R S AGGARWAL AND V AGGARWAL Solutions for Class 10 Maths Chapter 4 – Triangles

Chapter 4 – Triangles Exercise Ex. 4A

Question 1

D and E are points on the sides AB and AC respectively of a ABC such that DE || BC.

(i) If AD = 3.6 cm, AB = 10 cm and AE = 4.5 cm, find EC and AC

(ii) If AB = 13.3 cm, AC = 11.9 cm and EC = 5.1 cm, find AD

(iii) If and AC = 6.6 cm, find AE

(iv) If and EC = 3.5 cm, find AE

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

(iii) In ABC, DE || BC, AC = 6.6 cm,

?

Hence, AE = 2.4 cm

(iv) In ABC, DE || BC, Given

Hence AE = 4 cm

Question 2

D and E are points on the sides AB and AC respectively of a ABC such that DE || BC. Find the value of x, when

(i) AD= x cm, DB= (x – 2) cm, AE = (x + 2) cm and EC = (x – 1) cm

(ii) AD = 4 cm, DB = (x – 4) cm, AE = 8 cm and EC = (3x – 19) cm

(iii) AD = (7x – 4) cm, AE = (5x – 2) cm, DB = (3x + 4) cm and EC = 3x 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

Question 3

D and E are points on the sides AB and AC respectively of a ABC. In each of the following cases, determine whether DE || BC or not.

(i) AD = 5.7 cm, DB = 9.5 cm, AE = 4.8 cm and EC = 8 cm

(ii) AB = 11.7 cm, AC = 11.2 cm, BD = 6.5 cm and AE = 4.2 cm

(iii) AB = 10.8 cm, AD = 6.3 cm, AC = 9.6 cm and EC = 4 cm

(iv) AD = 7.2 cm, AE = 6.4 cm, AB = 12 cm and AC = 10 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.

Therefore, (each is equal to 1.4)

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

Question 4

In a ABC, AD is the bisector of A.

(i) If AB = 6.4 cm, AC = 8 cm and BD = 5.6 cm, find DC.

(ii) If AB = 10 cm, AC = 14 cm and BC = 6cm, find BD and DC

(iii) If AB = 5.6 cm, BD = 3.2 cm and BC = 6 cm, find AC.

(iv) If AB = 5.6 cm, AC = 4cm and DC = 3 cm, find 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

Question 5

M
is a point on the side BC of a parallelogram ABCD.DM when produced meets AB
product at N. Prove that

Solution 5

Question 6

Show that the line segment which joins the midpoints of the oblique sides of a trapezium is parallel to the parallel sides.

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

Question 7

In the adjoining figure, ABCD is a trapezium in which CD||AB and its diagonals intersect at O. If AO = (5x – 7) cm, OC = (2x + 1) cm, DO = (7x – 5) cm and OB = (7x + 1) cm, find the value of x.

Solution 7

We know that CD || AB in trap ABCD and its diagonals intersect at O.

Since the diagonals of a trapezium divides each other proportionally therefore, we have

Question 8

In
a ∆ABC,
M and N are points on the sides AB and AC respectively such that BM = CN. If ∠B
= ∠C
then show MN ∥
BC.

Solution 8

Question 9

ABC and DBC lie on the same side of BC, as shown in the figure. From a point P on BC, PQ || AB and PR||BD are drawn, meeting AC at Q and CD at R respectively. Prove that QR||AD.

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.

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.

Question 10

In the given figure, side BC of ABC is bisected at D and O is any point on AD. BO and CO produced meet AC and AB at E and F respectively, and AD is produced to X so that D is the midpoint of OX. Prove that AO : AX = AF : AB and show that EF||BC.

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

Question 11

ABCD is a parallelogram in which P is the midpoint of DC and Q is a point on AC such that . If PQ produced meets BC at R, prove that R is the midpoint of BC.

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.

Question 12

In the adjoining figure, ABC is a triangle in which AB = AC. If D and E are points on AB and AC respectively such that AD = AE, show that the points B, C, E and D are concyclic.

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)

Hence, the point B, C, E, D are concyclic

Question 13

In
∆ABC,
the bisector of ∠B
meets AC at D.A line PQ ∥ AC meets AB, BC and BD at P, Q and R
respectively.

Show
that PR ⨯ BQ = QR ⨯ BP.

