NCERT solutions for class 10 maths chapter 6 Triangles: We have studied about triangles and properties of the triangles in the previous classes. Here we are going to study similar triangles and their properties. Solutions of NCERT for class 10 maths chapter 6 Triangles is covering a detailed explanation to each and every question available in the practice exercises. This chapter starts by introducing the difference between similar and congruent figures. If two figures are the same in shape and size, the figures are said to be congruent. Two figures that have the same shape but not necessarily the same size are called similar figures. For example, all circles are similar but only those circles with the same radii are congruent which means all congruent figures are similar but converse may not be true. In this chapter of CBSE NCERT solutions for class 10 maths chapter 6 Triangles, you will get the solutions to the problems based on the similarity of triangles. The concept of similarity of triangles can be used to measure the height of objects. Below is a question solved in NCERT solutions for class 10 maths chapter 6 Triangles. This question can be solved using the concept of similarity of triangles.
Question: A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.
Answer:
Given:-
length of lamp post (AB) = 3.6m
Height of the girl = 90cm or 0.9m
Speed of the girl = 1.2m/sec
To find:- the length of the shadow DE.
Solution:- the girl walked BD distance in 4 seconds.
The distance traveled girl = length of BD = 4 x 1.2 => 4.8m
In Triangle ABE and Triangle CDE
Thus Triangle ABE and Triangle CDE are similar triangles.
From the theorem:- If two triangles are similar then the ratio of their sides is equal.
Hence the length of the shadow is 1.6m.
These types of examples are mentioned in this chapter. Class wise and chapter-wise NCERT solutions can be downloaded by clicking on the above-given link.
Similarity of triangles
Theorems based on similar triangles
Areas of similar triangles
Theorems related to Trapezium
Pythagoras theorem
All circles are similar.
Since all the circles have a similar shape. They may have different radius but the shape of all circles is the same.
Therefore, all circles are similar.
All squares are similar.
Since all the squares have a similar shape. They may have a different side but the shape of all square is the same.
Therefore, all squares are similar.
All equilateral triangles are similar.
Since all the equilateral triangles have a similar shape. They may have different sides but the shape of all equilateral triangles is the same.
Therefore, all equilateral triangles are similar.
Two polygons of the same number of sides are similar if their corresponding angles are equal and their corresponding sides are proportional.
Thus, (a) equal
(b) proportional
Q2 (1) Give two different examples of a pair of similar figures.
The two different examples of a pair of similar figures are :
1. Two circles with different radii.
2. Two rectangles with different breadth and length.
Q2 (2) Give two different examples of a pair of non-similar figures.
The two different examples of a pair of non-similar figures are :
1.Rectangle and circle
2. A circle and a triangle.
Q3 State whether the following quadrilaterals are similar or not:
Quadrilateral PQRS and ABCD are not similar as their corresponding sides are proportional i.e. but their corresponding angles are not equal.
Q1 In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
(i)
Let EC be x
Given: DE || BC
By using the proportionality theorem, we get
(ii)
Let AD be x
Given: DE || BC
By using the proportionality theorem, we get
(i)
Given :
PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
and
We have
Hence, EF is not parallel to QR.
(ii)
Given :
PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
and
We have
Hence, EF is parallel to QR.
Q2 (3) E and F are points on the sides PQ and PR respectively of a triangle PQR. For each of the following cases, state whether EF || QR : PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
(iii)
Given :
PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
and
We have
Hence, EF is parallel to QR.
Q3 In Fig. 6.18, if LM || CB and LN || CD, prove that
Answer:
Given : LM || CB and LN || CD
To prove :
Since , LM || CB so we have
Also, LN || CD
From equation 1 and 2, we have
Hence proved.
Q4 In Fig. 6.19, DE || AC and DF || AE. Prove that BF / FE = BE / EC
Given : DE || AC and DF || AE.
To prove :
Since , DE || AC so we have
Also,DF || AE
From equation 1 and 2, we have
Hence proved.
Q5 In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.
Given : DE || OQ and DF || OR.
To prove EF || QR.
Since DE || OQ so we have
Also, DF || OR
From equation 1 and 2, we have
Thus, EF || QR. (converse of basic proportionality theorem)
Hence proved.
