R S AGGARWAL AND V AGGARWAL Solutions for Class 10 Maths Chapter 4 – Triangles
Chapter 4 – Triangles Exercise Ex. 4A
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
(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
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
(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
_{ }_{ }
_{ }
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
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
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
(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
M
is a point on the side BC of a parallelogram ABCD.DM when produced meets AB
product at N. Prove that
Show that the line segment which joins the midpoints of the oblique sides of a trapezium is parallel to the parallel sides.
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
In the adjoining figure, ABCD is a trapezium in which CDAB 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.
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
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.
ABC and DBC lie on the same side of BC, as shown in the figure. From a point P on BC, PQ  AB and PRBD are drawn, meeting AC at Q and CD at R respectively. Prove that QRAD.
?
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.
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 EFBC.
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
_{}
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.
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.
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.
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
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.
Chapter 4 – Triangles Exercise Ex. 4B
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)
(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)
In the given figure, ODC _{}OBA, _{}BOC = 115^{o} and _{}CDO = 70^{o}. Find (i) _{}DOC (ii) _{}DCO(iii) _{}OAB(iv) _{}OBA
ODC ~ OBC
_{}BOC = 115^{o}
_{}CDO = 70^{o}
_{(i) }DOC = (180^{o} – _{}BOC)
= (180^{o} – 115^{o})
= 65^{o}
_{(ii) }OCD = 180^{o} – CDO – DOC
_{ }OCD = 180^{o} – (70^{o} + 65^{o})
= 45^{o}
(iii) Now, ABO ~ ODC
_{ }AOB = _{}COD (vert. Opp _{}s) = 65^{o}
_{ }OAB = _{}OCD = 45^{o}
_{(iv) }OBA = _{}ODC(alternate angles) = 70^{o}
So, _{}OAB = 45^{o} and _{}OBA = 70^{o}
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
Given: OAB _{} OCD
AB = 8 cm, BO = 6.4 cm, CD = 5 cm, OC = 3.5 cm
_{}
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
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
The perimeters of two similar triangles ABC and PQR are 32cm and 24cm respectively. If PQ = 12cm, find AB.
ABC and PQR are similar triangles, therefore corresponding sides of both the triangles are proportional.
_{}
Hence, AB = 16 cm
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.
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
In the given figure, _{}CAB = 90^{o} and AD _{}BC. Show that BDA ~ BAC. If AC = 75 cm, AB = 1m and BC = 1.25 m, find AD.
Given: AB = 100 cm, BC = 125 cm, AC = 75 cm
Proof:
In BAC and BDA
_{}BAC = _{}BDA = 90^{o}
_{}B = _{}B (common)
BAC _{}BDA(by AA similarities)
_{}
Therefore, AD = 60 cm
In the given figure, _{}ABC = 90^{o} and BD _{}AC. If AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, find BC.
Given that AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm
In CBA and CDB
_{}CBA = _{}CDB = 90^{o}
_{}C = _{}C (Common)
Therefore, CBA _{} CDB (by AA similarities)
_{}
Hence, BC = 8.1 cm
In the given figure, _{}ABC = 90^{o} and BD _{}AC. If BD = 8 cm, AD = 4 cm, find CD.
Given that BD = 8 cm, AD = 4 cm
In DBA and DCB, we have
_{}BDA = _{}CDB = 90^{o}
_{ }_{}DBA = _{}DCB [each = 90^{o} – _{}A]
_{}DBA _{}DCB (by AAA similarity)
_{}
Hence, CD = 16 cm
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
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.
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
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
_{A}FD = _{}EFB (vertically opposite _{}s)
DAF = BEF (Alternate _{}s)
_{}
Hence proved.
In the given figure, DB _{}BC, DE _{}AB and AC _{}BC.
Prove that _{}
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 = 90^{o}
_{}DBE = _{}BAC
_{}~ _{}[By AA similarity]
_{}
Hence proved.
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.
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.
In an isosceles ABC, the base AB is produced both ways in P and Q such that AP × BQ =AC^{2}. Prove that ACP _{}BCQ
In ACP and BCQ
CA = CB
_{}_{}CAB = _{}CBA
_{}
_{}ACP _{}BCQ
In the given figure, _{}1 = _{}2and _{}
Prove that ACB _{}DCE
_{}1 = _{}2 (given)
_{ } (given)_{ }
Also, _{}2 = _{}1 _{}
Therefore, by SAS similarity criterion ACB ~ DCE
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.
