Class 10 R S AGGARWAL AND V AGGARWAL Solutions Maths Chapter 7 - Triangles
Triangles Exercise Ex. 7A
Solution 1
(i)In ABC, DE || BC, AD = 3.6 cm, AB = 10 cm, AE = 4.5 cm
Hence, AC = 12.5 cm and EC = 8cm
(ii)In ABC, DE || BC, AB = 13.3 cm, AC = 11.9 cm and EC = 5.1 cm
Hence, AD = 7.7 cm
(iii)In ABC, DE || BC, AC = 6.6 cm,
?
Hence, AE = 2.4 cm
(iv)In ABC, DE || BC, Given
Hence AE = 4 cm
Solution 2
(i)D and E are points on the sides AB and AC respectively of a ABC such that DE || BC, AD = x cm, DB = (x - 2) cm,
AE = (x + 2) cm, EC = (x - 1) cm
Hence, x = 4
(ii)In ABC, DE || BC, AD = 4 cm, DB = (x - 4) cm, AE = 8 cm, EC = (3x - 19) cm
Hence, x = 11
(iii)In ABC, DE || BC, AD = (7x - 4) cm, AE = (5x - 2) cm, DB = (3x + 4)cm, EC = 3x cm
Solution 3
Given: A ABC in which D and E are points on the sides AB and AC respectively.
To prove: DE ||BC
Proof:
(i)AD = 5.7 cm, DB = 9.5 cm, AE = 4.8 cm and EC = 8 cm
Since D and E are the points on AB and AC respectively.
Hence, by the converse of Thales theorem DE || BC
(ii)AB = 11.7 cm, AC = 11.2 cm, BD = 6.5 cm, AE = 4.2 cm
Since D and E are points on AB and AC respectively.
Hence, by the converse of Thales theorem DE is not parallel to BC.
(iii) AB = 10.8 cm, AD = 6.3 cm, AC = 9.6 cm, EC = 4 cm
Since D and E are the points on AB and AC respectively.
Hence by the converse of Thales theorem DE || BC
(iv)AD = 7.2 cm, AE = 6.4 cm, AB = 12 cm, AC = 10 cm
Since D and E are points on the side AB and AC respectively.
Hence, by the converse of Thales theorem DB is not parallel to BC
Solution 4
(i) AB = 6.4 cm, AC = 8 cm, BD = 5.6 cm
Let BC = x
Now, DC = (BC - BD)
= (x - 5.6) cm
In ABC, AD is the base for of A
So, by the angle bisector theorem, We have
Hence, BC = 12.6 cm and DC = (12.6 - 5.6) cm = 7 cm
(ii) AB = 10 cm, AC = 14 cm, BC = 6cm
Let BD = x,
DC = (BC - BD) = (6 - x) cm
In ABC, AD is the bisector of ??A
So, By angle bisector theorem,
Hence, BD = 2.5 cm and DC = (6 - 2.5) cm = 3.5 cm
(iii)AB = 5.6 cm, BD = 3.2 cm and BC = 6 cm
DC = BC - BD = (6 - 3.2) cm = 2.8 cm
Let AC = x,
In ABC, AD is the bisector of A
So, by the angle bisector theorem we have
Hence, AC = 4.9 cm
(iv)AB = 5.6 cm, AC = 4 cm, DC = 3 cm
Let BD = x,
In ABC, AD is the bisector of A
So, by the angle bisector theorem, we have
Hence, BD = 4.2 cm
So BC = BD + AC = (4.2 + 3) cm
BC = 7.2 cm
Solution 5
Solution 6
Let ABCD be the trapezium and let E and F be the midpoints of AD and BC respectively.
Const: Produce AD and BC to meet at P
In PAB, DC || AB
Solution 7
CD || AB in trapezium ABCD and its diagonals intersect at O.
But, will make DO =
And, length cannot be negative.
Solution 8
Solution 9
?
Given: ABC and DBC lie on the same side of BC. P is a point on BC, PQ || AB and PR || BD are drawn meeting AC at Q and CD at R respectively.
To Prove: QR || AD
Proof: In ABC
Hence, in ACD, Q and R the points in AC and CD such that
QR || AD(by the converse of Thales theorem)
Hence proved.
