
1.Number System
14
Lecture1.1

Lecture1.2

Lecture1.3

Lecture1.4

Lecture1.5

Lecture1.6

Lecture1.7

Lecture1.8

Lecture1.9

Lecture1.10

Lecture1.11

Lecture1.12

Lecture1.13

Lecture1.14


2.Polynomials
10
Lecture2.1

Lecture2.2

Lecture2.3

Lecture2.4

Lecture2.5

Lecture2.6

Lecture2.7

Lecture2.8

Lecture2.9

Lecture2.10


3.Coordinate Geometry
8
Lecture3.1

Lecture3.2

Lecture3.3

Lecture3.4

Lecture3.5

Lecture3.6

Lecture3.7

Lecture3.8


4.Linear Equations
8
Lecture4.1

Lecture4.2

Lecture4.3

Lecture4.4

Lecture4.5

Lecture4.6

Lecture4.7

Lecture4.8


5.Euclid's Geometry
7
Lecture5.1

Lecture5.2

Lecture5.3

Lecture5.4

Lecture5.5

Lecture5.6

Lecture5.7


6.Lines and Angles
10
Lecture6.1

Lecture6.2

Lecture6.3

Lecture6.4

Lecture6.5

Lecture6.6

Lecture6.7

Lecture6.8

Lecture6.9

Lecture6.10


7.Triangles
11
Lecture7.1

Lecture7.2

Lecture7.3

Lecture7.4

Lecture7.5

Lecture7.6

Lecture7.7

Lecture7.8

Lecture7.9

Lecture7.10

Lecture7.11


8.Quadrilaterals
13
Lecture8.1

Lecture8.2

Lecture8.3

Lecture8.4

Lecture8.5

Lecture8.6

Lecture8.7

Lecture8.8

Lecture8.9

Lecture8.10

Lecture8.11

Lecture8.12

Lecture8.13


9.Area of Parallelogram
11
Lecture9.1

Lecture9.2

Lecture9.3

Lecture9.4

Lecture9.5

Lecture9.6

Lecture9.7

Lecture9.8

Lecture9.9

Lecture9.10

Lecture9.11


10.Constructions
7
Lecture10.1

Lecture10.2

Lecture10.3

Lecture10.4

Lecture10.5

Lecture10.6

Lecture10.7


11.Circles
11
Lecture11.1

Lecture11.2

Lecture11.3

Lecture11.4

Lecture11.5

Lecture11.6

Lecture11.7

Lecture11.8

Lecture11.9

Lecture11.10

Lecture11.11


12.Heron's Formula
8
Lecture12.1

Lecture12.2

Lecture12.3

Lecture12.4

Lecture12.5

Lecture12.6

Lecture12.7

Lecture12.8


13.Surface Area and Volume
16
Lecture13.1

Lecture13.2

Lecture13.3

Lecture13.4

Lecture13.5

Lecture13.6

Lecture13.7

Lecture13.8

Lecture13.9

Lecture13.10

Lecture13.11

Lecture13.12

Lecture13.13

Lecture13.14

Lecture13.15

Lecture13.16


14.Statistics
15
Lecture14.1

Lecture14.2

Lecture14.3

Lecture14.4

Lecture14.5

Lecture14.6

Lecture14.7

Lecture14.8

Lecture14.9

Lecture14.10

Lecture14.11

Lecture14.12

Lecture14.13

Lecture14.14

Lecture14.15


15.Probability
8
Lecture15.1

Lecture15.2

Lecture15.3

Lecture15.4

Lecture15.5

Lecture15.6

Lecture15.7

Lecture15.8

NCERT Solutions – Triangles Exercise 7.1 – 7.5
Exercise 7.1
Q.1 In quadrilateral ACBD, AC = AD and AB bisects ∠A (see figure). Show that ΔABC≅ΔABD. What can you say about BC and BD?
Now in Δs ABC and ABD, we have
AC = AD [Given]
∠CAB=∠BAD [Since AB bisects ∠A]
and AB = AB [Common]
Therefore by SAS congruence criterion, we have
ΔABC≅ΔABD
⇒ BC = BD [Since corresponding parts of congruent triangles are equal]
Q.2 ABCD is a quadrilateral in which AD = BC and ∠DAB=∠CBA (see figure). Prove that
(i) ΔABD≅ΔBAC
(ii) BD = AC
(iii) ∠ABD=∠BAC
In Δs ABD and BAC, we have
AD = BC [Given]
∠DAB=∠CBA [Given]
AB = AB [Common]
Therefore by SAS criterion of congruence, we have
ΔABD≅ΔBAC,, which proves (i)
⇒ BD = AC
and , ∠ABD=∠BAC, which proves (ii) and (iii)
[Since corresponding parts of congruent triangles are equal]
