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NCERT exemplar Class 10 Maths solutions chapter 6 provides the understanding for triangles and the similarity of triangles. It is the first chapter under the unit: Geometry of Class 10. The solutions available under NCERT exemplar Class 10 Maths chapter 6 solutions have been prepared and compiled by our subject matter experts of mathematics with cumulative experience of more than 10 years and equips the student to study NCERT Class 10 Maths effortlessly. These Class 10 Maths NCERT exemplar chapter 6 solutions provide step-by-step detailed solutions to enhance the learning of concepts based on Triangles. The CBSE Syllabus for Class 10 is the building block for these NCERT exemplar Class 10 Maths solutions chapter 6.
Question:1
In Fig., and . Then,
(a)
(b)
(c)
(d)
Answer:
Given :-
First of all we have find and in triangles and respectively
In and
In
(given)
Now let us find angle CAD from triangle ADC
Here
(sum of interior angles of triangle = 180^{o})
Now let us find angle DBA using triangle ABC
Here
{sum of interior angles of triangle = 180^{o}}
In and we get
{90^{o} each}
{common side}
{from equation (1) and (2) it is clear that both is }
(by ASA similarity)
{by cross multiplication}
Hence option (C) is correct.
Question:2
The lengths of the diagonals of a rhombus are 16 cm and 12 cm. Then, the length of the side of the rhombus is
(A) 9 cm
(B) 10 cm
(C) 8 cm
(D) 20 cm
Answer:
Answer : [B]
Here ABCD is a rhombus and we know that diagonals of a rhombus are perpendicular bisector of each other
Given: and
also and
Using Pythagoras theorem in we get
Hence option (B) is correct
Question:3
If and is not similar to , then which of the following is not true?
(A) (B)
(C) (D)
Answer:
Answer : [C]
Given :
Taking the first two terms we get
by cross multiply we get
Hence option D is correct
Now taking the last two terms we get
{By cross multiplication}
Hence option (A) is also correct
Now taking first and last terms, we get
{By cross multiplication}
Hence option (B) is also correct
By using congruence properties we conclude that only option C is not true.
Question:4
If in two triangles ABC and PQR, , then
(a)
(b)
(c)
(d)
Answer:
Answer: [A]
In and
(A) If
Here;
It matches the equation (1) Hence option A is correct.
(B) If
Here;
It does not match equation (1)
Therefore option (B) is not correct.
(c)
Here;
It does not match equation (1) Hence option (C) is not correct
(D)
Here;
It does not match the equation (1). Hence option (D) is not correct.
Question:5
In Fig., two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, and . Then, is equal to
(A) 50° (B) 30° (C) 60° (D) 100°
Answer:
Answer: [D]
In and .
{vertically opposite angles}
{By SAS similarity criterion}
{corresponding angles of similar triangles}
In
{Sum of interior angles of a triangle is 180°}
Hence option D is correct.
Question:6
If in two triangles DEF and PQR, , then which of the following is not true?
(a)
(b)
(c)
(d)
Answer:
Answer : [B]
In and
and
Now draw two triangles with the help of given conditions:-
Here and {Given}
{by AA similarity}
{corresponding angles of similar triangles}
Here option (A) is true because both the terms are derived from equation (1)
Here option (B) is not true because the first term is not derived from equation (1)
Here option (C) is true because both the terms are derived from equation (1)
Here option D i.e. is also true because both terms are derived from equation (1)
Hence option (B) is not true.
Question:7
In triangles . Then, the two triangles are
(A) congruent but not similar (B) similar but not congruent
(C) neither congruent nor similar (D) congruent as well as similar
Answer:
Answer : [B]
In and
In and
(given)
(given)
(third angle)
Therefore (by AAA similarity)
Also, it is given that
Hence and is not congruent because congruent figures have the same shape and same size.
Therefore the two triangles are similar but not congruent.
Question:8
It is given that , with then is equal to
(a)
(b)
(c)
(d)
Answer:
Answer : [A]
Given :
It is given that
[Q If triangles are similar their areas are proportional to the squares of the corresponding sides.]
Hence option A is correct.
Question:9
It is given theat and then the following is true ?
Answer:
Answer: [A]
Now equate first and the last term of equation (1) we get
Question:10
If in triangle ABC and DEF , then they will be similar when
Answer:
Answer : [C]
Given :
Here corresponding sides of given triangles ABC and DEF are equal in ratio, therefore, the triangles are similar
Also, we know that if triangles are similar then their corresponding angles are equal
Hence option C is correct.
