Chapter 6 : Triangles – Chapter-Wise Important Questions and Answers | Class 10 Mathematics CBSE Solutions
CBSE Class 10 Mathematics Chapter 6 : Triangles Important Questions with Solutions and Answers
Chapter 6 : TRIANGLES
SECTION = A
Question: DE is drawn parallel to the base BC of a 
ABC , meeting AB at D and AC at E . If 
and CE = 2 cm , then AE is :
(a) 5 cm (b) 4 cm (c) 6 cm (d) 7 cm
Solution: (c) 6 cm
[ We have, 
In ABC and DE
BC, then
]
Question: In figure , DE
BC ,then EC is equal to :
(a) 2 cm (b) 3 cm (c) 5 cm (d) 6 cm
Solution: (a) 2 cm
[ Here, 
In 
and 
, We have
]
Question: If ∆ABC~∆RPQ , AB = 3 cm , BC = 5 cm , AC = 6 cm , RP =6 cm and PQ = 10 cm , then RQ is :
(a) 6 cm (b) 12 cm (c) 10 cm (d) 3 cm
Solution: (b) 12 cm
[ Since, ∆ABC~∆RPQ , we have
So, 
]
Question: In DEW , ABEW . If AD = 4 cm , DE = 12 cm and DW =24 cm , then the value of DB is :
(a) 12 cm (b) 24 cm (c) 8 cm (d) 4 cm
Solution: (c) 8 cm
[ In DEW and ABEW, We have
cm ]
Question: In given figure , MNAB , AB = 7.5 cm , AM = 4 cm and MC = 2 cm , then the length of BN is :
(a) 5 cm (b) 4 cm (c) 2 cm (d) 8 cm
Solution: (a) 5 cm
[ In 
and
, then
]
Question: In the figure , D and E are points on AB and AC respectively such that DEBC . If 
and AE = 4.5 cm , then AC is equal to
(a) 17.5 cm (b) 18 cm (c) 17 cm (d) 16.5 cm
Solution: (c) 17 cm
[ Given, 
Since, DE∥BC , then 
]
Fill in the blank
Question: All circles are 
. [ congruent / similar]
Solution: Similar .
Question: All squares are 
. [ similar / congruent ]
Solution: Similar .
Question: All triangles are similar . [ isosceles / equilateral / acute triangle ]
Solution: Equilateral .
Question: Two polygons of the same number of sides are similar , if (a) their corresponding angles are
and (b) their corresponding sides are 
. [congruent / equal / proportional /Similar ]
Solution: Equal , Proportional .
SECTION = B
Question: In figure, 
. Prove that PQR is an isosceles triangle.
Solution: Given, 
To Prove: PQR is an isosceles triangle.
Proof : Since, 
So, 
and also, 
[Corresponding angles]
Given , 
So, 
Thus, PQR is an isosceles triangle. Prove.
Question: In figure , 
. Show that 
and 
.
Solution: Since , 
In 
and 
, we have
[ Vertical opposite angles ]
[ S.A.S ]
Therefore , 
and Proved .
Question: In given figure , AB∥DE and BC∥EF . Prove that AC∥DF .
Solution: In 
and AB∥DE , We have
In 
and BC∥EF , We have
From 
and 
, we get 
AC∥DF proved .
Question: 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]
Question: 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.
Question: In Fig. 6.37 , if 
, show that 
.
Solution : Given, 
.
Then we show that : 
.
Proof : Since,
and 
In 
, we have 
and 
[Common Angle]
[SAS] Proved
Section = C
Question: 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.
Question: 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.
Question: 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.
Question: 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.
SECTION = D
Question: ABCD is a trapezium with AB∥DC . E and F are points on non-parallel sides AD and BC respectively such that EF is parallel to AB . Show that
Solution: Given, ABCD is a trapezium with AB∥DC . E and F are points on non-parallel sides AD and BC respectively such that EF∥AB .
To Prove : 
Construction : Join AC to intersect EF at G .
Proof : AB∥DC and EF∥AB and Also EF∥DC
In ∆ADC and EG∥DC [
EF∥DC ]
In ∆ACB and GF∥AB [ GF∥AB ]
From (i) and (ii) , we get Proved.
SECTION = E
Question: Prove that a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio .
Solution: Given, a triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively .
To prove : 
.
Construction : Join BE and CD and also , draw 
and 
.
Proof : We know that , Area of triangle 
× Base × Height
and 
and 
Since, 
and 
are on the same base DE and between the same parallels BC and DE.
So, 
From 
, 
and 
, we have
Proved.
Question: 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.
Question: 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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