Solution 13

Chapter 4 – Triangles Exercise Ex. 4B

Question 1

In each of the given pairs of triangles, find which pair of triangles are similar. State the similarity criterion and write the similarity relation in symbolic form:

(i)

(ii)

(iii)

(iv)

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)

Question 2

In the given figure, ODC OBA, BOC = 115o and CDO = 70o. Find (i) DOC (ii) DCO(iii) OAB(iv) OBA

Solution 2

ODC ~ OBC

BOC = 115o

CDO = 70o

(i) DOC = (180oBOC)

= (180o – 115o)

= 65o

(ii)       OCD = 180oCDO – 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

Question 3

In the given figure, OAB ~ OCD. If AB = 8 cm, BO = 6.4 cm, OC = 3.5 cm and CD = 5cm, find (i) OA (ii) DO

Solution 3

Given: OAB OCD

AB = 8 cm, BO = 6.4 cm, CD = 5 cm, OC = 3.5 cm

Question 4

In the given figure, if ADE = B, show that ADE ~ ABC. If AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm and BC = 4.2 cm, find DE

Solution 4

Given: ADE = B, AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm, BC = 4.2 cm

Proof:

A = A                               (common)

Hence, DE = 2.8 cm

Question 5

The perimeters of two similar triangles ABC and PQR are 32cm and 24cm respectively. If PQ = 12cm, find AB.

Solution 5

ABC and PQR are similar triangles, therefore corresponding sides of both the triangles are proportional.

Hence, AB = 16 cm

Question 6

The corresponding sides of two similar triangles ABC and DEF are BC = 9.1 cm and EF = 6.5 cm. if the perimeter of DEF is 25 cm, find the perimeter of ABC.

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

Question 7

In the given figure, CAB = 90o and AD BC. Show that BDA ~ BAC. If AC = 75 cm, AB = 1m and BC = 1.25 m, find AD.

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)

Question 8

In the given figure, ABC = 90o and BD AC. If AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, find BC.

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

Question 9

In the given figure, ABC = 90o and BD AC. If BD = 8 cm, AD = 4 cm, find CD.

Solution 9

Given that BD = 8 cm, AD = 4 cm

In DBA and DCB, we have

BDA = CDB = 90o

DBA = DCB                                  [each = 90oA]

DBA DCB (by AAA similarity)

Hence, CD = 16 cm

Question 10

P and Q are points on the sides AB and AC respectively of a ABC. If AP = 2 cm, PB = 4cm, AQ = 3 cm and QC = 6 cm, show that BC = 3PQ

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.

Question 11

ABCD is a parallelogram and E is a point on BC. If the diagonal BD intersect AE at F, prove that AF × FB = EF × FD

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.

Question 12

In the given figure, DB BC, DE AB and AC BC.

Prove that

Solution 12

In the given figure: DB  AB, AC  BC and DB || AC

AB is the transversal

DBE = BAC [Alternate s]

In BDE and ABC

DEB = ACB = 90o

DBE = BAC

~ [By AA similarity]

Hence proved.

Question 13

A vertical stick of length 7.5m casts a shadow 5 m long on the ground and at the same time a tower casts a shadow 24 m long. Find the height of the tower.

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.

Question 14

In an isosceles ABC, the base AB is produced both ways in P and Q such that AP × BQ =AC2. Prove that ACP BCQ

Solution 14

In ACP and BCQ

CA = CB

CAB = CBA

ACP BCQ

Question 15

In the given figure, 1 = 2and

Prove that ACB DCE

Solution 15

1 = 2                                 (given)

(given)

Also, 2 = 1

Therefore, by SAS similarity criterion ACB ~ DCE

Question 16

ABCD is a quadrilateral in which AD = BC. If P, Q, R, S be the midpoints of AB, AC, CD and BD respectively, show that PQRS is a rhombus.

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.

Question 17

In
a circle, two chords AB and CD intersect at a point P inside the circle.
Prove that

a. PAC ∼ ∆PDB

b. PA .PB =PC.PD.

Solution 17

Question 18

Two
chords AB and CD of a circle intersect at a point P outside the circle. Prove
that

a. PAC ∼ ∆PDB

b. PA .PB =PC.PD.

Solution 18

Question 19

In
a right triangle ABC, right-angle at B, D is a point on hypotenuse such that
BD ⊥
AC. If DP ⊥
AB and DQ ⊥
BC then prove that

a. DQ2= DP.QC

b. DP2=DQ. AP

Solution 19

Chapter 4 – Triangles Exercise Ex. 4C

Question 1

ABC DEF and their areas are respectively and . If EF = 15.4 cm, find BC

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

Question 2

The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, find the length of QR.

Solution 2

Given: ?ABC PQR,

area of ABC = 9 and area of PQR = 16.