Solution 10
Given BD = CD and OD = DX
Join BX and CX
Thus, the diagonals of quad OBXC bisect each other
OBXC is a parallelogram
BX || CF and so, OF || BX
Similarly, CX || OE
In ABX, OF || BX
Solution 11
Given: ABCD is a parallelogram in which P is the midpoint of DC and Q is a point on AC such that. PQ produced meets BC at R.
To prove: R is the midpoint of BC
Construction: Join BD
Proof: Since the diagonals of a || gm bisect each other at S such that
Q is the midpoint of CS
So, PQ || DS.
Therefore, QR || SB.
In CSB, Q is the midpoint of CS and QR || SB.
So R is the midpoint of BC.
Solution 12
Given: ABC is a triangle in which AB = AC. D and E are points on AB and AC respectively such that AD = AE
To prove: The points B, C, E and D are concyclic.
Proof: AB = AC (given)
AD = AE (given)
Quad BCEA is cyclic
Hence, the point B, C, E, D are concyclic
Solution 13
Triangles Exercise Ex. 7B
Solution 1
(i)In ABC and PQR
A = Q = 50°
B = P = 60°
C = R = 70°
ABC ~QPR(by AAA similarity)
(ii)In ABC and EFD
A = D = 70°
SAS: Similarity condition is not satisfied as A and D are not included angles.
(iii)CAB QRP (SAS Similarity)
(iv) In EFD and PQR
FE = 2cm, FD = 3 cm, ED = 2.5 cm
PQ = 4 cm, PR = 6 cm, QR = 5 cm
FED ~PQR (SSS similarity)
(v)
In CAB and RMN
Solution 2
ODC ~ OBC
BOC = 115o
CDO = 70o
(i) DOC = (180o - BOC)
= (180o - 115o)
= 65o
(ii) OCD = 180o - CDO - DOC
OCD = 180o - (70o + 65o)
= 45o
(iii) Now, ABO ~ ODC
AOB = COD (vert. Opp s) = 65o
OAB = OCD = 45o
(iv) OBA = ODC(alternate angles) = 70o
So, OAB = 45o and OBA = 70o
Solution 3
Given: OAB OCD
AB = 8 cm, BO = 6.4 cm, CD = 5 cm, OC = 3.5 cm
Solution 4
Given: ADE = B, AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm, BC = 4.2 cm
Proof:
In ADE and ABC,
A = A (common)
ADE = B (given)
Therefore, ADE ABC (AA Criterion)
Hence, DE = 2.8 cm
Solution 5
ABC and PQR are similar triangles, therefore corresponding sides of both the triangles are proportional.
Hence, AB = 16 cm
Solution 6
ABC and DEF are two similar triangles, therefore corresponding sides of both the triangles are proportional.
Hence,
Let perimeter of ABC = x cm
Hence, perimeter of ABC = 35 cm
Solution 7
Given: AB = 100 cm, BC = 125 cm, AC = 75 cm
Proof:
In BAC and BDA
BAC = BDA = 90o
B = B (common)
BAC BDA(by AA similarities)
Therefore, AD = 60 cm
Solution 8
Given that AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm
In CBA and CDB
CBA = CDB = 90o
C = C (Common)
Therefore, CBA CDB (by AA similarities)
Hence, BC = 8.1 cm
Solution 9
Given that BD = 8 cm, AD = 4 cm
In DBA and DCB, we have
BDA = CDB = 90o
DBA = DCB [each = 90o - A]
DBA DCB (by AAA similarity)
Hence, CD = 16 cm
Solution 10
Given: P is a point on AB.
Then, AB = AP + PB = (2 + 4) cm = 6 cm
Also Q is a point on AC.
Then, AC = AQ + QC = (3 + 6) cm = 9 cm
Thus, in APQ and ABC
A = A (common)
And
APQ ~ ABC(by SAS similarity)
Hence proved.
Solution 11
Given: ABCD is a parallelogram and E is point on BC. Diagonal DB intersects AE at F.
To Prove: AF × FB = EF × FD
Proof: In AFD and EFB
AFD = EFB (vertically opposite s)
DAF = BEF (Alternate s)
Hence proved.
