NCERT Solutions for Class 10 Maths Chapter 6 Triangles
NCERT Solutions for Class 10 Maths Chapter 6 Triangles contains all the questions with detailed solutions. Students are advised to solve these questions for better understanding of the concepts in Chapter 6.
Ananya Gupta19 Mar, 2024
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NCERT Solutions for Class 10 Maths Chapter 6: NCERT Solutions for Class 10 Maths Chapter 6, which deals with Triangles, are important study materials for CBSE Class 10 students. This chapter covers key concepts and rules about triangles, helping students understand geometry better.
Sometimes, students find it hard to decide which theorem to use for solving different triangle problems. These solutions make things easier by explaining concepts step by step, making it simpler for students to understand and solve triangle-related questions confidently.NCERT Solutions for Class 10 Maths Chapter 6 Triangles
NCERT Solutions for Class 10 Maths Chapter 6 are a helpful resource for students studying for their CBSE board exams. This chapter focuses on triangles, covering essential concepts, theorems, and rules. With these solutions, students can easily understand and solve problems related to triangle geometry. The step-by-step explanations provided in the solutions help students build a strong foundation in geometry and improve their problem-solving skills.NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.1
1. Fill in the blanks using the correct word given in brackets:
(i) All circles are _______________. (congruent, similar)
(ii) All squares are _______________. (similar, congruent)
(iii) All _______________ triangles are similar. (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are _______________ and (b) their corresponding sides are _______________. (equal, proportional)
Answer:
(i) similar (ii) similar (iii) equilateral (iv) equal, proportional2. Give two different examples of pair of:
(i) similar figures
(ii) non-similar figures
Answer:
(i) Two different examples of pair of similar figures are: (a) Any two rectangles (b) Any two squares (ii) Two different examples of pair of non-similar figures are: (a) A scalene and an equilateral triangle (b) An equilateral triangle and a right angled triangleNCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2
1. In figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
Answer:
(i) Since DE || BC,
Let, EC = x cm
It is given that DE || BC
By using basic proportionality theorem, we obtain
.png)
x = 2
EC = 2 cm.
(ii) Let AD = x cm.
It is given that DE || BC.
By using basic proportionality theorem, we obtain
x = 2.4
AD = 2.4cm.
2. E and F are points on the sides PQ and PR respectively of a ∆ PQR. For each of the following cases, state whether EF II QR:
(i) PE = 3.9 cm, EQ = 3cm, 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
Answer:
(i)Given: PE = 3.9 cm, EQ = 3cm, PF = 3.6 cm and FR = 2.4 cm
Hence,
.png)
Therefore, EF is not parallel to QR.
(ii)Given: PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
.png)
Therefore,EF is parallel to QR.
(iii)Given: PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
And ER = PR – PF = 2.56 – 0.36 = 2.20 cm
.png)
.png)
Hence,
Therefore, EF is parallel to QR.
3. In figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD.
Answer:
In figure, LM || CB.png)
By using Basic Proportionality theorem
And in ∆ ACD, LN || CD
.png)
......i
Similalry, in ∆ ACD, LN || CD
.png)
.......ii
By using Basic Proportionality theorem
From eq. (i) and (ii), we have
4. In figure, DE || AC and DF || AE. Prove that BF/GE = BE/EC .
Answer:
In ∆ BCA, DE || AC
[Basic Proportionality theorem] ……….(i)
And in ∆ BEA, DF || AE
[Basic Proportionality theorem] ……….(ii)
From eq. (i) and (ii), we have
5. In figure, DE || OQ and DF || OR. Show that EF || QR.
.png)
Answer:
In ∆ PQO, DE || OQ
[Basic Proportionality theorem] ……….(i)
And in ∆ POR, DF || OR
[Basic Proportionality theorem] ……….(ii)
From eq. (i) and (ii), we have
Therefore, EF || QR [By the converse of Basic Proportionality Theorem]
6. In figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
Answer:
And in ∆ POQ, AB || PQ
[Basic Proportionality theorem] ……….(i)
And in ∆ OPR, AC || PR
[Basic Proportionality theorem] ……….(ii)
From eq. (i) and (ii), we have
Therefore, BC || QR (By the converse of Basic Proportionality Theorem)
7. 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).