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

Question 3

ABC PQR and ar(ABC) = 4 ar(PQR). If BC = 12 cm, find QR.

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

Question 4

The areas of two similar triangles are 169 and 121 respectively. If the longest side of the larger triangle is 26 cm, find the longest side of the smaller triangle.

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.

Question 5

ABC DEF and their areas are respectively 100 and 49 . If the altitude of ABC is 5 cm, find the corresponding altitude of DEF.

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

Question 6

The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas.

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

Question 7

The areas of two similar triangles are 81 and 49 respectively. If the altitude of the first triangle is 6.3 cm, find the corresponding altitude of the other.

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

Question 8

The areas of two similar triangles are 100 and 64 respectively. If a median of the smaller triangle is 5.6 cm, find the corresponding median of the other.

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

Question 9

In the given figure, ABC is a triangle and PQ is straight line meeting AB in P and AC in Q. If AP = 1cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 cm, prove that area of APQ is of the area of the ABC.

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.

Question 10

In the given figure, DE||BC. If DE = 3 cm, BC = 6cm and ar(ADE) =, find the area of ABC.

Solution 10

Given DE || BC

DE = 3 cm and BC = 6 cm

In ADE and ABC, we have

Question 11

ABC is right angled at A and AD BC. If BC = 13 cm and AC = 5 cm, find the ratio of the areas of ABC and ADC.

Solution 11

In BAC and ADC, we have

ACB = DCA            (common)

Therefore, the ratio of the areas of ABC and ADC = 169:25

Question 12

In the given figure, DE||BC and DE : BC = 3 : 5. Calculate the ratio of the areas of ADE and the trapezium BCED.

Solution 12

Let DE = 3x and BC = 5x

In ADE and ABC, we have

AED = ACB                  (corres. s)

Then, ar(ABC) = 25x2 units

Therefore, ratio of ar(ADE) to the ar(trap BCED) = 9:16

Question 13

In ABC, D and E are midpoints of AB and AC respectively. Find the ratio of the areas of ADE and ABC.

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

AED = ACB                       (corres. s)

Let AD = x and AB = 2x

Therefore, the ratio of the areas of ADE and ABC = 1:4

Chapter 4 – Triangles Exercise Ex. 4D

Question 1

The sides of certain triangles are given below. Determine which of them are right triangles:

(i) 9 cm, 16 cm, 18 cm

(ii) 7 cm, 24 cm, 25 cm

(iii) 1.4 cm, 4.8 cm, 5 cm

(iv) 1.6 cm, 3.8 cm, 4 cm

(v) (a-1)cm, cm, (a + 1)cm

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

Question 2

Aman goes 80 m due east and then 150 m due north. How far is he from the starting point?

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.

Question 3

A man goes 10 m due south and then 24 m due west. How far is he 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.

Question 4

A 13m long ladder reaches a window of a building 12 m above the ground. Determine the distance of the foot of the ladder from the building.

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.

Question 5

A ladder is placed in such a way that its foot is at a distance of 15m from a wall and its top reaches a window 20m above the ground. Find the length of the ladder.

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.

Question 6

Two vertical poles of height 9 m and 14 m stand on a plane ground. If the distance between their feet is 12m, find the distance between their tops.

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.

Question 7

A
guy wire attached to a vertical pole of height 18 m is 24 m long and has a
stake attached to the other end. How far from the base of the pole should the
stake be drive so that the wire will be taut?

Solution 7

Question 8

In the given figure, O is a point inside a PQR such that PQR = 90o, OP = 6 cm and OR = 8 cm. If PQ = 24 cm and QR = 26 cm, prove that PQR is right-angled.

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.

Question 9

ABC is an isosceles triangle with AB = AC = 13 cm. The length of altitude from A on BC is 5 cm. Find BC.

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

Question 10

Find the length of altitude AD of an isosceles ABC in which AB = AC = 2a units and BC = a units.

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

Question 11

ABC is an equilateral triangle of side 2a units. Find each of its altitudes.

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,

Hence, length of each altitude is

Question 12

Find the height of an equilateral triangle of side 12 cm.

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

Hence the height of the triangle is

Question 13

Find the length of a diagonal of a rectangle whose adjacent sides are 30 cm and 16 cm.

Solution 13

Let ABCD is the given rectangle, let BD is a diagonal making a ADB.

Using Pythagoras theorem:

Hence, length of diagonal DB is 34 cm.

Question 14

Find the length of each side of a rhombus whose diagonals are 24 cm and 10 cm long.

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

Question 15

In ABC, D is the midpoint of BC and AE  BC. If AC > AB, show that

Solution 15

Given: ABC in which D is the midpoint of BC. AE  BC and AC > AB.