Question:11
and BC = 15 cm then PR is equal to ?
Answer:
Answer : [A]
[Q If triangles are similar, their areas are proportional to the squares of the corresponding sides.]
[Using equation (1)]
Question:12
If S is a point on side PQ of a triangle PQR such that PS = QS = RS, then
Question:1
Is the triangle with sides 25 cm, 5 cm and 24 cm a right triangle? Give reasons for your answer.
Answer:
Answer : [False]
Sides of the triangle are 25cm, 5 cm and 24 cm
According to Pythagoras theorem, the square of the hypotenuse of a right-angle triangle is equal to the sum of the square of the remaining two sides.
Hence the given statement is false.
The triangle with sides 25 cm, 5 cm and 24 cm is not a right-angle triangle.
Question:2
Answer:
Answer : [False]
If two triangles are similar, then their corresponding angles are equal and their corresponding sides are in the same ratio
and their angles are also equal that is
.
Question:3
Answer:
Answer : [True]
Given: PQ = 12.5 cm, PA = 5 cm, BR = 6 cm and PB = 4 cm
QA = QP – PA = 12.5 – 5 = 7.5 cm
Here
According to Converse of basic proportionality theorem if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Question:4
In Fig 6.4, BD and CE intersect each other at the point P. Is PBC PDE? Why?
Answer:
Answer : [True]
Given:- BD and CE intersect each other at point P.
Two triangles are similar if their corresponding angles are equal and their corresponding sides are in the same ratio.
Hence given statement is true.
Question:5
In triangles PQR and MST, P = 55°, Q = 25°, M = 100° and S = 25°. Is QPR TSM? Why?
Answer:
Answer: [False]
Also are know that if all corresponding angles of two triangles are equal then triangles are similar.
Hence given statement is false because QPR is not similar to TSM.
Question:6
Answer:
Answer: [False]
Quadrilateral:- A quadrilateral can be defined as a closed, two-dimensional shape which has four straight sides. We can find the shape of quadrilaterals in various things around us, like in a chessboard, a kite etc. The given statement "Two quadrilaterals are similar if their corresponding angles are equal" is not true because. Two quadrilaterals are similar if and only if their corresponding angles are equal and the ratio of their corresponding sides are also equal.
Question:7
Answer:
Answer: True
According to question,
also the perimeter of ABC is three times the perimeter of PQR
(using eq. (1) and (2))
from equation (1), (2) and (3) we conclude that sides of both the triangles are in the same ratio.
As we know that if the corresponding sides of two triangles are in the same ratio, then the triangles are similar by SSS similarity criterion
Hence given statement is true.
Question:8
Answer:
Answer : [True]
Let two right angle triangles are ABC and PQR.
Given:- One of the acute angle of one triangle is equal to an acute angle of the other triangle.
In
[Sum of interior angles of a triangle is 180°]
Also in
from equation (1) and (2)
from equation (1), (2) and (3) are observed that corresponding angles of the triangles are equal therefore triangle are similar.
Question:9
Answer:
Answer: [False]
Given:- Ratio of corresponding altitudes of two similar triangles is 3/5
As we know that the ratio of the areas of two similar triangles is equal to the ratio of squares of any two corresponding altitudes.
Hence the given statement is false because ratio of areas of two triangles is 9/25 which is not equal to 6/5
Question:10
D is a point on side QR of PQR such that PD QR. Will it be correct to say that PQD ~ RPD? Why?
Answer:
Answer : [False]
In PQD and RPD
(common side)
(each 90^{o})
The given statement is false.
Because according to the definition of similarity two triangles are similar, if their corresponding angles are equal and their corresponding sides are in the same ratio
Question:11
In Fig., if D = C, then is it true that ? Why?
Answer:
Answer : [True]
In ADE and ACB
(given)
(common angle)
And we know that if two angles of one triangle are equal to the two angles of another triangle, then the two triangles are similar by AA similarity criteria.
{by AA similarity criterion}
Hence the given statement is true.
Question:12
Answer:
Answer. [False]
Given:- (1) An angle of one triangle is equal to an angle of another triangle.
(2) Two sides of one triangle are proportional to the two sides of the other triangle.
According to SAS similarity criterion if one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio, then the triangles are similar.