Q.3 AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.
Since AB and CD intersect at O. Therefore ,
∠AOD=∠BOC …. (1) [Vertically opp. angles]
In Δs AOD and BOC, we have
∠AOD=∠BOC [From (1)]
∠DAO=∠OBC [Each = 90º]
and , AD = BC [Given]
Therefore by AAS congruence criterion, we have
ΔAOD≅ΔBOC
⇒ OA = OB
[Since corresponding parts of congruent triangles are equal]
i.e., O is the mid point of AB.
Hence, CD bisects AB.
Q.4 l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that ΔABC≅ΔCDA.
Since l and m are two parallel lines intersected by another pair of parallel lines p and q. Therefore , AD  BC and AB  CD
⇒ ABCD is a parallelogram.
i.e. , AB = CD
and BC = AD
Now, in Δs ABC and CDA, we have
AB = CD [Proved above]
BC = AD [Proved above]
and AC = AC [Common]
Therefore By SSS criterion of congruence.
ΔABC≅ΔCDA
Q.5 Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see figure). Show that :
(i) ΔAPB≅ΔAQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.
In Δs APB and AQB we have
∠APB=∠AQB [Since Each = 90º]
∠PAB=∠QAB [Since AB bisects ∠PAQ]
AB = AB [Common]
By AAS congruence criterion, we have
ΔAPB≅ΔAQB, which proves (i)
⇒ BP = PQ
[Since Corresponding parts of congruent triangles are equal]
i.e., B is equidistant from the arms of ∠A, which proves (ii).
Q.6 In figure AC = AE, AB = AD and ∠BAD=∠EAC. Show that BC = DE.
In Δs ABC and ADE, we have
AB = AD [Given]
∠BAC=∠DAE
[Since ∠BAD=∠EAC ⇒ ∠BAD+∠DAC=∠EAC+∠DAC ⇒
⇒ ∠BAD=∠DAE]
and , AC = AE [Given ]
Therefore By SAS criterion of congruence, we have
ΔABC≅ΔADE
⇒ BC = DE
[Since Corresponding parts of congruent triangles are equal]
Q.7 AB is a line segment and P is its mid point. D and E are points on the same side of AB such that ∠BAD=∠ABE and ∠EPA=∠DPB (see figure). Show that –
(i) ΔDAP≅ΔEBP
(ii) AD = BE.
We have , ∠EPA=∠DPB
⇒ ∠EPA+∠DPE=∠DPB+∠DPE
⇒ ∠DPA=∠EPB
Now , in Δs EBP and DAP, we have
∠EPB=∠DPA [From (1)]
BP = AP [Given]
and , ∠EBP=∠DAP [Given]
So, by ASA criterion of congruence, we have
ΔEBP≅ΔDAP
⇒ BE = AD i.e., AD = BE
[Since Corresponding parts of congruent triangles are equal. ]
Q.8 In right triangle ABC, right angled at C, M is the mid point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see figure). Show that :
(i) ΔAMC≅ΔBMD
(ii) ∠DBC is a right angle
(iii) ΔDBC≅ΔACB
(iv) CM=12AB
(i) In Δs AMC and BMD, we have
AM = BM [Since M is the mid point of AB]
∠AMC=∠BMD [Vertically opp. ∠s]
and CM = MD [Given]
Therefore By SAS criterion of congruence , we have
ΔAMC≅ΔBMD
(ii) Now, ΔAMC≅ΔBMD
⇒ BD=CAand∠BDM=∠ACM … (1)
[Since corresponding parts of congruent triangles are equal]
Thus, transversal CD cuts CA and BD at C and D respectively such that the alternate angles ∠BDMand∠ACM are equal. Therefore , BD  CA.
⇒ ∠CBD+∠BCA = 180º
[Since sum of consecutive interior angles are supplementary ]
⇒ ∠CBD+ 90º = 180º [Since ∠BCA = 90º]
⇒ ∠CBD = 180º – 90º
⇒ ∠DBC= 90º
(iii) Now, in Δs DBC and ACB, we have
BD = CA [From (1)]
∠DBC=∠ACB [Since Each = 90º]
BC = BC [Common]
Therefore by SAS criterion of congruence, we have
ΔDBC≅ΔACB
(iv) CD = AB [Since corresponding parts of congruent triangles are equal]
⇒ 12CD=12AB ⇒ CM=12AB
Exercise 7.2