Given : AB || PQ and AC || PR
To prove: BC || QR
Since, AB || PQ so we have
Also, AC || PR
From equation 1 and 2, we have
Therefore, BC || QR. (converse basic proportionality theorem)
Hence proved.
Let PQ is a line passing through the midpoint of line AB and parallel to line BC intersecting line AC at point Q.
i.e. and .
Using basic proportionality theorem, we have
Since
Q is the midpoint of AC.
Let P is the midpoint of line AB and Q is the midpoint of line AC.
PQ is the line joining midpoints P and Q of line AB and AC, respectively.
i.e. and .
we have,
From equation 1 and 2, we get
By basic proportionality theorem, we have
Draw a line EF passing through point O such that
To prove :
In , we have
So, by using basic proportionality theorem,
In , we have
So, by using basic proportionality theorem,
Using equation 1 and 2, we get
Hence proved.
Draw a line EF passing through point O such that
Given :
In , we have
So, by using basic proportionality theorem,
However, its is given that
Using equation 1 and 2 , we get
(By basic proportionality theorem)
Therefore, ABCD is a trapezium.
(i)
(By AAA)
So ,
(ii) As corresponding sides of both triangles are proportional.
(By SSS)
(iii) Given triangles are not similar because corresponding sides are not proportional.
(iv) by SAS similarity criteria.
(v) Given triangles are not similar because the corresponding angle is not contained by two corresponding sides
(vi) In , we know that
In , we know that
( By AAA)
Given : , and
(DOB is a straight line)
In
Since , , so
( Corresponding angles are equal in similar triangles).
In , we have
( Alternate interior angles as )
( Alternate interior angles as )
( Vertically opposite angles are equal)
( By AAA)
( corresponding sides are equal)
Hence proved.
Q4 In Fig. 6.36, and . Show that
Given : and
To prove :
In ,
(Given)
In ,
(Common)
( By SAS)
Q5 S and T are points on sides PR and QR of PQR such that P = RTS. Show that RPQ ~ RTS.
Given : P = RTS
To prove RPQ ~ RTS.
In RPQ and RTS,
P = RTS (Given )
R = R (common)
RPQ ~ RTS. ( By AA)
Q6 In Fig. 6.37, if ABE ACD, show that ADE ~ ABC.
Answer:
Given :
To prove ADE ~ ABC.
Since
( By CPCT)
(By CPCT)
In ADE and ABC,
( Common)
and
( and )
Therefore, ADE ~ ABC. ( By SAS criteria)
Q7 (1) In Fig. 6.38, altitudes AD and CE of intersect each other at the point P. Show that:
To prove :
In ,
( Both angles are right angle)
(Vertically opposite angles )
( By AA criterion)
Q7 (2) In Fig. 6.38, altitudes AD and CE of intersect each other at the point P. Show that:
To prove :
In ,
( Both angles are right angle)
(Common )
( By AA criterion)
Q7 (3) In Fig. 6.38, altitudes AD and CE of intersect each other at the point P. Show that:
To prove :
In ,
( Both angles are right angle)
(Common )
( By AA criterion)
Q7 (4) In Fig. 6.38, altitudes AD and CE of intersect each other at the point P. Show that:
To prove :
In ,
( Both angles are right angle)
(Common )
( By AA criterion)
Q8 E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that
To prove :
In ,
( Opposite angles of a parallelogram are equal)
( Alternate angles of AE||BC)
( By AA criterion )
Q9 (1) In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
To prove :
In ,
( Each )
( common)
( By AA criterion )
Q9 In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that
To prove :
In ,
( Each )
( common)
( By AA criterion )
( corresponding parts of similar triangles )
Hence proved.
Q10 (1) CD and GH are respectively the bisectors of and such that D and H lie on sides AB and FE of respectively. If , show that:
To prove :
Given :
( CD and GH are bisectors of equal angles)
( CD and GH are bisectors of equal angles)
In
( proved above)
( proved above)
( By AA criterion)
Hence proved.
