Chapter 6: Triangles — Class 10 Mathematics NCERT Solutions | CBSE
Chapter 6: Triangles – NCERT Class 10 Mathematics Solutions | CBSE
Chapter 6: TRIANGLES
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)
Solution : (i) similar (ii) similar (iii) equilateral (iv) equal , proportional .
2. Given two different examples of pair of : (i) similar figures (ii) non-similar figures .
Solution : (i) Similar figures :
(ii) Non-similar figures :
3. State whether the following quadrilaterals are similar or not :
Fig. 6.8
Solution: In quadrilateral PQRS and ABCD , we get
But, 
Therefore, PQRS and ABCD quadrilaterals are not similar .
EXERCISE 6.2
1. In Fig. 6.17, (i) and (ii) , 
. Find EC in (i) and AD in (ii)
Solution: (i) In 
and , we have
(ii) Here, AD = ? , DB = 7.2 cm , AE = 1.8 cm and EC = 5.4 cm
In and we have,
2.4 cm
2. E and F are points on the sides PQ and PR respectively of a 
. For each of the following cases, state whether 
:
(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: In given figure :
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
We have, 
So, 
and 
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
So, and 
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
and 
So, and .
3. In Fig. 6.18, if 
and 
, prove that 
.
Solution : Given, and
To Prove:
Proof: In
and 
we have , 
Again, 
and 
we have , 
and 
we have , 
Proved.
4. In Fig. 
and 
. Prove that 
.
Solution: In and 
we get, 
In 
and 
we get , 
From and 
we get , Proved.
5. In Fig. 6.20, 
and 
. Show that 
.
Solution: Given, and . Then we show that EF||QR .
In ∆OPQ and 
, we get 
In 
and 
, we get 
From and , we get , 
Thus, Proved.
6. In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that 
and 
. Show that 
.
Solution: A, B and C are points on OP, OQ and OR respectively such that and .
Then we show that .
Proof: In ∆OPQ and AB∥PQ , we get 
In 
and 
, we get 
From 
and
we get, 
Therefore, 
Proved.
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).
Solution: Given, ABC is a triangle and 
. D is a mid-point of the side AB .
To prove : 
Proof: Since, D is a mid-point of the side AB .
In and we get , 
From and
we get, 
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).
Solution: Given, D and E are the mid-point of the sides AB and AC of the triangle ABC respectively.
To prove:
Proof: since, D is a mid-point of the side AB , then
Again, E is a mid-point of the side AC , then
From i 
and (ii) 
we get, 
Therefore, Proved.
9. ABCD is a trapezium in which 
and its diagonal intersect each other at the point O . Show that 
.
Solution: ABCD is a trapezium in which and its diagonal intersect each other at the point O . Then we show that .
Construction: We join OP such that 
Proof: Since, 
and also 
In 
and 
we get, 
In and
we get , 
From and 
we get , 
Proved.
10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that 
⋅ Show that ABCD is a trapezium.
Solution: The diagonals of a quadrilateral ABCD intersect each other at the point O such that ⋅ Then show that ABCD is a trapezium.
Construction: We join OP such that .
Proof: In and we get,
But, 
and we get, 
So, and
Therefore, 
Thus , ABCD is a trapezium.
EXERCISE 6.3
1. State which pairs of triangle in Fig. 6.34 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 :
Solution: (i) In and
, we have
[AAA]
Yes , Angle-angle-angle (AAA similarity criterion) , 
(ii) In 
and 
, we have
, 
and 
∆ABC~∆QRP [SSS]
Yes , side-side-side (similarity criterion) , 
(iii) In 
and
, we have
, 
and 
No , and are not similar .
(iv) In 
and , we have
and 
[SAS]
Yes , side-angle-side (similarity criterion) , [SAS]
(v) In and 
, we have
and are not similar , because the 80° angle in is not positioned between the two known proportional sides.
(vi) Here, 
In 
and , we have
[AAA]
Yes , angle-angle-angle (similarity criterion) , [AAA]
2. In Fig. 6.35 , 
and 
. Find 
and 
.
Solution : Given, 
and 
Since, BD is a straight line .
In 
, we have 
Again, ∆ODC~∆OBA
Therefore, 
, 
and 
.
3. Diagonals AC and BD of a trapezium ABCD with 
intersect each other at the point O . Using a similarity criterion for two triangles , show that 
Solution : Given, Diagonals AC and BD of a trapezium ABCD with intersect each other at the point O . Then we show that .
Proof : In 
and 
, we have
[ Vertically opposite angle]
[ Alternative interior angle]
[ Alternative interior angle]
[ A.A.A. rule ]
Proved .
4. In Fig. 6.36, 
and 
. Show that 
.
Solution : Given, 
So, 
PQR is an isosceles triangle .