Q.1 In an isosceles triangle ABC, with AB = AC, the bisectors of ∠Band∠C intersect each other at O. Join A to O. Show that :
(i) OB = OC (ii) AO bisects ∠A.
Sol.
(i) In Δ ABC, we have
AB = AC
⇒ ∠C=∠B [Since angles opposite to equal sides are equal]
⇒ 12∠B=12∠C
⇒ ∠OBC=∠OCB
[Since OB and OC bisect ∠s B and C respectively. Therefore ∠OBC=12∠Band∠OCB=12∠C]
⇒ OB = OC [Since sides opp. to equal ∠s are equal]
(ii) Now, in Δs ABO and ACO , we have
AB = AC [Given]
∠OBC=∠OCB [From (1)]
OB = OC [From (2)]
Therefore By SAS criterion of congruence, we have
ΔABO≅ΔACO
⇒ ∠BAO=∠CAO
[Since corresponding parts of congruent triangles are equal]
⇒ AO bisects ∠A.
Q.2 In Δ ABC, AD is the perpendicular bisector of BC (see figure). Show that Δ ABC is an isosceles triangle in which AB = AC.
In Δs ABD and ACD, we have
DB = DC [Given]
∠ADB=∠ADC [since AD ⊥ BC]
AD = AD [Common]
Therefore by SAS criterion of congruence, we have.
ΔABD≅ΔACD
⇒ AB = AC
[Since corresponding parts of congruent triangles are equal]
Hence, Δ ABC is isosceles.
Q.3 ABC is an isosceles triangle in which altitudes BE and CF are drawn to sides AC and AB respectively (see figure). Show that these altitudes are equal.
In Δs ABE and ACF, we have
∠AEB=∠AFC [Since Each = 90º]
∠BAE=∠CAF [Common]
and BE = CF [Given]
Therefore By AAS criterion of congruence, we have
ΔABE≅ΔACF
⇒ AB = AC
[Since Corresponding parts of congruent triangles are equal]
Hence, Δ ABC is isosceles.
Q.4 ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that
(i) ΔABE≅ΔACF
(ii) AB = AC , i.e., ABC is an isosceles triangle.
Let BE ⊥ AC and CF ⊥ AB.
In Δs ABE and ACF, we have
∠AEB=∠AFC [Since Each = 90º]
∠A=∠A [Common]
and , AB = AC [Given]
Therefore By AAS criterion of congruence,
ΔABE≅ΔACF
⇒ BE = CF
[Since corresponding parts of congruent triangles are equal]
Q.5 ABC and DBC are two isosceles triangles on the same base BC (see figure). Show that ∠ABD=∠ACD.
In Δ ABC, we have
AB = AC
⇒ ∠ABC=∠ACB … (1)
[Since angles opposite to equal sides are equal]
In ΔBCD, we have
BD = CD
⇒ ∠DBC=∠DCB … (2)
[Since angles opposite to equal sides are equal]
Adding (1) and (2), we have
∠ABC+∠DBC=∠ACB+∠DCB
⇒ ∠ABC=∠ACD
Q.6 Δ ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that ∠BCD is a right angle.
In Δ ABC, we have
AB = AC [given]
⇒ ∠ACB=∠ABC … (1)
[Since angles opp. to equal sides are equal]
Now, AB = AD [Given]
Therefore AD = AC [Since AB = AC]
Thus , in Δ ADC, we have
AD = AC
⇒ ∠ACD=∠ADC … (2)
[Since angles opp. to equal sides are equal]
Adding (1) and (2) , we get
∠ACB+∠ACD=∠ABC+∠ADC
⇒ ∠BCD=∠ABC+∠BDC [Since ∠ADC=∠BDC]
⇒ ∠BCD+∠BCD=∠ABC+∠BDC+∠BCD [Adding ∠BCD on both sides]
⇒ 2∠BCD=180∘ [Angle sum proprty]
⇒ ∠BCD= 90º
Hence , ∠BCD is a right angle.
Q.7 ABC is a right angled triangle in which ∠A= 90º and AB = AC. Find ∠Band∠C.
Sol.
We have,
∠A= 90º
AB = AC
⇒ ∠C=∠B
[Since angles opp. to equal sides of a triangle are equal]
Also ∠A+∠B+∠C= 180º [Angle – sum property]
⇒ 90º + 2∠B= 180º [Since ∠C=∠B]
⇒ 2∠B= 180º – 90º = 90º
⇒ ∠B=90o2=45o
Therefore ∠C=∠B=45o
Q.8 Show that the angles of an equilateral triangle are 60º each.
Sol.
Let Δ ABC be an equilateral triangle so that AB = AC = BC.
