NCERT Math Solutions for Class 10 Maths Chapter 6 Triangle(English Medium) || CBSE Class 10 Maths Chapter 6 Triangle Ex 6.2

Explore the NCERT Solutions for Class 10 Maths Chapter 6: Triangle (English Medium), with detailed explanations for Exercise 6.2. This resource is designed to simplify complex Triangle, helping Class 10 Maths students grasp the subject effortlessly. For any queries regarding NCERT Solutions for Class 10 Maths Chapter 6 Triangle Ex 6.2, feel free to leave a comment, and we’ll respond as soon as possible. || Class 10 Maths Chapter 6 Triangle Ex 6.2 || NCERT Solutions for Class 10 Maths Chapter 6: Triangle (English Medium)

NCERT Math Solutions for Class 10 Maths Chapter 6 Triangle(English Medium) || CBSE Class 10 Maths Chapter 6 Triangle Ex 6.2
Exercise - 6.2

NCERT Math Solutions for Class 10 Maths Chapter 6 Triangle(English Medium) || CBSE Class 10 Maths Chapter 6 Triangle Ex 6.2

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1. In Fig. 6.17, (i) and (ii), $ \mathrm{DE} \| \mathrm{BC} $. Find EC in (i) and AD in (ii)

Answer

(i) Let us take $ \mathrm{EC}=\mathrm{x} $ cm
Given: DE || BC
Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.
Now, using basic proportionality theorem, we get:
$\frac{A D}{D B}=\frac{A E}{E C}$
$\frac{1.5}{3}=\frac{1}{x}$
$X=\frac{3 \times 1}{1.5} $
$ \mathrm{x}=2 \mathrm{~cm} $
Hence, $ \mathrm{EC}=2 \mathrm{~cm} $
(ii) Let us take $ \mathrm{AD}=\mathrm{x} \mathrm{~cm} $
Given: $ \mathrm{DE} \| \mathrm{BC} $
Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.
Now, using basic proportionality theorem, we get
$ \frac{A D}{D B}=\frac{A E}{E C} $
$ \frac{x}{7.2}=\frac{1.8}{5.4} $
$ x=\frac{1.8 \times 7.2}{5.4} $
Hence, $ \mathrm{AD}=2.4 \mathrm{~cm} $

NCERT Math Solutions for Class 10 Maths Chapter 6 Triangle(English Medium) || CBSE Class 10 Maths Chapter 6 Triangle Ex 6.2

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2. E and F are points on the sides PQ and PR respectively of a $ \Delta $PQR . For each of the following cases, state whether EF$ \| \mathrm{QR} $ :

(i) $ \mathrm{PE}=3.9 \mathrm{~cm}, \mathrm{EQ}=3 \mathrm{~cm}, \mathrm{PF}=3.6 \mathrm{~cm} $ and $ \mathrm{FR}=2.4 \mathrm{~cm} $

Answer

Given :
$\mathrm{PE}=3.9 \mathrm{~cm}$
$\mathrm{EQ}=3 \mathrm{~cm}$
$\mathrm{PF}=3.6 \mathrm{~cm}$
$\mathrm{FR}=2.4 \mathrm{~cm} $
Now we know,
Triangle Proportionality Theorem: If a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides these two sides proportionally.
So, if the lines EF and QR are to be parallel, then ratio $\mathrm{PE}:\mathrm{EQ}$ should be proportional to $\mathrm{PF}:\mathrm{PR}$
$\frac{P E}{E Q}=\frac{3.9}{3}=1.3$
$\frac{P F}{F R}=\frac{3.6}{2.4}=1.5 $
Hence,
$ \frac{P E}{E Q} \neq \frac{P F}{F R} $
Therefore, EF is not parallel to QR

(ii) $ \mathrm{PE}=4 \mathrm{~cm}, \mathrm{QE}=4.5 \mathrm{~cm}, \mathrm{PF}=8 \mathrm{~cm} $ and $ \mathrm{RF}=9 \mathrm{~cm} $

Answer

We know that, Triangle Proportionality Theorem: If a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides these two sides proportionally.
So, if the lines EF and QR are to be parallel, then ratio $ \mathrm{PE}: \mathrm{EQ} $ should be proportional to $ \mathrm{PF}: \mathrm{PR} $
$\frac{P E}{E Q}=\frac{4}{4.5}=\frac{8}{9}$
$\frac{P F}{F R}=\frac{8}{9} $
Hence, $ \frac{P E}{E Q}=\frac{P F}{F R} $
Therefore, EF is parallel to QR

NCERT Math Solutions for Class 10 Maths Chapter 6 Triangle(English Medium) || CBSE Class 10 Maths Chapter 6 Triangle Ex 6.2

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(iii) $ \mathrm{PQ}=1.28 \mathrm{~cm}, \mathrm{PR}=2.56 \mathrm{~cm}, \mathrm{PE}=0.18 \mathrm{~cm} $ and $ \mathrm{PF}=0.36 \mathrm{~cm} $