Then, BD = CD and AED = 90o,

In AED,

Putting value of from (1) in (2), we get

Question 16

In ABC, AB = AC. Side BC is produced to D. Prove that

Solution 16

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.

Question 17

In the given figure, D is the midpoint of side BC and AE BC. If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that:

(i)

(ii)

(iii)

(iv)

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

Question 18

In
∆ABC
,AB = AC. Side BC is produced to D. Prove that

CD.

Solution 18

Question 19

ABC is an isosceles triangle, right-angled at B. Similar triangles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of ABE and ACD.

Solution 19

ABC is an isosceles triangle right angled at B,

Let AB = BC = x cm

By Pythagoras theorem,

Question 20

Solution 20

Question 21

In
a ∆ABC,
AD is a median and AL ⊥ BC. Prove that:

Solution 21

Question 22

Naman is doing fly-fishing in a
stream. The tip of this fishing rod is 1.8 above the surface of the water and
the fly at the end of the string rests on the water 3.6 m away from him and
2.4 m from the point directly under the tip of the rod. Assuming that the
string (from the tip of his rod to the fly) is taut, how much string does he
have out) see the adjoining figure)? If he pulls in the string at the rate of
5 cm per second, what will be the; horizontal distance of the fly from him
after 12 seconds?

Solution 22

Chapter 4 – Triangles Exercise Ex. 4E

Question 1

State
the two properties which are necessary for given two triangles to be similar.

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.

Question 2

State
the basic proportionality theorem.

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.

Question 3

State
the converse of ‘Thales’ theorem.

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.

Question 4

State
the midpoint theorem.

Solution 4

The line segment joining the mid-points of any two sides of a
triangle is parallel to the third side.

Question 5

State
the AAA-similarity criterion.

Solution 5

If in any two triangles, the corresponding angles are equal, then
their corresponding sides are proportional and hence the triangles are
similar.

Question 6

State
the AA-similarity criterion.

Solution 6

If two angles of one triangle are respectively equal to two angles of
another triangle then the two triangles are similar.

Question 7

State
the SSS-criterion for similarity of triangles.

Solution 7

If the corresponding sides of two triangles are proportional then
their corresponding angles are equal, and hence the two triangles are
similar.

Question 8

State
the SAS-similarity criterion.

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.

Question 9

State
Pythagoras’ theorem.

Solution 9

In a right triangle, the square of the hypotenuse is equal to the sum
of the squares of the other two sides.

Question 10

State
the converse of Pythagoras’ theorem.

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.

Question 11

If
D,E and F are respectively the midpoints of sides
AB,BC and CA of ∆ABC then what is
the ratio of the areas of ∆DEF and ∆ABC?

Solution 11

Question 12

Two
triangle ABC and PQR are such that AB =3 cm, AC =6 cm, ∠A
=700 ,PR =9 cm, ∠P = 700 and
PQ = 4.5 cm. Show that ∆ABC ∼ ∆PQR
and state the similarity criterion.

Solution 12

Question 13

If
∆ABC

∆DEF
such that 2 AB = DE and BC = 6 cm, find EF.

Solution 13

Question 14

In
the given figure, DE ∥ BC such that AD =
x cm , DB
=(3x+4) cm, AE = (x+3) cm and EC = (3x +19) cm. Find the value of x.

Solution 14

Question 15

A
ladder 10 m long reaches the window of a house 8 m above the ground. Find the
distance of the foot of the ladder from the base of the wall.

Solution 15

Question 16

Find
the length of the altitude of an equilateral triangle of side 2a
cm.

Solution 16

Question 17

∆ABC ∼

DEF such that ar (∆ABC) =64 cm2
and or (∆DEF)
= 169 cm2. If BC = 4 cm , find EF.

Solution 17

Question 18

In
a trapezium ABCD, it is given that AB ∥ CD and AB = 2 CD.Its diagonals AC and BD intersect at the point O such
that ar (∆AOB) = 84 cm2.
Find ar (∆COD)

Solution 18

Question 19

The
corresponding sides of two similar triangles are in the ratio 2:3. If the
area of the smaller triangle is 48 cm2, find the area of the
larger triangle.

Solution 19

Question 20

Solution 20

Question 21

Find
the length of each side of a rhombus whose diagonals are 24 cm and 10 cm
long.

Solution 21

Question 22

Two
triangles DEF and GHK are such that ∠D = 480
and ∠H
= 570. If ∆DEF ∼ ∆ GHK then find the measure of ∠F.