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 _{ }(Midpoint theorem)
Similarly,
_{}SP  RQ and PQ  SR and PQ = RQ = SP = SR
Hence,PQRS is a rhombus.
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.
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.
In
a right triangle ABC, rightangle at B, D is a point on hypotenuse such that
BD ⊥
AC. If DP ⊥
AB and DQ ⊥
BC then prove that
a. DQ^{2}= DP.QC
b. DP^{2}=DQ. AP
Chapter 4 – Triangles Exercise Ex. 4C
ABC _{} DEF and their areas are respectively _{} and _{}. If EF = 15.4 cm, find BC
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
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.
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
ABC _{} PQR and ar(ABC) = 4 ar(PQR). If BC = 12 cm, find QR.
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
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.
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.
ABC _{} DEF and their areas are respectively 100_{} and 49 _{}. If the altitude of ABC is 5 cm, find the corresponding altitude of DEF.
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
The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas.
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
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.
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
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.
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
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.
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.
In the given figure, DEBC. If DE = 3 cm, BC = 6cm and ar(ADE) =_{}, find the area of ABC.
Given DE  BC
DE = 3 cm and BC = 6 cm
ar(ADE) = 15
In ADE and ABC, we have
_{}
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.
In BAC and ADC, we have
_{}BAC = _{}ADC = 90^{o} (AD _{}BC)
_{}ACB = _{}DCA (common)
BAC _{}ADC
_{}
Therefore, the ratio of the areas of ABC and ADC = 169:25
In the given figure, DEBC and DE : BC = 3 : 5. Calculate the ratio of the areas of ADE and the trapezium BCED.
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) = 9x^{2} units
Then, ar(ABC) = 25x^{2} units
_{}
Therefore, ratio of ar(ADE) to the ar(trap BCED) = 9:16
In ABC, D and E are midpoints of AB and AC respectively. Find the ratio of the areas of ADE and ABC.
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
Chapter 4 – Triangles Exercise Ex. 4D
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) (a1)cm, _{}cm, (a + 1)cm
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
Aman goes 80 m due east and then 150 m due north. How far is he from the starting point?
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 and_{}ABC = 90^{o}
From right ABC, we have
_{}
Hence, the man is 170m northeast from the starting point.
A man goes 10 m due south and then 24 m due west. How far is he from the starting point?
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 = 90^{o}
Using Pythagoras theorem:
_{}
Hence, the man is 26 m southwest from the starting position.
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.
Let AB be the building and CB be the ladder.
Then,
AB = 12 m, CB = 13 m and CAB = 90^{o}
By Pythagoras theorem, we have
Hence, the distance of the foot of the ladder from the building is 5 m.
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.
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 = 90^{o}
By Pythagoras theorem, we have
Hence, the length of ladder is 25 m.
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.
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.
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?
In the given figure, O is a point inside a PQR such that PQR = 90^{o}, OP = 6 cm and OR = 8 cm. If PQ = 24 cm and QR = 26 cm, prove that PQR is rightangled.
In PQR, _{}QPR = 90^{o}, PQ = 24 cm, and QR = _{}
In POR, PO = 6 cm, QR = 8cm and _{}POR = 90^{o}
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.
ABC is an isosceles triangle with AB = AC = 13 cm. The length of altitude from A on BC is 5 cm. Find BC.
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 = 90^{o}
In ALC,
_{ }ALC = 90^{o}
Find the length of altitude AD of an isosceles ABC in which AB = AC = 2a units and BC = a units.
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
_{}
ABC is an equilateral triangle of side 2a units. Find each of its altitudes.
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 = 90^{o}
Hence, length of each altitude is _{}
Find the height of an equilateral triangle of side 12 cm.
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 _{}
Find the length of a diagonal of a rectangle whose adjacent sides are 30 cm and 16 cm.
Let ABCD is the given rectangle, let BD is a diagonal making a ADB.
_{}BAD = 90^{o}
Using Pythagoras theorem:
_{}
Hence, length of diagonal DB is 34 cm.
Find the length of each side of a rhombus whose diagonals are 24 cm and 10 cm long.
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
In ABC, D is the midpoint of BC and AE BC. If AC > AB, show that _{}
Given: ABC in which D is the midpoint of BC. AE BC and AC > AB.
Then, BD = CD and AED = 90^{o},
Then, _{}ADE < 90^{o} and ADC > 90^{o}
In AED,
Putting value of _{}from (1) in (2), we get
_{}
In ABC, AB = AC. Side BC is produced to D. Prove that
_{}
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.