Answer

In this we know that,
Triangle Proportionality Theorem: If a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides these two sides proportionally.
So, if the lines EF and QR are to be parallel, then ratio $\mathrm{PE}:{EQ}$ should be proportional to $\mathrm{PF}:{PR}$
$\frac{P E}{P Q}=\frac{0.18}{1.28}=\frac{8}{128}=\frac{9}{64}$
$\frac{P F}{P R}=\frac{0.36}{2.56}=\frac{9}{64} $
Hence,
$ \frac{P E}{P Q}=\frac{P F}{P R} $
EF is parallel to QR

3. In Fig. 6.18, if $ \mathrm{LM} \| \mathrm{CB} $ and $ \mathrm{LN} \| \mathrm{CD} $, prove that $ \frac{A M}{A B}=\frac{A N}{A D} $

Answer

To Prove: $ \frac{A M}{A B}=\frac{A N}{A D} $
Given: $LM || CB$ and $LN || CD$ From the given figure: In $ \triangle \mathrm{ALM} $ and $ \Delta \mathrm{ABC} $
$ \mathrm{LM} \| \mathrm{CB} $
Proportionality theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion
Now, using basic proportionality theorem that the corresponding sides will have same proportional lengths, we get:
$ \frac{A M}{A B}=\frac{A L}{A C} \quad\ldots \text{(i)}$
Similarly, LN parallel to CD
Therefore,
$ \frac{A N}{A D}=\frac{A L}{A C}\quad\ldots \text{(ii)} $
From (i) and (ii), we obtain
$ \frac{A M}{A B}=\frac{A N}{A D} $
Hence, proved.

NCERT Math Solutions for Class 10 Maths Chapter 6 Triangle(English Medium) || CBSE Class 10 Maths Chapter 6 Triangle Ex 6.2

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4. In Fig. 6.19, DE $ \| \mathrm{AC} $ and $ \mathrm{DF} \| \mathrm{AE} $. Prove that

$\frac{B F}{F E}=\frac{B E}{E C}$

Answer

To Prove $ \frac{B F}{F E}=\frac{B E}{E C} $
Given:
Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion. In triangle $ \mathrm{ABC}, \mathrm{DE} $ is parallel to AC
Therefore,
By Basic proportionality theorem
$ \frac{B F}{F E}=\frac{B E}{E C} \quad\ldots \text{(1)}$
In triangle $ \mathrm{BAE}, \mathrm{DF} $ is parallel to AE
In triangle $ \mathrm{BAE}, \mathrm{DF} $ is parallel to AE
Therefore, By Basic proportionality theorem
$ \frac{B D}{D A}=\frac{B F}{F E}\quad\ldots \text{(2)} $
From (1) and (2), we get
$ \frac{B E}{E C}=\frac{B F}{F E} $
Hence, Proved.

5. In Fig. 6.20, $DE \| OQ$ and $ \mathrm{DF} \| OR$. Show that $ E F \| $ QR.

Answer

To Prove: $EF \| QR$
Given: In triangle POQ, DE parallel to OQ Proof:
In triangle $ \mathrm{POQ}, \mathrm{DE} $ parallel to OQ
Hence,
Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.
$ \frac{P E}{E q}=\frac{P D}{D O} \quad $ (Basic proportionality theorem) (i)
Now,
In triangle POR, DF parallel OR
Hence,
$ \frac{P F}{F R}=\frac{P D}{D O} \quad $ (Basic proportionality theorem) (ii)
From (i) and (ii), we get
$ \frac{P E}{E Q}=\frac{P F}{F R} $
Therefore,
EF is parallel to QR (Converse of basic proportionality theorem)
Hence, Proved.

NCERT Math Solutions for Class 10 Maths Chapter 6 Triangle(English Medium) || CBSE Class 10 Maths Chapter 6 Triangle Ex 6.2

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6. In Fig. 6.21, A, B and C are points on $ \mathrm{OP}, \mathrm{OQ} $ and OR respectively such that $ A B \| P Q $ and $ A C \| P R $. Show that $ B C \| Q R $.

Answer

To Prove: $BC || QR$
Given that in triangle $ \mathrm{POQ}, \mathrm{AB} $ parallel to PQ
Hence,
Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.
$ \frac{O A}{A P}=\frac{O B}{B Q} \ ( $Basic proportionality theorem $ ) $
Now,
Therefore,
Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.
Using Basic proportionality theorem, we get:
$ \frac{O A}{A P}=\frac{O C}{C R} $
From above equations, we get
$ \frac{O B}{B Q}=\frac{O C}{C R} $
BC is parallel to QR (By the converse of Basic proportionality theorem)
Hence, Proved.

7. Using Theorem 6.1, prove that a line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX)

Answer

Consider the given figure in which PQ is a line segment drawn through the mid-point $ P $ of line $ A B $, such that $ P Q $ is parallel to BC.