Solution 12
In the given figure: DB BC, AC BC and DB || AC
AB is the transversal
DBE = BAC [Alternate s]
In BDE and ABC
DEB = ACB = 90o
DBE = BAC
BDE ~ ABC
~ [By AA similarity]
Hence proved.
Solution 13
Let AB be the vertical stick and let AC be its shadow.
Then, AB = 7.5 m and AC = 5 m
Let DE be the vertical tower and let DF be its shadow
Then,DF = 24 m, Let DE = x meters
Now, in BAC and EDF,
BAC ~ EDF by SAS criterion
Therefore, height of the vertical tower is 36 m.
Solution 14
In ACP and BCQ
CA = CB
CAB = CBA
ACP BCQ
Solution 15
1 = 2 (given)
(given)
Also, 2 = 1
Therefore, by SAS similarity criterion ACB ~ DCE
Solution 16
Given: ABCD is a quadrilateral in which AD = BC. P, Q , R, S are the midpoints of AB, AC, CD and BD.
To prove: PQRS is a rhombus
Proof: In ABC,
Since P and Q are mid points of AB and AC
Therefore, PQ || BC and (Mid-point theorem)
Similarly,
SP || RQ and PQ || SR and PQ = RQ = SP = SR
Hence,PQRS is a rhombus.
Solution 17
Solution 18
Solution 19
Solution 20
Consider
Therefore, by SAS criterion of similarity,
Solution 21
Consider
Then, we have
Triangles Exercise Ex. 7C
Solution 1
Given: ABC DEF,
area of ABC= and area of DEF = 121
We know that the ratio of the area of two similar triangles is equal to the ratio of the squares of their corresponding sides.
Hence, BC = 11.2 cm
Solution 2
Given: ABC PQR,
area of ABC = 9 cm2 and area of PQR = 16 cm2.
We know that the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
Hence, QR = 6 cm
Solution 3
Given: ABC ~PQR,
area of ABC = 4 area of PQR.
Let area of PQR = x. Then area of ABC = 4x.
We know that the ratio of the areas of two similar triangle is equal to the ratio of the square of their corresponding sides.
Hence. QR = 6 cm
Solution 4
Given: ABC DEF such that ar(ABC) = 169 and ar(DEF) = 121
We know that the ratio of the area of similar triangles is equal to the ratio of the square of their corresponding sides.
Hence, the longest side of smallest triangle side is 22 cm.
Solution 5
Given: ABC DEF
ar(ABC) = 100 and ar(DEF) = 49
Let AL and DM be the corresponding altitude of ABC and DEF respectively such that AL = 5 cm and let DM = x cm
We know that the ratio of the area of two similar triangles is equal to the ratio of the square of corresponding altitudes.
Therefore, the required altitude is 3.5 cm
Solution 6
Given: ABC DEF
Let AL and DM be the corresponding altitudes of ABC and DEF respectively such that AL = 6 cm and DM = 9 cm.
We know that the ratio of squares of altitudes of two similar triangles is equal to the ratio of the corresponding areas.
Hence, ratio of their areas = 4 : 9
Solution 7
Given: ABC DEF such that
ar(ABC) = 81 and ar(DEF) = 49
Let AL and DM be the corresponding altitudes of ABC and DEF respectively, such that AL = 6.3 cm and Let DM= x cm
We know that the ratio of the area of two similar triangles is equal to the ratio of the square of corresponding altitudes:
Hence, the required altitude 4.9 cm
Solution 8
Given: ABC DEF such that ar(ABC) = 100 cm and ar(DEF) = 64
Let AP and DQ be the corresponding medians of ABC and DEF respectively such that DQ = 5.6cm.
Let AP = x cm.
We know that the ratio of the areas of two similar triangle is equal be the ratio of the squares of their corresponding medians.
Hence, AP = 7 cm
Solution 9
Given: AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 cm
AB = AP + PB = (1 + 3) cm = 4 cm
AC = AQ + QC = (1.5 + 4.5) cm = 6 cm
In APQ and ABC, we have
APQ = ABC (corresponding s)
AQP = ACB (corresponding s)
APQ ABC [by AA similarity]
Hence proved.