Q 10 (2) CD and GH are respectively the bisectors of such that D and H lie on sides AB and FE of respectively. If , show that:
To prove :
Given :
In ,
( CD and GH are bisectors of equal angles)
( )
( By AA criterion )
Q10 (3) CD and GH are respectively the bisectors of such that D and H lie on sides AB and FE of respectively. If , show that:
To prove :
Given :
In ,
( CD and GH are bisectors of equal angles)
( )
( By AA criterion )
To prove :
Given: ABC is an isosceles triangle.
In ,
()
( Each )
( By AA criterion)
AD and PM are medians of triangles. So,
Given :
In
(SSS similarity)
( Corresponding angles of similar triangles )
In
(proved above)
Therefore,. ( SAS similarity)
Q13 D is a point on the side BC of a triangle ABC such that . Show that
In,
( given )
(common )
( By AA rule)
( corresponding sides of similar triangles )
(given)
Produce AD and PM to E and L such that AD=DE and PM=DE. Now,
join B to E,C to E,Q to L and R to L.
AD and PM are medians of a triangle, therefore
QM=MR and BD=DC
AD = DE (By construction)
PM=ML (By construction)
So, diagonals of ABEC bisecting each other at D,so ABEC is a parallelogram.
Similarly, PQLR is also a parallelogram.
Therefore, AC=BE ,AB=EC and PR=QL,PQ=LR
(Given )
(SSS similarity)
...................1 (Corresponding angles of similar triangles)
Similarity,
........................2
Adding equation 1 and 2,
............................3
In
( Given )
( From above equation 3)
( SAS similarity)
CD = pole
AB = tower
Shadow of pole = DF
Shadow of tower = BE
In
( Each )
(Angle of sun at same place )
(AA similarity)
cm
Hence, the height of the tower is 42 cm.
Q16 If AD and PM are medians of triangles ABC and PQR, respectively where , prove that
Answer:
( Given )
............... ....1( corresponding sides of similar triangles )
....................................2
AD and PM are medians of triangle.So,
..........................................3
From equation 1 and 3, we have
...................................................................4
In
(From equation 2)
(From equation 4)
(SAS similarity)
Q1 Let and their areas be, respectively, 64 and 121 . If EF = 15.4 cm, find BC.
( Given )
ar(ABC) = 64 and ar(DEF)=121 .
EF = 15.4 cm (Given )
Given: Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O.
AB = 2 CD ( Given )
In
(vertically opposite angles )
(Alternate angles)
(Alternate angles)
(AAA similarity)
Q3 In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that
Answer:
Let DM and AP be perpendicular on BC.
In
(Each )
(Vertically opposite angles)
(AA similarity)
Since
Q4 If the areas of two similar triangles are equal, prove that they are congruent.
Let , therefore,
(Given )
(SSS )
D, E, and F are respectively the mid-points of sides AB, BC and CA of . ( Given )
and DE||AC
In ,
(corresponding angles )
(corresponding angles )
(By AA)
Let be x.
Similarly,
and
Let AD and PS be medians of both similar triangles.
Purring these value in 1,
In
(proved above)
(proved above)
(SAS )
Therefore,
From 1 and 4, we get
Let ABCD be a square of side units.
Therefore, diagonal =
Triangles form on the side and diagonal are ABE and DEF, respectively.
Length of each side of triangle ABE = a units
Length of each side of triangle DEF = units
Both the triangles are equilateral triangles with each angle of .
( By AAA)
Using area theorem,
(A) 2: 1 (B) 1: 2 (C) 4 : 1 (D) 1: 4
Given: ABC and BDE are two equilateral triangles such that D is the mid-point of BC.
All angles of the triangle are .
ABC BDE (By AAA)
Let AB=BC=CA = x
then EB=BD=ED=
Option C is correct.
Q9 Sides of two similar triangles are in the ratio 4: 9. Areas of these triangles are in the ratio
(A) 2 : 3 (B) 4: 9 (C) 81: 16 (D) 16: 81
Sides of two similar triangles are in the ratio 4: 9.
Let triangles be ABC and DEF.
We know that
Option D is correct.
NCERT solutions for class 10 maths chapter 6 Triangles Excercise: 6.5
In the case of a right triangle, the length of its hypotenuse is highest.
hypotenuse be h.
Taking, 7 cm, 24 cm
By Pythagoras theorem,
= given third side.
Hence, it is the right triangle with h=25 cm.