Solution 22

Question 23

Solution 23

Question 24

In
triangles BMP and CNR it is given that PB = 5 cm, MP = 6 cm, BM = 9 cm and NR
= 9 cm. If ∆BMP ∼ ∆CNR
then find the perimeter of ∆CNR.

Solution 24

Question 25

Each
of the equal sides of an isosceles triangle is 25 cm. Find the length of it
altitude if the base is 14 cm.

Solution 25

Question 26

A
man goes 12 m due south and then 35 m due west. How far is he from the
starting point?

Solution 26

Question 27

If
the lengths of the sides BC, CA and AB of a ∆ABC
are a ,
b and c respectively and AD is the bisector of ∠A then find the
lengths of BD and DC.

Solution 27

Question 28

In
the given figure, ∠AMN = ∠MBC
= 760. If a,b
and c are the lengths of AM, MB and BC respectively then express the length
of MN in terms of a,b and c.

Solution 28

Question 29

The
lengths of the diagonals of a rhombus are 40 cm and 42 cm. Find the length of
each side of the rhombus.

Solution 29

Question 30

For
each of the following statements state whether true (T) or false (F):

Two
circles with different radii are similar.

Solution 30

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.

Question 31

For
each of the following statements state whether true (T) or false (F):

Any
two rectangles are similar.

Solution 31

Two rectangles are similar if their corresponding sides are
proportional.

So, the statement is false.

Question 32

For
each of the following statements state whether true (T) or false (F):

If
two triangles are similar then their corresponding angles are equal and their
corresponding sides are equal.

Solution 32

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.

Question 33

For
each of the following statements state whether true (T) or false (F):

The
length of the line segment joining the midpoints of any two sides of a
triangle is equal to half the length of the third side.

Solution 33

Question 34

For
each of the following statements state whether true (T) or false (F):

In
a ∆ABC,
AB = 6 cm, ∠A
= 45° and AC = 8 cm and in a ∆DEF, DF = 9 cm, ∠D
= 45° and DE = 12 cm, then ∆ABC ∼ ∆DEF.

Solution 34

Question 35

For
each of the following statements state whether true (T) or false (F):

The
polygon formed by joining the midpoints of the sides of a quadrilateral is a
rhombus.

Solution 35

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.

Question 36

For
each of the following statements state whether true (T) or false (F):

The
ratio of the areas of two similar triangles is equal to the ratio of their
corresponding angle-bisector segments.

Solution 36

Question 37

For
each of the following statements state whether true (T) or false (F):

The
ratio of the perimeters of two similar triangles is the same as the ratio of
their corresponding medians.

Solution 37

Question 38

For
each of the following statements state whether true (T) or false (F):

If
O is any point inside a rectangle ABCD then OA2 + OC2 =
OB2 + OD2.

Solution 38

Question 39

For
each of the following statements state whether true (T) or false (F):

The
sum of the squares on the sides of a rhombus is equal to the sum of the
squares on its diagonals.

Solution 39

Chapter 4 – Triangles Exercise FA

Question 1

∆ABC ∆DEF and their
perimeters are 32 cm and 24 cm respectively. If AB = 10 cm then DE = ?

(a) 8 cm

(b) 7.5 cm

(c) 15 cm

(d)

Solution 1

Question 2

In the given figure, DE BC. If DE = 5 cm, BC
= 8 cm and AD = 3.5 cm then AB =?

(a) 5.6 cm

(b) 4.8 cm

(c) 5.2 cm

(d) 6.4 cm

Solution 2

Question 3

Two poles of height 6 m
and 11 m stand vertically upright on a plane ground. If the distance between
their feet is 12 m then the distance between their tops is

(a) 12 m

(b) 13 m

(c) 14 m

(d) 15 m

Solution 3

Question 4

The areas of two similar
triangles are 25 cm2 and 36 cm2 respectively. If the
altitude of the first triangle is 3.5 cm then the corresponding altitude of
the other triangle is

(a) 5.6 cm

(b) 6.3 cm

(c) 4.2 cm

(d) 7 cm

Solution 4

Question 5

If ∆ABC ∆DEF such that 2AB =
DE and BC = 6 cm, find EF.

Solution 5

Question 6

In the given figure, DE BC such that AD = x
cm, DB = (3x+4) cm, AE = (x+3) cm and EC = (3x+19) cm. Find the value of x.

Solution 6

Question 7

A 10 m long reaches the
window of a house 8 m above the ground. Find the distance of the foot of the
ladder from the base of the wall.

Solution 7

Question 8

Find the length of the
altitude of an equilateral triangle of side 2a cm.

