(OA)/(AP) = (OB)/(BQ) ----------- equation 1
2) In ∆ OPR, AC ‖ PR, a) Using basic proportionality theorem,
(OA)/(AP) = (OC)/(CR) ----------- equation 2
3) From equations 1 and 2, we have (OB)/(BQ) = (OC)/(CR).4) So, BC ‖ QR. Hence proved.
Q7. Using Theorem 6.1, 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).
1) In ∆ OPQ, AB ‖ PQ,
a) Using basic proportionality theorem,
(OA)/(AP) = (OB)/(BQ) ----------- equation 1
2) In ∆ OPR, AC ‖ PR,
a) Using basic proportionality theorem,
(OA)/(AP) = (OC)/(CR) ----------- equation 2
3) From equations 1 and 2, we have (OB)/(BQ) = (OC)/(CR).
4) So, BC ‖ QR. Hence proved.
Q7. Using Theorem 6.1, 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).
Solution:
1) In ∆ ABC, point P is the midpoint of AB,a) So we have,
(AP)/(PB) = 1 ----------- equation 1
2) In ∆ ABC, PQ ‖ BC, a) Using basic proportionality theorem,
(AP)/(PB) = (AQ)/(QC) ----------- equation 2
3) From equations 1 and 2, we have (AQ)/(QC) = 1.4) Here, AQ = QC, so Q is the midpoint of AC.5) So, line PQ bisects the third side AC of ∆ ABC. Hence proved.
Q8. Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it inClass IX).
1) In ∆ ABC, point P is the midpoint of AB,
a) So we have,
(AP)/(PB) = 1 ----------- equation 1
2) In ∆ ABC, PQ ‖ BC,
a) Using basic proportionality theorem,
(AP)/(PB) = (AQ)/(QC) ----------- equation 2
3) From equations 1 and 2, we have (AQ)/(QC) = 1.
4) Here, AQ = QC, so Q is the midpoint of AC.
5) So, line PQ bisects the third side AC of ∆ ABC. Hence proved.
Q8. Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in
Class IX).
Solution:1) In ∆ ABC, point P is the midpoint of AB,a) So we have,
(AP)/(PB) = 1 ----------- equation 1
2) In ∆ ABC, point Q is the midpoint of AC, a) So we have,
(AQ)/(QC) = 1 ----------- equation 2
3) From equations 1 and 2, we have (AP)/(PB) = (AQ)/(QC) = 1.4) So, PQ ‖ BC. Hence proved.
Q9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at point O. Show that AO/BO = CO/DO.
1) In ∆ ABC, point P is the midpoint of AB,
a) So we have,
(AP)/(PB) = 1 ----------- equation 1
2) In ∆ ABC, point Q is the midpoint of AC,
a) So we have,
(AQ)/(QC) = 1 ----------- equation 2
3) From equations 1 and 2, we have (AP)/(PB) = (AQ)/(QC) = 1.
4) So, PQ ‖ BC. Hence proved.
Q9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at point O. Show that AO/BO = CO/DO.
Solution:
1) In trapezium ABCD, AB ‖ CD, and diagonals AC and BD intersect at O.2) Construct line PQ such that PQ ‖ AB and PQ ‖ DC.3) In ∆ ABC, OQ ‖ AB,a) Using basic proportionality theorem,
(BQ)/(CQ) = (AO)/(CO) ----------- equation 1
4) In ∆ BDC, OQ ‖ DC, a) Using basic proportionality theorem,
(BQ)/(CQ) = (BO)/(DO) ----------- equation 2
3) From equations 1 and 2, we have (AO)/(CO) = (BO)/(DO).4) So, we have (AO)/(BO) = (CO)/(DO). Hence proved.
Q10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that (AO)/(BO) = (CO)/(DO). Show that ABCD is a trapezium.
1) In trapezium ABCD, AB ‖ CD, and diagonals AC and BD intersect at O.
2) Construct line PQ such that PQ ‖ AB and PQ ‖ DC.
3) In ∆ ABC, OQ ‖ AB,
a) Using basic proportionality theorem,
(BQ)/(CQ) = (AO)/(CO) ----------- equation 1
4) In ∆ BDC, OQ ‖ DC,
a) Using basic proportionality theorem,
(BQ)/(CQ) = (BO)/(DO) ----------- equation 2
3) From equations 1 and 2, we have (AO)/(CO) = (BO)/(DO).
4) So, we have (AO)/(BO) = (CO)/(DO). Hence proved.
Q10. The diagonals of a quadrilateral ABCD intersect each other at
the point O such that (AO)/(BO) = (CO)/(DO). Show that ABCD is a trapezium.
Solution:
1) In quadrilateral ABCD, diagonals AC and BD intersect at O.2) Construct line PQ such that PQ ‖ AB.3) In ∆ ABC, OQ ‖ AB,a) Using basic proportionality theorem,
(BQ)/(CQ) = (AO)/(CO) ----------- equation 1
4) As per the problem,(AO)/(BO) = (CO)/(DO)
(AO)/(CO) = (BO)/(DO) ----------- equation 2
5) From equations 1 and 2, we have,(BO)/(DO) = (BQ)/(CQ) ----------- equation 3
6) So, from equation 3 and the converse of the basic proportionality theorem,PQ ‖ AB, so AB ‖ CD.
7) Therefore, quadrilateral ABCD is the trapezium.
1) In quadrilateral ABCD, diagonals AC and BD intersect at O.
2) Construct line PQ such that PQ ‖ AB.
3) In ∆ ABC, OQ ‖ AB,
a) Using basic proportionality theorem,
(BQ)/(CQ) = (AO)/(CO) ----------- equation 1
4) As per the problem,
(AO)/(BO) = (CO)/(DO)
(AO)/(CO) = (BO)/(DO) ----------- equation 2
5) From equations 1 and 2, we have,
(BO)/(DO) = (BQ)/(CQ) ----------- equation 3
6) So, from equation 3 and the converse of the basic proportionality theorem,
PQ ‖ AB, so AB ‖ CD.
7) Therefore, quadrilateral ABCD is the trapezium.
Conclusion
In conclusion, mastering the topic of triangles is essential for building a strong foundation in geometry. By understanding concepts like congruence, similarity, and the Pythagoras theorem, you can easily solve a wide range of problems. As you continue practicing, these geometric principles will become second nature, helping you excel in exams and real-life applications.
Keep exploring, stay curious, and remember—the beauty of mathematics lies in its logical simplicity!
Stay tuned for more insights and solutions to other important topics from the Class 10 NCERT syllabus.
#Class10Maths #NCERTSolutions #Triangles #GeometryBasics #MathForStudents #MathMadeEasy #PythagorasTheorem #MathBlog #StudentLife #LearningIsFun #MathConcepts
In conclusion, mastering the topic of triangles is essential for building a strong foundation in geometry. By understanding concepts like congruence, similarity, and the Pythagoras theorem, you can easily solve a wide range of problems. As you continue practicing, these geometric principles will become second nature, helping you excel in exams and real-life applications.
Keep exploring, stay curious, and remember—the beauty of mathematics lies in its logical simplicity!
Stay tuned for more insights and solutions to other important topics from the Class 10 NCERT syllabus.
#Class10Maths #NCERTSolutions #Triangles #GeometryBasics #MathForStudents #MathMadeEasy #PythagorasTheorem #MathBlog #StudentLife #LearningIsFun #MathConcepts
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