But here, the given statement is false because one angle and two sides of two triangles are equal but these sides not including equal angle.
Question:1
Given :
To prove :
Proof : (given)
Because PQR holds Pythagoras theorem, therefore, PQR is right-angled triangle right angle at Q.
{each angle is 90°}
[each equal to 90° – angle R]
As we know that if the two angles of one triangle are equal to the two angles of another triangle, then the two triangles are similar by AA similarity criterion.
Now using the property of area of similar triangles
Question:2
Find the value of x for which DE || AB in Fig.
Answer:
Answer : [x=2]
Given: DE || AB
According to basic proportionality theorem. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct point, then the other two sides are divided in the same ratio.
Hence the required value of x is 2.
Question:3
In Fig., if angle1 = angle 2 and NSQ MTR, then prove that PTS PRQ.
Answer:
Given :
To prove:-
Proof:-
We know that sides opposite to equal angles are also equal
from equation (1) and (2)
According to Converse of basic proportionality theorem if a line divides any two sides of a triangle in the same ratio, then the line is parallel to third side.
In and
We know that if the corresponding angles of two triangles are equal, then the triangles are similar by AAA similarity criterion
Hence proved
Question:4
Answer:
Answer: 9:1
Given:- PQRS is a trapezium in which PQ || RS and PQ = 3 RS.
(vertically opposite angles)
(alternate angle)
As we know that if two angles of one triangle are equal to the two angles of another triangle, then the two triangles are similar by AA similarity criterion
By the property of area of similar triangle
Hence the required ratio is 9: 1.
Question:5
Answer:
Given:- AB || DC and AC and PQ intersect each other at the point O.
To prove:- OA.CQ = OC.AP
Proof:-
Since AB || OC and PQ is transversal
As we know that if the two angles of one triangle are equal to the two angles of another triangle, then the two triangles are similar by AA similarity criterion
Then
By cross multiply we get
OA.CQ = AP.OC
Hence proved.
Question:6
Find the altitude of an equilateral triangle of side 8 cm.
Answer:
Let ABC be an equilateral triangle of side 8 cm
i.e. AB = BC = AC = 8 cm and AD BC
Then D is the mid-point of BC.
Apply Pythagoras theorem in triangle ABD we get
Question:7
Answer:
Answer: 18 cm
Given :
and AB = 4cm , DE = 6 cm
EF = 9 cm, FD = 12 cm
Here (given)
Taking first two terms we get
Taking first and last terms we get
Question:8
In Fig., if DE || BC, find the ratio of ar (ADE) and ar (DECB).
Answer:
Answer : 1: 3
Given:- DE || BC and DE = 6 cm, BC = 12 cm
As we know that if two angles of one triangle are equal to the two angles of another triangle, then the two triangles are similar by AA similarity criterion.
Then
Question:9
Answer:
Answer : [AD=60cm]
Given:- ABCD is a trapezium in which AB || DC, P and Q are points on AD and BC. Such that PQ || DC.
PD = 18 cm, BQ = 35, QC = 15 cm
To prove:- Find AD
Proof:-
Construction:- Join BD
In ABD, PO || AB
And we know that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then according to basic proportionality theorem the other two sides are divided in the same ratio.
In BDC, OQ || DC
Similarly by using basic proportionality theorem.
from equation (1) and (2) we get
Question:10
Answer:
Answer : [108 cm^{2}]
Given:- Corresponding sides of two similar triangles are in the ratio of 2 : 3.
Area of smaller triangle = 48 cm^{2}
We know that the ratio of the areas of two similar triangles is equal to the ratio of the square of any two corresponding sides.
i.e
Question:11
In a triangle PQR, N is a point on PR such that QN PR . If PN. NR = QN^{2}, prove that PQR = 90° .
Answer:
Given:- In a triangle PQR, N is a point on PR such that QN PR and PN.NR = QN^{2}
To prove:- PQR = 90°
Proof:
We have PN.NR = QN^{2}
and PNQ = RNQ (each equal to 90°)
we know that if one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio, then the triangles are similar by SAS similarity criterion.