Solution 8

Question 9

∆ABC ∆DEF such that
ar(∆ABC)=64 cm2 and ar(∆DEF)=169 cm2. If BC=4 cm, find
EF.

Solution 9

Question 10

In a trapezium ABCD, it is given that AB CD and AB = 2CD. Its diagonals AC and BD intersect at the point O such that ar(∆AOB)=84 cm2. Find ar(∆COD).

Solution 10

Question 11

The corresponding sides
of two similar triangles are in the ratio 2 : 3. If the area of the smaller
triangle is 48 cm2, find the area of the larger triangle.

Solution 11

Question 12

Solution 12

Question 13

Prove that the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

Solution 13

Question 14

Solution 14

Question 15

Find the length of each
side of a rhombus whose diagonals are 24 cm and 10 cm long.

Solution 15

Question 16

Prove that the ratio of
the perimeters of two similar triangles is the same as the ratio of their
corresponding sides.

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

In the given figure,
∆ABC is an obtuse triangle, obtuse-angled at B. If AD
CB (produced) prove
that AC2=AB2+BC2+2BC.BD.

Solution 19

Question 20

Solution 20

Chapter 4 – Triangles Exercise MCQ

Question 1

A man goes 24 m due west and then 10 m due north. How far is he from the starting point?

(a) 34 m

(b) 17 m

(c) 26 m

(d) 28 m

Solution 1

Question 2

Two poles of height 13 in and 7 m respectively stand vertically on a plane ground at a distance of 8 m from each other. The distance between their tops is

(a) 9 m

(b) 10 m

(c) 11 m

(d) 12 m

Solution 2

Question 3

A vertical stick 1.8 m long casts a shadow 45 cm long on the ground. At the same time, what is the length of the shadow of a pole 6 m high?

(a) 2.4 in

(b) 1.35 m

(c) 1.5 m

(d) 13.5 m

Solution 3

Question 4

A vertical pole 6 m long casts a shadow of length 3.6 m on the ground. What is the height of a tower which casts a shadow of length 18 m at the same time?

(a) 10.8 m

(b) 28.8 m

(c) 324 m

(d) 30 m

Solution 4

Question 5

The shadow of a 5-m-long stick is 2 m long. At the same time the length of the shadow of a 12.5-m-high tree (in m) is

(a) 3.0

(b) 3.5

(c) 4.5

(d) 5.0

Solution 5

Question 6

A ladder 25 m long just reaches the top of a building 24 m high from the ground. What is the distance of the foot of the ladder from the building?

(a) 7 m

(b) 14 m

(c) 21 m

(d) 24.5 m

Solution 6

Question 7

In the given figure, O is a point inside a ∆MNP such that MOP = 90°, OM = 16 cm and OP = 12 cm. If MN = 21 cm and NMP = 90° then NP = ?

(a) 25 cm

(b) 29 cm

(c) 33 cm

(d) 35 an

Solution 7

Question 8

The hypotenuse of a right triangle is 25 cm. The other two sides are such that one is 5 an longer than the other. The lengths of these sides are

(a) 10 cm, 15 cm

(b) 15 cm, 20 cm

(c) 12 cm, 17 cm

(d) 13 cm, 18 cm

Solution 8

Question 9

The height of an equilateral triangle having each side 12cm, is

Solution 9

Question 10

∆ABC is an isosceles triangle with AB =AC = 13 cm and the length of altitude from A on BC is 5 cm. Then, BC=?

(a) 12 cm

(b) 16 cm

(c) 18 cm

(d) 24 cm

Solution 10

Question 11

In a ∆ABC it is given that AB = 6 cm, AC = 8 cm and AD is the bisector of A. Then, BD: DC  = ?

(a) 3 : 4

(b) 9 : 16

(c) 4 : 3

(d)

Solution 11

Question 12

In a ∆ABC it is given that AD is the internal bisector of A. If BD=4cm, DC=5cm and AB=6 cm, then AC=?

(a) 4.5 cm

(b) 8 cm

(c) 9 cm

(d) 7.5 cm

Solution 12

Question 13

In a ∆ABC, it is given that AD is the internal bisector of A. If AB = 10 cm, AC = 14 cm and BC = 6 cm, then CD = ?

(a) 4.8 cm

(b) 3.5 cm

(c) 7 cm

(d) 10.5 cm

Solution 13

Question 14

In a triangle, the perpendicular from the vertex to the base bisects the base. The triangle is

(a) right-angled

(b) isosceles

(c) scalene

(d) obtuse-angled

Solution 14

Question 15

In an equilateral triangle ABC, if AD ⊥  BC then which of the following is true?