To Prove: PQ bisects AC
Given: $ P Q || \mathrm{BC} $ and PQ bisects AB
Proof:
According to Theorem 6.1 : If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion. Now, using basic proportionality theorem, we get
$\frac{A Q}{Q C}=\frac{A P}{P B}$
$\frac{A Q}{Q C}=\frac{1}{1} $
[As $ \mathrm{AP}=\mathrm{PB} \operatorname{coz} \mathrm{P} $ is the mid-point of AB ]
Hence,
$ \mathrm{AQ}=\mathrm{QC} $
Or, Q is the mid-point of AC
Hence proved.

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

To Prove: $ P Q \| B C $
Given: $ P $ and $ Q $ are midpoints of $ A B $ and $ A C $
Proof:
Let us take the given figure in which PQ is a line segment which joins the mid-points P and Q of line AB and AC respectively
i.e., $ \mathrm{AP}=\mathrm{PB} $ and $ \mathrm{AQ}=\mathrm{QC} $
We observe that,
$ \frac{A P}{P B}=\frac{1}{1} $
And,
$ \frac{A Q}{Q C}=\frac{1}{1} $
Therefore,
$ \frac{A P}{P B}=\frac{A Q}{Q C} $
Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.
Hence, using basic proportionality theorem we get:
PQ parallel to BC
Hence, Proved.

NCERT Math Solutions for Class 10 Maths Chapter 6 Triangle(English Medium) || CBSE Class 10 Maths Chapter 6 Triangle Ex 6.2

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9. ABCD is a trapezium in which $ \mathrm{AB} \| \mathrm{DC} $ and its diagonals intersect each other at the point O. Show that $ \frac{A O}{B O}=\frac{C O}{D O} $.

Answer

The figure is given below:

Given: $ A B C D $ is a trapezium
$ \mathrm{AB} \| \mathrm{CD} $
Diagonals intersect at O
To Prove $ =\frac{A O}{B O}=\frac{C O}{D O} $
Construction: Construct a line EF through point O, such that EF is parallel to CD.

Proof:
In $ \triangle \mathrm{ADC}, \mathrm{EO} $ is parallel to CD
According to basic proportionality theorem, if a side is drawn parallel to any side of the triangle then the corresponding sides formed are proportional.
Now, using basic proportionality theorem in $ \triangle \mathrm{ABD} $ and $ \triangle \mathrm{ADC} $, we obtain
$ \frac{A E}{E D}=\frac{A O}{O C} \quad\ldots \text{(i)}$
In $ \triangle \mathrm{ABD}, \mathrm{OE} $ is parallel to AB
So, using basic proportionality theorem in $ \triangle \mathrm{EOD} $ and $ \triangle \mathrm{ABD} $, we get
$ \frac{E D}{A E}=\frac{O D}{B O} $
$ \frac{A E}{E D}=\frac{B O}{O D}\quad\ldots \text{(ii)} $
From (i) and (ii), we get
$ \frac{A O}{O C}=\frac{B O}{O D} $
Therefore by cross multiplying we get,
$ \frac{A O}{B O}=\frac{O C}{O D} $
Hence, Proved.

10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that $ \frac{A O}{B O}=\frac{C O}{D O} $. Show that ABCD is a trapezium

Answer

The quadrilateral ABCD is shown below, BD and AC are the diagonals.

Construction: Draw a line OE parallel to AB
Given: In $ \triangle \mathrm{ABD}, \mathrm{OE} $ is parallel to AB
To prove: $ A B C D $ is a trapezium
According to basic proportionality theorem, if in a triangle another line is drawn parallel to any side of triangle, then the sides so obtain are proportional to each other.
Now, using basic proportionality theorem in $ \triangle \mathrm{DOE} $ and $ \triangle \mathrm{ABD} $, we obtain
$ \frac{A E}{E D}=\frac{B O}{O D} \quad\ldots \text{(i)}$
It is given that,
$ \frac{A O}{O C}=\frac{O B}{O D} \quad\ldots \text{(ii)}$
From (i) and (ii), we get
$ \frac{A E}{E D}=\frac{A O}{O C} \quad\ldots \text{(iii)}$
Now for $ A B C D $ to be a trapezium $ A B $ has to be parallel of $ C D $
Now From the figure we can see that If eq(iii) exists then,
$EO || DC$ (By the converse of basic proportionality theorem)
Now if,
$ \Rightarrow \mathrm{AB}\|\mathrm{OE}\| \mathrm{DC} $
Then it is clear that
$ \Rightarrow \mathrm{AB} \| \mathrm{CD} $
Thus the opposite sides are parallel and therefore it is a trapezium.
Hence,
ABCD is a trapezium.

NCERT Math Solutions for Class 10 Maths Chapter 6 Triangle(English Medium) || CBSE Class 10 Maths Chapter 6 Triangle Ex 6.2

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NCERT Math Solutions for Class 10 Maths Chapter 6 Triangle(English Medium) || CBSE Class 10 Maths Chapter 6 Triangle Ex 6.2

Download the Math Ninja App Now