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)_{}
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 = 90^{o}
(i) In AEC, AEC = 90^{o}
_{}
_{}
_{}
(ii)In ABE, ABE = 90^{o}
_{}
(iii)Adding (1) and (2), we get
_{}
(iv)Subtracting (2) from (1), we get
_{}
In
∆ABC
,AB = AC. Side BC is produced to D. Prove that
(AD^{2}AC^{2})=BD.
CD.
ABC is an isosceles triangle, rightangled at B. Similar triangles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of ABE and ACD.
ABC is an isosceles triangle right angled at B,
Let AB = BC = x cm
By Pythagoras theorem,
_{}
In
a ∆ABC,
AD is a median and AL ⊥ BC. Prove that:
Naman is doing flyfishing 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?
Chapter 4 – Triangles Exercise Ex. 4E
State
the two properties which are necessary for given two triangles to be similar.
Two triangles are said to be similar to each other if:
(i) their corresponding angles are equal, and
(ii)
their corresponding sides are proportional.
State
the 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 sides are divided in
the same ratio.
State
the converse of ‘Thales’ theorem.
If a line divides any two sides of a triangle in the same ratio then
the line must be parallel to the third side.
State
the midpoint theorem.
The line segment joining the midpoints of any two sides of a
triangle is parallel to the third side.
State
the AAAsimilarity criterion.
If in any two triangles, the corresponding angles are equal, then
their corresponding sides are proportional and hence the triangles are
similar.
State
the AAsimilarity criterion.
If two angles of one triangle are respectively equal to two angles of
another triangle then the two triangles are similar.
State
the SSScriterion for similarity of triangles.
If the corresponding sides of two triangles are proportional then
their corresponding angles are equal, and hence the two triangles are
similar.
State
the SASsimilarity criterion.
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.
State
Pythagoras’ theorem.
In a right triangle, the square of the hypotenuse is equal to the sum
of the squares of the other two sides.
State
the converse of Pythagoras’ theorem.
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.
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?
Two
triangle ABC and PQR are such that AB =3 cm, AC =6 cm, ∠A
=70^{0 },PR =9 cm, ∠P = 70^{0 }and
PQ = 4.5 cm. Show that ∆ABC ∼ ∆PQR
and state the similarity criterion.
If
∆ABC
∼
∆DEF
such that 2 AB = DE and BC = 6 cm, find EF.
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.
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.
Find
the length of the altitude of an equilateral triangle of side 2a
cm.
∆ABC ∼
∆
DEF such that ar (∆ABC) =64 cm^{2}
and or (∆DEF)
= 169 cm^{2}. If BC = 4 cm , find EF.
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 cm^{2}.
Find ar (∆COD)
The
corresponding sides of two similar triangles are in the ratio 2:3. If the
area of the smaller triangle is 48 cm^{2}, find the area of the
larger triangle.
Find
the length of each side of a rhombus whose diagonals are 24 cm and 10 cm
long.
Two
triangles DEF and GHK are such that ∠D = 48^{0}
and ∠H
= 57^{0}. If ∆DEF ∼ ∆ GHK then find the measure of ∠F.
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.
Each
of the equal sides of an isosceles triangle is 25 cm. Find the length of it
altitude if the base is 14 cm.
A
man goes 12 m due south and then 35 m due west. How far is he from the
starting point?
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.
In
the given figure, ∠AMN = ∠MBC
= 76^{0}. 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.
The
lengths of the diagonals of a rhombus are 40 cm and 42 cm. Find the length of
each side of the rhombus.
For
each of the following statements state whether true (T) or false (F):
Two
circles with different radii are similar.
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.
For
each of the following statements state whether true (T) or false (F):
Any
two rectangles are similar.
Two rectangles are similar if their corresponding sides are
proportional.
So, the statement is false.
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.
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.
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.
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.
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.
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.
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 anglebisector segments.
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.
For
each of the following statements state whether true (T) or false (F):
If
O is any point inside a rectangle ABCD then OA^{2} + OC^{2} =
OB^{2} + OD^{2}.
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.
Chapter 4 – Triangles Exercise FA
∆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)
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
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
The areas of two similar
triangles are 25 cm^{2} and 36 cm^{2} 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
If ∆ABC ∼ ∆DEF such that 2AB =
DE and BC = 6 cm, find EF.
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.