Answer:
.png)
Consider the given figure in which l is a line drawn through the mid-point P of line segment AB meeting AC at Q, such that PQ || BC
By using Basic Proportionality theorem, we obtain,
(P is the midpoint of AB ∴ AP = PB)
⇒ AQ = QC
Or, Q is the mid-point of AC.
8. 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).
Answer:
Given: A triangle ABC, in which P and Q are the mid-points of sides AB and AC respectively..png)
Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and AC respectively.
i.e., AP = PB and AQ = QC
It can be observed that
.png)
and
Therefore,
Hence, by using basic proportionality theorem, we obtain
PQ || BC.
9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO .
Answer:
Given: A trapezium ABCD, in which AB || DC and its diagonals AC and BD intersect each other at O.
Draw a line EF through point O, such that EF || CD
In ΔADC, EO || CD
By using basic proportionality theorem, we obtain
...i
In ΔABD, OE || AB
So, by using basic proportionality theorem, we obtain
⇒
...ii
From eq. (i) and (ii), we get
⇒
10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO Such that ABCD is a trapezium.
Answer:
Given: A quadrilateral ABCD, in which its diagonals AC and BD intersect each other at O such that , i.e.
Quadrilateral ABCD is a trapezium.
Construction: Through O, draw OE || AB meeting AD at E.
In ∆ ADB, we have OE || AB [By construction] By Basic Proportionality theorem
.....i
However, it is given that
.....ii
From eq. (i) and (ii), we get
⇒ EO || DC [By the converse of basic proportionality theorem]
⇒ AB || OE || DC
⇒ AB || CD
∴ ABCD is a trapezium.
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NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.3
1. State which pairs of triangles in figure, are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
Answer:
(i) In ΔABC and ΔPQR, we have ∠A = ∠P = 60° (Given) ∠B = ∠Q = 80° (Given) ∠C = ∠R = 40° (Given) ∴ ΔABC ~ ΔPQR (AAA similarity criterion) (ii) In ΔABC and ΔPQR, we have AB/QR = BC/RP = CA/PQ ∴ ΔABC ~ ΔQRP (SSS similarity criterion) (iii) In ΔLMP and ΔDEF, we have LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6 MP/DE = 2/4 = 1/2 PL/DF = 3/6 = 1/2 LM/EF= 2.7/5 = 27/50 Here, MP/DE = PL/DF ≠ LM/EF Hence, ΔLMP and ΔDEF are not similar. (iv) In ΔMNL and ΔQPR, we have MN/QP = ML/QR = 1/2 ∠M = ∠Q = 70° ∴ ΔMNL ~ ΔQPR (SAS similarity criterion) (v) In ΔABC and ΔDEF, we have AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80° Here, AB/DF = 2.5/5 = 1/2 And, BC/EF = 3/6 = 1/2 ⇒ ∠B ≠ ∠F Hence, ΔABC and ΔDEF are not similar. (vi) In ΔDEF,we have ∠D + ∠E + ∠F = 180° (sum of angles of a triangle) ⇒ 70° + 80° + ∠F = 180° ⇒ ∠F = 180° - 70° - 80° ⇒ ∠F = 30° In PQR, we have ∠P + ∠Q + ∠R = 180 (Sum of angles of Δ) ⇒ ∠P + 80° + 30° = 180° ⇒ ∠P = 180° - 80° -30° ⇒ ∠P = 70° In ΔDEF and ΔPQR, we have ∠D = ∠P = 70° ∠F = ∠Q = 80° ∠F = ∠R = 30° Hence, ΔDEF ~ ΔPQR (AAA similarity criterion) 2. In the figure, ΔODC ∝ ¼ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.
Answer:
DOB is a straight line. ∴ ∠DOC + ∠COB = 180° ⇒ ∠DOC = 180° − 125° = 55° In ΔDOC, ∠DCO + ∠CDO + ∠DOC = 180° (Sum of the measures of the angles of a triangle is 180º.) ⇒ ∠DCO + 70º + 55º = 180° ⇒ ∠DCO = 55° It is given that ΔODC ∼ ΔOBA. ∴ ∠OAB = ∠ OCD [Corresponding angles are equal in similar triangles.] ⇒ ∠OAB = 55°. 3. 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 AO/OC = OB/ODAnswer:
Given: ABCD is a trapezium in which AB DC.