Then QNP and RNQ are equiangular.
i.e.
adding equation (2) and (3) we get
In triangle PQR
[Using equation (4)]
Hence proved
Question:12
Answer:
Answer : [12 cm ]
Given:- area of smaller triangle = 36 cm^{2}
area of larger triangle = 100 cm^{2}
length of the side of larger triangle = 20 cm
let the length of the corresponding side of the smaller triangle = x cm
According to the property of area of similar triangles,
Question:13
In Fig., if , AC = 8 cm and AD = 3 cm, find BD.
Answer:
Answer :
AC = 8 cm and AD = 3 cm
We know that if two angles of one triangle are equal to the two angles of another triangle, then the two triangles are similar by AA similarity criterion.
Question:14
Answer:
Answer : [10 m]
Let BC = 15 m, AB = 24 m in ABC and A = Q
Again let DE = 16m and EDF = Q in DEF
In ABC and DEF
We know that if two angles of one triangle are equal to the two angles of another triangle, then the two triangles are similar by AA similarity criterion.
Hence the height of the point on the wall where the top of the laden reaches is 10 m.
Question:15
Answer:
Answer : [8 m]
Here AC = 10 m is a ladder
BC = 6 m distance from the base of the wall.
In right-angle triangle ABC use Pythagoras theorem
Hence, the height of the point on the wall where the top of the ladder reaches is 8 m.
Question:1
, then find the lengths of PD and CD.
Answer:
Answer: [PD=5cm, CD=2cm]
Given
:
AP = 12 cm and CP = 4 cm
on taking first and second term we get
On taking first and last term we get
Question:2
such that and .
Find the lengths of the remaining sides of the triangles.
Answer:
Answer:
Given :
sides of both triangles are in the same ratio
i.e
Put all these values in equation (1)
Taking the first and second term
Taking the first and the last term
Question:3
Answer:
Let ABC is a triangle in which line DE parallel to BC which interest line AB and AC at D and E.
To prove:- line DE divides both sides in the same ratio.
Construction:-
Proof :-
Similarly
Also
from equation (1), (2) and (3)
Hence proved
Question:4
In Fig, if PQRS is a parallelogram and then prove that
Answer:
To prove:-
from equation (1) and (2)
Subtracting 1 from both sides
We know that if a line divides any two sides of a triangle in the same ratio then by converse of basic proportionality theorem the line is parallel to the third side.
Hence proved.
Question:5
Answer:
Answer: [0.8 m]
Here, AC = 5m, BC = 4 m, AD = 1.6 m
Let CE = x m
In use Pythagoras theorem
Similarly, in triangle EBD use Pythagoras theorem
Hence the top of the ladder would slide up words on the wall at distance is 0.8 m.
Question:6
Answer:
Answer: [8 km]
Given :-
In use, Pythagoras theorem
Dividing by 8 we get
x = – 12 is not possible because distance cannot be negative.
Now
Distance covered to reach city B from A via city C = AC + CB
Distance covered to reach city B from city A after the construction of highway
Saved distance
Question:7
Answer:
Answer: 20.4 m
height of flag pole BC = 18 m
shadow AB = 9.6 m
In triangle BC using Pythagoras theorem
Hence distance of the top of the pole from the far end of the shadow is 20.4 m
Question:8
Answer:
Answer : [9 m]
height of street light bulb = 6 m
woman's height = 1.5 m
woman's shadow = 3m
Let the distance between pole and women = x m
Then
Question:9
In Fig., ABC is a triangle right angled at B and . If AD = 4 cm, and CD = 5 cm, find BD and AB.
Answer:
Answer:
Given :- and
AD = 4 cm and CD = 5 cm
In and
(each equal to 90^{o})
(each equal to 90^{o}-C)
(by AA similarity criterion)
By cross multiply we get
In use Pythagoras theorem
We know that
Question:10
Answer:
Given: PQR is a triangle
PQ = 6 cm, PS = 4 cm
In
(common angles and each angle is 90°)
(each equal to )
( by AA similarity criterion)
By cross multiply we get
In use Pythagoras theorem.
Put in equation (1)
In use Pythagoras theorem
Question:12
In a quadrilateral ABCD, . Prove that
Answer:
Given : ABCD is a quadrilateral,
To prove :-
Proof : In
(given)
for find use sum of angle of a triangle is equal to 180°
In use Pythagoras theorem we get
In use Pythagoras theorem we get
Add equation (2) and (3) we get
In use Pythagoras theorem we get
In use Pythagoras theorem we get
Now add equation (5) and (6) we get
From equation (4) and (7) we get
Hence proved.
