CBSE Class 10 Maths Chapter-6 Triangles – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 10 Maths Chapter 6 – Triangles prepared by expert Maths teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 10 Maths Chapter-6 Triangles Important Questions
CBSE Class 10 Maths Important Questions Chapter 6 – Triangles
3 Mark Questions
1. In the given figure, 
and 
Prove that
Ans. Since 
Since 
[In 
sides opposites to opposite angles are equal]
Now in 
and 
From (iii), 
And 
[Common]
[By S.A.S. Rule of similarity]
2. In the given figure, PA, QB and RC are each perpendicular to AC. Prove that 
Ans. In 
and 
Similarly,
Adding (i) and (ii), we get
3. In the given figure, DE||BC and AD:DB = 5:4, find 
Ans. In 
and 
[Common]
[Corresponding
]
[By A.A Rule]
Again in 
and 
[Alternate ]
[Vertically opposite ]
[By A.A Rule]
[From (i)]
4. Determine the length of an altitude of an equilateral triangle of side ‘2a’ cm.
Ans. In right triangles 
and 
In right 
(By Pythagoras Theorem)
5. In the given figure, if 
and 
Then prove that 
Ans. Since 
(in )
Again [given in
]
Thus, in 
and 
(Common)
6. In the given figure the line segment XY||AC and XY divides triangular region ABC into two points equal in area, Determine 
Ans. Since 
[Corresponding angles]
[A.A. similarity]
But ar
7. BL and CM are medians of 
right angled at A. Prove that 
Ans. BL and CM are medians of a in which 
From 
From right angled 
i.e. ,
From right-angled
i.e. 
[Mis mid-point]
Adding (ii) and (iii), we get
i.e. 
[From (i)]
8. ABC is a right triangle right angled at C. Let BC = a, CA = b, AB = c and let p be the length of perpendicular from C on AB, prove that
(i) cp = ab
(ii) 
Ans. (i) Draw 
Then,
Now ar of 
Also area of 
Then,
(ii) Since is a right-angled triangle with 
Thus
9. In figure, a triangle ABC is right-angled at B. side BC is trisected at points D and E, prove that 8
Ans. Given:is right-angled at B. Side BC is trisected at D and E.
To Prove: 
Proof: D and E are the paints of trisection of BC
and
In right-angled triangle ABD
[Using Pythagoras theorem]
In 
In
From (ii) and (iii), we have
From (iii) and (iv), we have
From (v) and (vi), we get
10. In figure, DEFG is a square and 
, show that 
Ans. Given: is right-angled at A and DEFG is a square
To Prove: 
Proof: Let 
Then,
Also 
is right-angled at D.
From (i) and (ii), we get
Consider 
[
is square]
[From (iii)]
[By AA similarity]
But 
[
side of a square]
11. In a quadrilateral ABCD, P,Q,R,S are the mid-points of the sides AB, BC, CD and DA respectively. Prove that PQRS is a parallelogram.
Ans. To Prove: PQRS is a parallelogram
Construction: Join AC
Proof: In 
[S and R are mid-points of AD and DC]
[by converse of B.P.T]
In 
[P and Q are mid points of AB and BC]
[By converse of B.P.T]
From (i) and (ii), we get
Similarly, join B to D and PS||QR
is a parallelogram.
12. Triangle ABC is right-angled at C and CD is perpendicular to AB, prove that 
Ans. Given: A right angled at C and
To Prove: 
Proof: Consider 
Let 
Then 
[
is right angled]
[
is right angled]
In 
[By AA similarity]
13. Triangle ABC is right angled at C and CD is perpendicular to AB. Prove that
Ans. Given: right-angled at C and
To prove:
Proof: Consider 
and 
Let
Then 
In
14. In figure, ABC and DBC are two triangles on the same base BC. If AD intersect EC at O, prove that 
Ans. Given: ABC and DBC are two triangles on the same base BC but on the opposite sides of BC, AD intersects BC at O.
Construction: Draw 
and 
To prove: 
Proof: In 
15. In figure, ABC is a right triangle right-angled at B. Medians AD and CE are of respective lengths 5 cm and 
, find length of AC.
Ans. Given: with 
, AD and CE are medians
To find: Length of AC
Proof: In 
right-angled at B,
In 
right-angled at B
Given that AD = 5 and 
16. In the given figure, 
and , show that
Ans. Given: 
Proof: As
Also 
In 
and TQR, we have
Also 
17. Given a triangle ABC. O is any point inside the triangle ABC, X,Y,Z are points on OA, OB and OC, such that XY||AB and XZ||AC, show that YZ||AC.
Ans. Given: A 
is a point inside 
are points on OA, OB and OC respectively such that XY||AB and XZ||AB and XZ||AC
To show: YZ||BC
Proof: In 
In 
From (i) and (ii), we get 
Now in 
[Converse of B.P.T]
18. PQR is a right triangle right angled at Q. If QS = SR, show that 
Ans. Given: PQR is a right Triangle, right-angled at Q
Also QS = SR
To prove:
Proof: In right-angled triangle PQR right angled at Q.
[By Pythagoras theorem]
Also 
In right-angled triangle PQS, right angled at Q.
From (i) and (iii), we get
19. A ladder reaches a window which is 12 m above the ground on one side of the street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 9 m high. Find the width of the street if the length of the ladder is 15 m.
Ans. Let AB be the width of the street and C be the foot of ladder.
Let D and E be the windows at heights 12m and 9m respectively from the ground.
In 
, right angled at A, we have
In 
right angled at B, we have
Hence, width of the street AB=AC+BC=9+12=21m
20. In figure,
, if the area of 
is 
then find the area of the quadrilateral PYZQ.
Ans. Given 
[By converse of B.P.T]
In 
and 
we have
[
[From (i) corresponding angles]
[common]
[By AA similarity]
We have 
Substituting in (i), we get
Area of quadrilateral