In ΔDOC and ΔBOA,
∠CDO = ∠ABO [Alternate interior angles as AB || CD]
∠DCO = ∠BAO [Alternate interior angles as AB || CD]
∠DOC = ∠BOA [Vertically opposite angles]
∴ ΔDOC ∼ ΔBOA [AAA similarity criterion]
∴
..... [Corresponding sides are proportional]
Hence,
4. In the figure, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.
Answer:
We have,
In ΔPQR, ∠PQR = ∠PRQ
∴ PQ = PR .....(i)
Given,
Using (i), we get,
In ΔPQS & In ΔTQR
∠Q = ∠Q
∴ ΔPQS ~ ΔTQR [SAS similarity criterion]
5. S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.
Answer:
Answer:
It is given that ΔABE ≅ ΔACD. ∴ AB = AC [By cpct] ...(i) And, AD = AE [By cpct] ...(ii) In ΔADE and ΔABC, AD/AB = AE/AC [Dividing equation (ii) by (i)] ∠A = ∠A [Common angle] ∴ ΔADE ~ ΔABC [By SAS similarity criterion] 7. In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that: (i) ΔAEP ~ ΔCDP (ii) ΔABD ~ ΔCBE (iii) ΔAEP ~ ΔADB (iv) ΔPDC ~ ΔBEC
Answer:
(i) In ΔAEP and ΔCDP,
∠AEP = ∠CDP (Each 90°)
∠APE = ∠CPD (Vertically opposite angles)
Hence, by using AA similarity criterion,
ΔAEP ~ ΔCDP
(ii) In ΔABD and ΔCBE,
∠ADB = ∠CEB (Each 90°)
∠ABD = ∠CBE (Common)
Hence, by using AA similarity criterion,
ΔABD ~ ΔCBE
(iii) In ΔAEP and ΔADB,
∠AEP = ∠ADB (Each 90°)
∠PAE = ∠DAB (Common)
Hence, by using AA similarity criterion,
ΔAEP ~ ΔADB
(iv) In ΔPDC and ΔBEC,
∠PDC = ∠BEC (Each 90°)
∠PCD = ∠BCE (Common angle)
Hence, by using AA similarity criterion,
ΔPDC ~ ΔBEC
8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.
Answer:
In ΔABE and ΔCFB,we have,
∠A = ∠C (Opposite angles of a parallelogram)
∠AEB = ∠CBF (Alternate interior angles as AE || BC)
∴By AA-criterion of similarity, we have
∴ ΔABE ~ ΔCFB (By AA similarity criterion)
9. In the figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:
(i) ΔABC ~ ΔAMP
(ii) CA/PA = BC/MP
Answer:
(i) In Δ ABC and AMP, we have, ∠ABC = ∠AMP =90 0 [Given] ∠BAC = ∠MAP [Common angles] ∴ Δ ABC ~ Δ AMP [By AA-criterion of similarity, we have] ⇒ CA/PA = BC/MP ..... (Corresponding sides of similar trianlges are proportional)NCERT Solutions for Class 10 Maths Chapter 6 FAQs
What are the basic properties of a triangle?
A triangle is a closed figure with three sides and three angles. The sum of the angles in a triangle is always 180 degrees.
What is the Pythagorean theorem and how is it used in triangles?
The Pythagorean theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. It is used to find the length of one side of a right triangle if the lengths of the other two sides are known.
What are the different types of triangles based on their sides and angles?
Based on sides, triangles can be classified as equilateral (all sides are equal), isosceles (two sides are equal), or scalene (no sides are equal). Based on angles, triangles can be classified as acute (all angles are less than 90 degrees), obtuse (one angle is greater than 90 degrees), or right-angled (one angle is exactly 90 degrees).
What are the congruence criteria for triangles?
Triangles are said to be congruent if their corresponding sides and angles are equal. The congruence criteria include SSS (side-side-side), SAS (side-angle-side), ASA (angle-side-angle), RHS (right-angle-hypotenuse-side), and AAS (angle-angle-side).
How do we find the area of a triangle?
The area of a triangle can be calculated using the formula: Area = 1/2 * base * height, where the base is the length of the bottom side of the triangle and the height is the perpendicular distance from the base to the opposite vertex.
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