Question:13
In fig. , l || m and line segments AB, CD and EF are concurrent at point P. Prove tha
t.
Answer:
Given: l || m and line segments AB, CD and EF are concurrent at point P
To prove:-
Proof :In
(vertically opposite angles)
(alternate angles)
and we know that if two angles of one triangle are equal to the two angles of another triangle then the two triangles are similar by AA similarity criterion
Then,
In
(vertically opposite angle)
(alternate angle)
(by AA similarity criterion)
Then
In
(vertically opposite angle)
(alternate angle)
(by AA similarity criterion)
Then
From equation (1), (2) and (3) we get
Hence proved.
Question:14
Answer:
Answer:
Given :- PA, QB, RC and SD are all perpendiculars to a line l, AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm
We know that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
(parallel lines)
Then according to basic proportionality theorem
length of PS = 36 (given)
Question:15
Answer:
ABCD is a trapezium and O is the point of intersection of the diagonals AC and BD.
Proof :-
(Common angle)
(corresponding angles)
(by AA similarity criterion)
Then
(Common angle)
(corresponding angles)
(by AA similarity criterion)
Then
We know that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
(by basic proportionality theorem)
from equation (3) and (4)
Add 1 on both sides we get
(use equation (1))
(use equation (2))
Hence proved
Question:16
Answer:
Given: Line segment DF intersect the side AC of a triangle ABC at the point E such that E is the mid-point of CA and
To prove :-
Construction:- Take point G on AB such that
Proof:- E is mid-point of CA (given)
In and E is mid-point of CA
Then according to mid-point theorem
In
According to basic proportionality theorem.
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
(use equation (1))
Hence proved.
Question:17
Answer:
Let PQR is a right angle triangle which is right angle at point Q.
Three semicircles are drawn on the sides of having diameters PQ, QR and PR respectively.
Let and are the areas of semicircles respectively.
To prove:-
Proof: In use Pythagoras theorem we get
Now area of semi-circle drawn on side PR is
area of semi-circle drawn a side QR is
area of semicircle drawn on side PQ is
add equation (2) and (3) we get
Hence
Hence proved
Question:18
Answer:
Let PQR is a right angle triangle which is a right angle at point R.
Three equilateral triangles are drawn on the sides of triangle PQR that is PRS, RTQ and PUQ
Let and are the areas of equilateral triangles respectively
To prove:-
Using Pythagoras theorem in we get
The formula of area of an equilateral triangle is
Area of equilateral
Area of equilateral
Area of equilateral
add equation (1) and (2) we get
Hence
Hence proved
These Class 10 Maths NCERT exemplar chapter 6 solutions provide a detailed knowledge of triangles and the similarity of triangles along with the theorems such as Pythagoras theorem. Most of the question types related to Triangles for Class 10 can be practiced by the students through the exemplar.
The NCERT exemplar Class 10 Maths solutions chapter 6 Triangles, due to its plentiful question bank makes it easier for the student to attempt other books such as NCERT Class 10 Maths, A textbook of Mathematics by Monica Kapoor, RD Sharma Class 10 Maths, RS Aggarwal Class 10 Maths et cetera.
Students get a very resourceful feature of NCERT exemplar Class 10 Maths solutions chapter 6 pdf download through which one can view/download the pdf version of these solutions and use them in case of low/no internet connectivity while practicing NCERT exemplar Class 10 Maths chapter 6 solutions.
Chapter No. | Chapter Name |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 | |
Chapter 15 |
We know that if two similar triangles have a sides ratio equal to R then the ratio of their areas will be R2. Therefore, in the given problem the ratio of area of two triangles will be 1:9.
Two triangles are said to be similar if their corresponding angles are equal and the ratio of the corresponding sides is consistent. The basic difference between congruent triangles and similar triangles is that congruent triangles have the same shape and size however similar triangles have different sizes and the same shapes.
The chapter Triangles is extremely important for Board examinations as it holds around 10-13% weightage of the whole paper.
Generally, the paper consists of 3-4 questions from the chapter of Triangles, and those are mainly distributed between Very Short, Short, and Long Answer questions. NCERT exemplar Class 10 Maths solutions chapter 6 explores all of the above-mentioned type-based questions in detail.
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Hello Aspirant, Hope your doing great, your question was incomplete and regarding what exam your asking.
Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.
hello Zaid,
Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.
best of luck!
According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.
You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.
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