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.
Find the length of the
altitude of an equilateral triangle of side 2a cm.
∆ABC ∼ ∆DEF such that
ar(∆ABC)=64 cm^{2} and ar(∆DEF)=169 cm^{2}. If BC=4 cm, find
EF.
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 cm^{2}. Find ar(∆COD).
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.
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.
Find the length of each
side of a rhombus whose diagonals are 24 cm and 10 cm long.
Prove that the ratio of
the perimeters of two similar triangles is the same as the ratio of their
corresponding sides.
In the given figure,
∆ABC is an obtuse triangle, obtuseangled at B. If AD ⊥ CB (produced) prove
that AC^{2}=AB^{2}+BC^{2}+2BC.BD.
Chapter 4 – Triangles Exercise MCQ
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
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
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
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
The shadow of a 5mlong stick is 2 m long. At the same time the length of the shadow of a 12.5mhigh tree (in m) is
(a) 3.0
(b) 3.5
(c) 4.5
(d) 5.0
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
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
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
The height of an equilateral triangle having each side 12cm, is
∆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
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)
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
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
In a triangle, the perpendicular from the vertex to the base bisects the base. The triangle is
(a) rightangled
(b) isosceles
(c) scalene
(d) obtuseangled
In an equilateral triangle ABC, if AD ⊥ BC then which of the following is true?
(a) 2AB^{2}=3AD^{2}
(b) 4AB^{2}=3AD^{2}
(c) 3AB^{2}=4AD^{2}
(d) 3AB^{2}=2AD^{2}
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
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
If the diagonals of a quadrilateral divide each other proportionally then it is a
(a) parallelogram
(b) trapezium
(c) rectangle
(d) square
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.
In the given figure, ABCD is a trapezium whose diagonals AC and BD intersect at O such that OA=(3x1) cm, OB=(2x+1)cm, OC=(5x3)cm and OD=(6x5)cm. Then, x=?
(a) 2
(b) 3
(c) 2.5
(d) 4
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
Correct option: (a)
The line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram as shown below.
If the bisector of an angle of a triangle bisects the opposite side them the triangle is
(a) scalene
(b) equilateral
(c) isosceles
(d) rightangled
(a) 30°
(b) 40°
(c) 45°
(d) 50°
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
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
In ∆ABC, DE ∥ BC so that AD = (7x4) cm, AE = (5x2) 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
(a) 4.2 cm
(b) 3.1 cm
(c) 2.8 cm
(d) 2.1 cm
∆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
∆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
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
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
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°
In the given figure, ∠BAC = 90° and AD ⊥ BC. Then,
(a) BC·CD = BC^{2}
(b) AB·AC = BC^{2}
(c) BD·CD = AD^{2}
(d) AB·AC = AD^{2}
(a) 45°
(b) 60°
(c) 90°
(d) 120°
(a) ∠B = ∠E
(b) ∠A = ∠D
(c) ∠B = ∠D
(d) ∠A = ∠F
In ∆DEF and ∆PQR, it is given that ∠D = ∠Q and ∠R = ∠E, then which of the following is not true?
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
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
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°
(a) ∆PQR ∼ ∆CAB
(b) ∆PQR ∼ ∆ABC
(c) ∆CBA ∼ ∆PQR
(d) ∆BCA ∼ ∆PQR
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
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
(a) 5 : 7
(b) 25 : 49
(c) 49 : 25
(d) 125 : 343
∆ABC ∼ ∆DEF such that ar(∆ABC) = 36 cm^{2} and ar(∆DEF)=49 cm^{2}. Then, the ratio of their corresponding sides is
(a) 36 : 49
(b) 6 : 7
(c) 7 : 6
(d)
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
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
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.
(a) 8 cm
(b) 10 cm
(c) 12 cm
(d)
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
In an isosceles ∆ABC, if AC = BC and AB^{2} = 2AC^{2} then ∠C =?
(a) 30°
(b) 45°
(c) 60°
(d) 90°
In ∆ABC, if AB = 16 cm, BC = 12 cm and AC = 20 cm, then ∆ABC is
(a) acuteangled
(b) rightangled
(c) obtuseangled
(d) not possible
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.
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.
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.
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 = (9x21) cm, OC = (2x1) cm and OD = 3 cm. Then x =?  s  2.1 
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 …….. cm^{2}.  r  
(d)  The length of diagonal of a rectangle having length 8 m and breadth 6 m is …… m.  s  10 