Again, 
[from (i) ]
In 
, we have
[Common angle]
[ given]
[S.A.S]
5. S and T are points on sides PR and QR of ∆PQR such that ∠P=∠RTS . Show that ∆RPQ~∆RTS .
Solution: Given, S and T are points on sides PR and QR of ∆PQR such that 
.
Then we show that ∆RPQ~∆RTS .
Proof : In 
and ∆RTS , we have
[Given]
[ Common angle]
[Third angle ]
[ A.A.A.] Proved.
6. In Fig. 6.37 , if 
, show that 
.
Solution : Given, 
.
Then we show that : 
.
Proof : Since,
and 
In 
, we have 
and 
[Common Angle]
[SAS] Proved
7. In Fig. 6.38, altitudes AD and CE of intersect each other at the point P . Show that :
(i) ∆AEP~∆CDP (ii) 
(iii) 
(iv) 
Solution : Given, altitudes AD and CE of intersect each other at the point P . Then we show that :
(i) ∆AEP~∆CDP (ii) (iii) (iv)
Proof: (i) In 
and 
, we have
[ Vertically opposite angle]
[Third angle]
[A.A.A. rule]
(ii) In 
and 
, we have
[Common angle]
[Third angle]
[A.A.A. Rule]
(iii) In and , we have
[Common angle]
[Third angle]
[AAA rule]
(iv) In 
and 
, we have
[Common angle]
[Third angle]
[A.A.A. rule]
8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F . Show that 
.
Solution : Given, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F .
To prove : .
Proof : In and 
, we have
[ Opposite angle of the parallelogram ]
[ Alternative interior angle]
[A.A rule] proved.
9. In Fig. 6.39 , ABC and AMP are two right triangles , right angled at B and M respectively . Prove that : (i) 
(ii) 
Solution: Given , ABC and AMP are two right triangles , right angled at B and M respectively .
To prove : (i) 
(ii) 
Proof : (i) In ∆ and 
,we have
[Common angle]
[Third angle]
[A.A.A. rule]
(ii) Since , [A.A.A. rule]
10. CD and GH are respectively the bisectors of 
and 
such that D and H lie on sides AB and FE of and 
respectively . If 
, show that :
(i) 
(ii) 
(iii) 
Solution : Given, CD and GH are respectively the bisectors of and such that D and H lie on sides AB and FE of and respectively and . Then we show that :
(i) (ii) (iii)
Proof : (i) Since, CD and GH are the bisectors of and respectively .
and 
In 
and 
we have
[ 
]
[ 
]
[A.A.]
Proved.
(ii) 
Proof: In 
and 
, we have
[ 
]
[ 
]
[Third angle]
[AAA similarity criterion ]
(iii) In
and 
, we have
[ ]
[ ]
[ Third angle ]
[AAA Similarity criterion]
11. In Fig. 6.40 , E is a point on side CB produced of an isosceles triangle ABC with 
. If 
and
, prove that 
.
Solution: Given, E is a point on side CB produced of an isosceles triangle ABC with . If and .
To prove that .
Proof : In , we have
i. e. 
In and 
, we have
[ Given]
[Third angle]
[ AAA rule ] Proved.
12. 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 Fig. 6.41) . Show that ∆ABC~∆PQR .
Solution : Given , Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆PQR . Then we show that ∆ABC~∆PQR .
Proof: Since, AD and PM be the median of the ABC and PQR respectively .
We have, 
In and 
, we have
[SSS]
So, 
In and ∆PQR , we have
[SAS] Proved.
13. D is a point on the sides BC of a triangle ABC such that 
. Show that 
.
Solution : Given, D is a point on the sides BC of a triangle ABC such that . Then we show that .
Proof: In and 
, we have
[Given]
[common angle]
[third angle]
[A.A.A.]
Proved.
14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR . Show that ∆ABC~∆PQR .
Solution: Given, Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR .
Then we have to show that ∆ABC~∆PQR .
Construction : We draw 
and 
.
Proof : Since , and , then
and 
[Using mid-point theorem]
Also , X and Y is the mid-point of the sides AB and QR , then
and 
We have , 
In 
and 
, We have
[SSS]
[CPCT] ………………… (i)
Similarly, we show that 
…………………. (ii)
Adding (i) and (ii) , we get 
In and , we have,
[SAS] Proved .
15. 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 .
Solution : Let and are the two triangle .
In given figure,
Here, AB = 6 m , BC = 4 m , PQ = ? and QR = 28 cm
In and , we have
[Same inclination]
[Third angle]
[A.A.A.]
Therefore, the height of the tower is 42 m .
16. If AD and PM are medians of triangle ABC and PQR , respectively where , prove that 
Solution: Given, AD and PM are medians of triangle ABC and PQR respectively and .
To prove :
Proof : Since , D and M are mid-point of the sides BC and QR .
So, 
and 
Given,
Then, 
In and , we have
[ 
]
[S.A.S.]
Proved.
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