In the case of a right triangle, the length of its hypotenuse is highest.
hypotenuse be h.
Taking, 3 cm, 6 cm
By Pythagoras theorem,
Hence, it is not the right triangle.
In the case of a right triangle, the length of its hypotenuse is highest.
hypotenuse be h.
Taking, 50 cm, 80 cm
By Pythagoras theorem,
Hence, it is not a right triangle.
In the case of a right triangle, the length of its hypotenuse is highest.
hypotenuse be h.
Taking, 5cm, 12 cm
By Pythagoras theorem,
= given third side.
Hence, it is a right triangle with h=13 cm.
Q2 PQR is a triangle right angled at P and M is a point on QR such that . Show that
Let be x
In ,
Similarly,
In ,
In
(By AAA)
Hence proved.
Q3 (1) In Fig. 6.53, ABD is a triangle right angled at A and AC BD. Show that
In
(common )
(By AA)
, hence prooved.
Q3 (2) In Fig. 6.53, ABD is a triangle right angled at A and AC BD. Show that
Answer:
Let be x
In ,
Similarly,
In ,
In
( Each right angle)
(By AAA)
Hence proved
Q3 (3) In Fig. 6.53, ABD is a triangle right angled at A and AC BD. Show that
In
(common )
(By AA)
Hence proved.
Q4 ABC is an isosceles triangle right angled at C. Prove that
Given: ABC is an isosceles triangle right angled at C.
Let AC=BC
In ABC,
By Pythagoras theorem
(AC=BC)
Hence proved.
Q5 ABC is an isosceles triangle with AC = BC. If , prove that ABC is a right triangle.
Given: ABC is an isosceles triangle with AC=BC.
In ABC,
(Given )
(AC=BC)
These sides satisfy Pythagoras theorem so ABC is a right-angled triangle.
Hence proved.
Q6 ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Given: ABC is an equilateral triangle of side 2a.
AB=BC=AC=2a
AD is perpendicular to BC.
We know that the altitude of an equilateral triangle bisects the opposite side.
So, BD=CD=a
In ADB,
By Pythagoras theorem,
The length of each altitude is .
In AOB, by Pythagoras theorem,
In BOC, by Pythagoras theorem,
In COD, by Pythagoras theorem,
In AOD, by Pythagoras theorem,
Adding equation 1,2,3,4,we get
(AO=CO and BO=DO)
Hence proved.
Q8 (1) In Fig. 6.54, O is a point in the interior of a triangle ABC, OD BC, OE AC and OF AB. Show that
Join AO, BO, CO
In AOF, by Pythagoras theorem,
In BOD, by Pythagoras theorem,
In COE, by Pythagoras theorem,
Adding equation 1,2,3,we get
Hence proved
Q8 (2) In Fig. 6.54, O is a point in the interior of a triangle ABC, OD BC, OE AC and OF AB.
Join AO, BO, CO
In AOF, by Pythagoras theorem,
In BOD, by Pythagoras theorem,
In COE, by Pythagoras theorem,
Adding equation 1,2,3,we get
OA is a wall and AB is a ladder.
In AOB, by Pythagoras theorem
Hence, the distance of the foot of the ladder from the base of the wall is 6 m.
OB is a pole.
In AOB, by Pythagoras theorem
Hence, the distance of the stack from the base of the pole is m.
Distance travelled by the first aeroplane due north in hours.
Distance travelled by second aeroplane due west in hours.
OA and OB are the distance travelled.
By Pythagoras theorem,
Thus, the distance between the two planes is .
Let AB and CD be poles of heights 6 m and 11 m respectively.
CP=11-6=5 m and AP= 12 m
In APC,
By Pythagoras theorem,
Hence, the distance between the tops of two poles is 13 m.
In ACE, by Pythagoras theorem,
In BCD, by Pythagoras theorem,
From 1 and 2, we get
In CDE, by Pythagoras theorem,
In ABC, by Pythagoras theorem,
From 3,4,5 we get
In ACD, by Pythagoras theorem,
In ABD, by Pythagoras theorem,
From 1 and 2, we get
Given : 3DC=DB, so
From 3 and 4, we get
Hence proved.
Q15 In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that
Given: An equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC.