Solution 10
Given DE || BC
DE = 3 cm and BC = 6 cm
ar(ADE) = 15
In ADE and ABC, we have
Solution 11
In BAC and ADC, we have
BAC = ADC = 90o (AD BC)
ACB = DCA (common)
BAC ADC
Therefore, the ratio of the areas of ABC and ADC = 169:25
Solution 12
Let DE = 3x and BC = 5x
In ADE and ABC, we have
ADE = ABC (corres. s)
AED = ACB (corres. s)
ADE ABC (by AA similarity)
Let, ar(ADE) = 9x2 units
Then, ar(ABC) = 25x2 units
Therefore, ratio of ar(ADE) to the ar(trap BCED) = 9:16
Solution 13
In ABC, D and E are midpoint of AB and AC respectively.
So, DE|| BC and
Now, in ADE and ABC, we have
ADE = ABC (corres. s)
AED = ACB (corres. s)
ADE ABC (by AA similarity)
Let AD = x and AB = 2x
Therefore, the ratio of the areas of ADE and ABC = 1:4
Triangles Exercise Ex. 7D
Solution 1
For a given triangle to be a right angled, the sum of the squares of the two sides must be equal to the square of the largest side.
(i)Let a = 9cm, b = 16 cm and c = 18 cm. Then
Hence the given triangle is not right angled.
(ii)Let a = 7cm, b = 24 cm and c = 25 cm, Then
Hence, the given triangle is a right triangle.
(iii)Let a = 1.4 cm, b = 4.8 cm, and c = 5 cm
Hence, the given triangle is a right triangle
(iv)Let a = 1.6 cm, b = 3.8 cm and c = 4 cm
Hence, the given triangle is not a right triangle
(v)Let p = (a - 1) cm, q = cm and r = (a + 1)
Hence, the given triangle is a right triangle
Solution 2
Starting from A, let the man goes from A to B and from B to C, as shown in the figure.
Then,
AB = 80 m, BC = 150 m andABC = 90o
From right ABC, we have
Hence, the man is 170m north-east from the starting point.
Solution 3
Starting from O, let the man goes from O to A and then A to B as shown in the figure.
Then,
OA = 10 m, AB = 24 m and OAB = 90o
Using Pythagoras theorem:
Hence, the man is 26 m south-west from the starting position.
Solution 4
Let AB be the building and CB be the ladder.
Then,
AB = 12 m, CB = 13 m and CAB = 90o
By Pythagoras theorem, we have
Hence, the distance of the foot of the ladder from the building is 5 m.
Solution 5
Let AB be the wall where window is at B, CB be the ladder and AC be the distance between the foot of the ladder and wall.
Then,
AB = 20 m, AC = 15 m, and CAB = 90o
By Pythagoras theorem, we have
Hence, the length of ladder is 25 m.
Solution 6
Let AB and CD be the given vertical poles.
Then,
AB = 9 m, CD = 14 m and AC = 12 m
Const: Draw, BE || AC.
Then,
CE = AB = 9m and BE = AC = 12 m
DE = (CD - CE)
= (14 - 9)
= 5 m
In right BED, we have
Hence, the distance between their tops is 13 m.
Solution 7
Solution 8
In PQR, QPR = 90o, PQ = 24 cm, and QR =
In POR, PO = 6 cm, QR = 8cm and POR = 90o
In POR,
In PQR,
By Pythagoras theorem, we have
Hence,
(sum of square of two sides equal to square of greatest side)
Hence, PQR is a right triangle which is right angled at P.
Solution 9
Given: ABC is an isosceles triangle with AB = AC = 13
Const: Draw altitude from A to BC (AL BC).
Now, AL = 5 cm
In ALB,
ALB = 90o
In ALC,
ALC = 90o
Solution 10
Given: ABC in which AB = AC = 2a units and BC = a units
Const: Draw AD BC then D is the midpoint of BC.
In ABC
Solution 11
In an equilateral triangle all sides are equal.
Then, AB = BC = AC = 2a units
Const: Draw an altitude AD BC
Given BC = 2a. Then, BD = a
In ABD,
ADB = 90o
Hence, the length of each altitude is
Solution 12
ABC is an equilateral triangle in which all side are equa.
Therefore, AB = BC = AC = 12 cm
If BC = 12 cm
Then, BD = DC = 6 cm
In ADB,
Hence the height of the triangle is
Solution 13
Let ABCD is the given rectangle, let BD is a diagonal making a ADB.