Solution 15

Question 16

In a rhombus of side 10 cm, one of the diagonals is 12 cm long. The length of the second diagonal is

(a) 20 cm

(b) 18 cm

(c) 16 cm

(d) 22 cm

Solution 16

Question 17

The lengths of the diagonals of a rhombus are 24 cm and 10 cm. The length of each side of the rhombus is

(a) 12 cm

(b) 13 cm

(c) 14 cm

(d) 17 cm

Solution 17

Question 18

If the diagonals of a quadrilateral divide each other proportionally then it is a

(a) parallelogram

(b) trapezium

(c) rectangle

(d) square

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.

Question 19

In the given figure, ABCD is a trapezium whose diagonals AC and BD intersect at O such that OA=(3x-1) cm, OB=(2x+1)cm, OC=(5x-3)cm and OD=(6x-5)cm. Then, x=?

(a) 2

(b) 3

(c) 2.5

(d) 4

Solution 19

Question 20

The line segments joining the midpoints of the adjacent sides of a quadrilateral form

(a) a parallelogram

(b) a rectangle

(c) a square

(d) a rhombus

Solution 20

Correct option: (a)

The line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram as shown below.

Question 21

If the bisector of an angle of a triangle bisects the opposite side them the triangle is

(a) scalene

(b) equilateral

(c) isosceles

(d) right-angled

Solution 21

Question 22

(a) 30°

(b) 40°

(c) 45°

(d) 50°

Solution 22

Question 23

In ∆ABC, DE BC so that AD = 2.4 cm, AE= 3.2 cm and EC = 4.8 cm. Then, AB =?

(a) 3.6 cm

(b) 6 cm

(c) 6.4 cm

(d) 7.2 cm

Solution 23

Question 24

In a ∆ABC, if DE is drawn parallel to BC, cutting AB and AC at D and E respectively such that AB = 7.2 cm, AC = 6.4 cm and AD = 4.5 cm. Then, AE =?

(a) 5.4 cm

(b) 4 cm

(c) 3.6 cm

(d) 3.2 cm

Solution 24

Question 25

In ∆ABC, DE BC so that AD = (7x-4) cm, AE = (5x-2) cm, DB = (3x+4) cm and EC = 3x cm. Then, we have

(a) x = 3

(b) x = 5

(c) x = 4

(d) x = 2.5

Solution 25

Question 26

(a) 4.2 cm

(b) 3.1 cm

(c) 2.8 cm

(d) 2.1 cm

Solution 26

Question 27

∆ABC ∆DEF and the perimeters of ∆ABC and ∆DEF are 30 cm and 18 cm respectively. If BC = 9 cm then EF=?

(a) 6.3 cm

(b) 5.4 cm

(c) 7.2 cm

(d) 4.5 cm

Solution 27

Question 28

∆ABC ∆DEF such that AB = 9.1 cm and DE = 6.5 cm. If the perimeter of ∆DEF is 25cm, what is the perimeter of ∆ABC?

(a) 35 cm

(b) 28 cm

(c) 42 cm

(d) 40 cm

Solution 28

Question 29

In ∆ABC, it is given that AB=9 cm, BC = 6cm and CA = 7.5 cm. Also, ∆DEF is given such that EF = 8cm and ∆DEF ∆ABC. Then, perimeter of ∆DEF is

(a) 22.5 cm

(b) 25 cm

(c) 27 cm

(d) 30 cm

Solution 29

Question 30

ABC and BDE are two equilateral triangles such that D is the midpoint of BC. Ratio of the areas of triangles ABC and BDE is

(a) 1 : 2

(b) 2 : 1

(c) 1 : 4

(d) 4 : 1

Solution 30

Question 31

It is given that ∆ABC ∆DFE. If A = 30°, C=50°, AB = 5cm, AC = 8 cm and DF = 7.5 cm then which of the following is true?

(a) DE = 12 cm, F = 50°

(b) DE = 12 cm, F = 100°

(c) EF = 12 cm, D = 100°

(d) EF = 12 cm, D = 30°

Solution 31

Question 32

In the given figure, BAC = 90° and AD BC. Then,

(a) BC·CD = BC2

(b) AB·AC = BC2

Solution 32

Question 33

(a) 45°

(b) 60°

(c) 90°

(d) 120°

Solution 33

Question 34

(a) B = E

(b) A = D

(c) B = D

(d) A = F

Solution 34

Question 35

In ∆DEF and ∆PQR, it is given that D = Q and R = E, then which of the following is not true?