To prove :
Let AB=BC=CA=a
Draw an altitude AE on BC.
So,
In AEB, by Pythagoras theorem
Given : BD = 1/3 BC.
In ADE, by Pythagoras theorem,
Given: An equilateral triangle ABC.
Let AB=BC=CA=a
Draw an altitude AE on BC.
So,
In AEB, by Pythagoras theorem
Q17 Tick the correct answer and justify : In AB = cm, AC = 12 cm and BC = 6 cm.
The angle B is :
(A) 120°
(B) 60°
(C) 90°
(D) 45°
In AB = cm, AC = 12 cm and BC = 6 cm.
It satisfies the Pythagoras theorem.
Hence, ABC is a right-angled triangle and right-angled at B.
Option C is correct.
Q1 In Fig. 6.56, PS is the bisector of . Prove that
A line RT is drawn parallel to SP which intersect QP produced at T.
Given: PS is the bisector of .
By construction,
(as PS||TR)
(as PS||TR)
From the above equations, we get
By construction, PS||TR
In QTR, by Thales theorem,
Hence proved.
Join BD
Given : D is a point on hypotenuse AC of D ABC, such that BD AC, DM BC and DN AB.Also DN || BC, DM||NB
In CDM,
In DMB,
From equation 1 and 2, we get
From equation 1 and 3, we get
In
(By AA)
(BM=DN)
Hence proved
Q2 (2) In Fig. 6.57, D is a point on hypotenuse AC of D ABC, such that BD AC, DM BC and DN AB. Prove that:
Answer:
In DBN,
In DAN,
BD AC,
From equation 1 and 3, we get
From equation 2 and 3, we get
In
(By AA)
(NB=DM)
Hence proved.
Q3 In Fig. 6.58, ABC is a triangle in which ABC > 90° and AD CB produced. Prove that
In ADB, by Pythagoras theorem
In ACD, by Pythagoras theorem
(From 1)
Q4 In Fig. 6.59, ABC is a triangle in which ABC < 90° and AD BC. Prove that
In ADB, by Pythagoras theorem
In ACD, by Pythagoras theorem
(From 1)
Q5 (1) In Fig. 6.60, AD is a median of a triangle ABC and AM BC. Prove that :
Given: AD is a median of a triangle ABC and AM BC.
In AMD, by Pythagoras theorem
In AMC, by Pythagoras theorem
(From 1)
(BC=2 DC)
Q5 (2) In Fig. 6.60, AD is a median of a triangle ABC and AM BC. Prove that :
In ABM, by Pythagoras theorem
(BC=2 BD)
Q5 (3) In Fig. 6.60, AD is a median of a triangle ABC and AM BC. Prove that:
In ABM, by Pythagoras theorem
In AMC, by Pythagoras theorem
..................................2
Adding equation 1 and 2,
In parallelogram ABCD, AF and DE are altitudes drawn on DC and produced BA.
In DEA, by Pythagoras theorem
In DEB, by Pythagoras theorem
....................................2
In ADF, by Pythagoras theorem
In AFC, by Pythagoras theorem
Since ABCD is a parallelogram.
SO, AB=CD and BC=AD
In
(AE||DF)
AD=AD (common)
(ASA rule)
Adding 2 and, we get
(From 4 and 6)
\
Q7 (1) In Fig. 6.61, two chords AB and CD intersect each other at point P. Prove that :
Join BC
In
( vertically opposite angle)
(Angles in the same segment)
(By AA)
Q7 (2) In Fig. 6.61, two chords AB and CD intersect each other at point P. Prove that :
Join BC
In
( vertically opposite angle)
(Angles in the same segment)
(By AA)
(Corresponding sides of similar triangles are proportional)
In
(Common)
(Exterior angle of a cyclic quadrilateral is equal to opposite interior angle)
So, ( By AA rule)
In
(Common)
(Exterior angle of a cyclic quadrilateral is equal to opposite interior angle)
So, ( By AA rule)
24440 (Corresponding sides of similar triangles are proportional)
Q9 In Fig. 6.63, D is a point on side BC of D ABC such that Prove that AD is the bisector of BAC.
Answer:
Produce BA to P, such that AP=AC and join P to C.
(Given )