BAD = 90o
Using Pythagoras theorem:
Hence, length of diagonal DB is 34 cm.
Solution 14
Let ABCD be the given rhombus whose diagonals intersect at O.
Then AC = 24 cm and BD = 10 cm
We know that the diagonals of a rhombus bisect each other at right angles.
From right AOB, we have
Hence, each side of a rhombus 13 cm
Solution 15
Given: ABC in which D is the midpoint of BC. AE BC and AC > AB.
Then, BD = CD and AED = 90o,
Then, ADE < 90o and ADC > 90o
In AED,
Putting value of from (1) in (2), we get
Solution 16
Solution 17
Given: D is the midpoint of side BC, AE BC, BC = a, AC = b, AB = c, ED = x, AD = p and AE = h
In AEC, AEC = 90o
(i) In AEC, AEC = 90o
(ii)In ABE, ABE = 90o
(iii)Adding (1) and (2), we get
(iv)Subtracting (2) from (1), we get
Solution 18
Const: Draw a perpendicular AE from A
Thus, AE BC
Proof:
In ABC,AB = AC
And AE is a bisector of BC
Then,BE = EC
In right angle triangles AED and ACE
Hence proved.
Solution 19
ABC is an isosceles triangle right angled at B,
Let AB = BC = x cm
By Pythagoras theorem,
Solution 20
Solution 21
(a)
Solution 22
Triangles Exercise Ex. 7E
Solution 1
Two triangles are said to be similar to each other if:
(i) their corresponding angles are equal, and
(ii) their corresponding sides are proportional.
Solution 2
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct point, then the other sides are divided in the same ratio.
Solution 3
If a line divides any two sides of a triangle in the same ratio then the line must be parallel to the third side.
Solution 4
The line segment joining the mid-points of any two sides of a triangle is parallel to the third side.
Solution 5
If in any two triangles, the corresponding angles are equal, then their corresponding sides are proportional and hence the triangles are similar.
Solution 6
If two angles of one triangle are respectively equal to two angles of another triangle then the two triangles are similar.
Solution 7
If the corresponding sides of two triangles are proportional then their corresponding angles are equal, and hence the two triangles are similar.
Solution 8
If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional then the two triangles are similar.
Solution 9
In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Solution 10
In a triangle, if the square of one side is equal to the sum of the squares of the other two sides then the angle opposite to the first side is a right angle.
Solution 11
Solution 12
Solution 13
Hence, EF = 12 cm
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Let
AC = 24 cm
BD = 10 cm
Solution 22
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
Solution 28
Solution 29
Solution 30 (i)
Similar figures have the same shape but need not have the same size.
Since all circles irrespective of the radii will have the same shape, all will be similar.
So, the statement is true.
Solution 30 (ii)
Two rectangles are similar if their corresponding sides are proportional.
So, the statement is false.
Solution 30 (iii)
Two triangles are said to be similar to each other if:
(i) their corresponding angles are equal, and
(ii) their corresponding sides are proportional.
So, the statement is false.
Solution 30 (iv)
Solution 30 (v)
Solution 30 (vi)
The line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram as shown.
It may or may not be a rhombus.
So, the statement is false.
Solution 30 (vii)
Solution 30 (viii)
Solution 30 (ix)
Solution 30 (x)
Triangles Exercise MCQ
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Correct option: (b)
Recall that the diagonals of a trapezium divide each other proportionally.
Note that this happens even in a parallelogram, square and rectangle, but without additional information it is not possible to be sure.
Solution 19
Solution 20
Correct option: (a)
The line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram as shown below.
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
Solution 28
Solution 29
Solution 30
Solution 31
Solution 32
Solution 33
Solution 34
Solution 35
Solution 36
Solution 37
Solution 38
Solution 39
Solution 40
Solution 41
Solution 42
Solution 43
Solution 44
Solution 45
Solution 46
Correct option: (b)
The line segments joining the midpoints of the sides of a triangle form four triangles, each of which is similar to the original triangle.
Solution 47
Solution 48
Solution 49
Solution 50
Solution 51
Solution 52
Correct option: (b)
Clearly, option (b) is false since the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
Solution 53
Solution 54
Triangles Exercise Test Yourself
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20