Solution 35

Question 36

If ∆ABC ∆EDF and ∆ABC is not similar to ∆DEF then which of the following is not true?

(a) BC·EF = AC·FD

(b) AB·EF = AC·DE

(c) BC·DE = AB·EF

(d) BC·DE = AB·FD

Solution 36

Question 37

In ∆ABC and ∆DEF, it is given that B = E, F = C and AB = 3DE, then the two triangles are

(a) congruent but not similar

(b) similar but not congruent

(c) neither congruent nor similar

(d) similar as well as congruent

Solution 37

Question 38

In the given figure, 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, APB = 50° and CDP=30° then PBA=?

(a) 50°

(b) 30°

(c) 60°

(d) 100°

Solution 38

Question 39

(a) ∆PQR ∆CAB

(b) ∆PQR ∆ABC

(c) ∆CBA ∆PQR

(d) ∆BCA ∆PQR

Solution 39

Question 40

Corresponding 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) 9 : 4

(d) 16 : 81

Solution 40

Question 41

Solution 41

Question 42

In an equilateral ∆ABC, D is the midpoint of AB and E is the midpoint of AC. Then, ar(∆ABC) : ar(∆ADE)=

?

(a) 2 : 1

(b) 4 : 1

(c) 1 : 2

(d) 1 : 4

Solution 42

Question 43

(a) 5 : 7

(b) 25 : 49

(c) 49 : 25

(d) 125 : 343

Solution 43

Question 44

∆ABC ∆DEF such that ar(∆ABC) = 36 cm2 and ar(∆DEF)=49 cm2. Then, the ratio of their corresponding sides is

(a) 36 : 49

(b) 6 : 7

(c) 7 : 6

(d)

Solution 44

Question 45

Two isosceles triangles have their corresponding angles equal and their areas are in the ratio 25:36. The ratio of their corresponding heights is

(a) 25 : 36

(b) 36 : 25

(c) 5 : 6

(d) 6 : 5

Solution 45

Question 46

The line segments joining the midpoints of the sides of a triangle form four triangles, each of which is

(a) congruent to the original triangle

(b) similar to the original triangle

(c) an isosceles triangle

(d) an equilateral triangle

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.

Question 47

(a) 8 cm

(b) 10 cm

(c) 12 cm

(d)

Solution 47

Question 48

In the given figure, O is the point of intersection of two chords AB and CD such that OB = OD and AOC=45°. Then, ∆OAC and ∆ODB are

(a) equilateral and similar

(b) equilateral but not similar

(c) isosceles and similar

(d) isosceles but not similar

Solution 48

Question 49

In an isosceles ∆ABC, if AC = BC and AB2 = 2AC2 then C =?

(a) 30°

(b) 45°

(c) 60°

(d) 90°

Solution 49

Question 50

In ∆ABC, if AB = 16 cm, BC = 12 cm and AC = 20 cm, then ∆ABC is

(a) acute-angled

(b) right-angled

(c) obtuse-angled

(d) not possible

Solution 50

Question 51

Which of the following is a true statement?

(a) Two similar triangles are always congruent.

(b) Two figures are similar if they have the same shape and size.

(c) Two triangles are similar if their corresponding sides are proportional.

(d) Two polygons are similar if their corresponding sides are proportional.

Solution 51

Question 52

Which of the following is a false statement?

(a) If the areas of two similar triangles are equal then the triangles are congruent.

(b) The ratio of the areas of two similar triangles is equal to the ratio of their corresponding sides.

(c) The ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding medians.

(d) The ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

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.

Question 53

Match the following columns

 Column I Column II (a) p 6 (b) If ∆ABC ∼ ∆DEF such that 2AB = 3DE and BC = 6 cm then EF = ……. cm. q 4 (c) If ∆ABC ∼ ∆PQR such that ar(∆ABC) : ar(∆PQR) = 9 : 16 and BC = 4.5 cm then QR = …….. cm r 3 (d) In the given figure, AB ∥ CD and OA = (2x+4) cm, OB = (9x-21) cm, OC = (2x-1) cm and OD = 3 cm. Then x =? s 2.1

Solution 53

Question 54

Match the following columns

 Column I Column II (a) A man goes 10m due east and then 20 m due north. His distance from the starting point is ……..m. p (b) In an equilateral triangle with each side 10 cm, the altitude is ………cm. q (c) The area of an equilateral triangle having each side 10 cm is …….. cm2. r (d) The length of diagonal of a rectangle having length 8 m and breadth 6 m is …… m. s 10

Solution 54

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