Now Since AB = AC
⇒ ∠C=∠B … (1) [Since angles opp. to equal sides are equal]
Also since, CB = CA
⇒ ∠A=∠B …. (2) [Since angles opp. to equal sides are equal]
From (1) and (2), we have
∠A=∠B=∠C
Also ∠A+∠B+∠C= 180º [Angle – sum property]
Thererfore ∠A+∠A+∠A= 180º
⇒ 3∠A= 180º
⇒∠A=180∘3
⇒∠A= 60º
Therefore ∠A=∠B=∠C= 60º
Thus , each angle of an equilateral triangle is 60º.
Exercise 7.3
Q.1 Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that
(i) ΔABD≅ΔACD
(ii) ΔABP≅ΔACP
(iii) AP bisects ∠A as well as ∠D
(iv) AP is the perpendicular bisector of BC.
Sol.
(i) In Δs ABD and ACD, we have
AB = AC [Given]
BD = DC [Given]
and AD = AD [Common]
Therefore by SSS criterion of congruence, we have ΔABD≅ΔACD
(ii) In Δs ABP and ACP , we have
AB = AC [Given]
∠BAP=∠PAC
[Since ΔABD≅ΔACD
⇒ ∠BAD=∠DAC ⇒ ∠BAP=∠PAC
and AP = AP [Common]
Therefore By SAS criterion of congruence , we have ΔABP≅ΔACP
(iii) Since ΔABD≅ΔACD. Therefore , ∠BAD=∠DAC
⇒ AD bisects ∠A ⇒ AP bisects ∠A … (1)
In Δs BDP and CDP, we have
BD = CD [Given]
BP = PC [Since ΔABP≅ΔACP ⇒ BP = PC]
and DP = DP [Common]
By SSS criterion of congruence, we have
ΔBDP≅ΔCDP
⇒ ∠BDP=∠PDC
⇒ DP bisects ∠D ⇒ AP bisects ∠D … (2)
Combining (1) and (2), we get
AP bisects ∠A as well as ∠D
(iv) Since AP stands on BC
Therefore ∠APB+∠APC= 180º [Linear Pairs]
But ∠APB=∠APC [Proved above]
Therefore ∠APB=∠APC=180o2=90o
Also BP = PC [Proved above]
⇒ AP is perpendicular bisector of BC.
Q.2 AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC (ii) AD bisects ∠A.
Sol.
AD is the altitude drawn from vertex A of an isosceles Δ ABC to the opposite base BC so that AB = AC, ∠ADC=∠ADB= 90º.
Now, in Δs ADB and ADC, we have
Hyp. AB = Hyp. AC [Given]
AD = AD [Common]
and ∠ADC=∠ADB [Since Each = 90º]
Therefore By RHS criterion of congruence, we have
ΔADB≅ΔADC
⇒ BD = DC and ∠BAD=∠DAC
[Since corresponding parts of congruent triangle are equal]
Hence, AD bisects BC, which proves (i) and AD bisects ∠A, which proves (ii).
Q.3 Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of Δ PQR (see figure). Show that :
(i) ΔABM≅ΔPQN
(ii) ΔABC≅ΔPQR
Two Δs ABC and PQR in which AB = PQ, BC = QR and AM = PN.
Since AM and PN are median of Δs ABC and PQR respectively.
Now, BC = QR [Given]
⇒ 12BC=12QR (Median divides opposite sides in two equal parts)
⇒ BM = QN … (1)
Now , in Δs ABM and PQN we have
AB = PQ [Given]
BM = QN [From (i)]
and AM = PN [Given]
Therefore By SSS criterion of congruence, we have
ΔABM≅ΔPQN, which proves (i)
⇒ ∠B=∠Q … (2)
[Since, corresponding parts of congruent triangle are equal]
Now, in Δs ABC and PQR we have
AB = PQ [Given]
∠B=∠Q [From (2)]
BC = QR [Given]
Therefore By SAS criterion of congruence, we have
ΔABC≅ΔPQR, which proves (ii)
Q.4 BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Sol.
In Δs BCF and CBE, we have
∠BFC=∠CEB [Since Each = 90º]
Hyp. BC = Hyp. BC [Common]
FC = EB
Therefore By R.H.S. criterion of congruence, we have
ΔBCF≅ΔCBE
⇒ ∠FBC=∠ECB
[Since corresponding parts of congruent triangles are equal]
Now, in Δ ABC
∠ABC=∠ACB [Since ∠FBC=∠ECB]
⇒ AB = AC
[Since sides opposite to equal angles of a triangle are equal]
Therefore ΔABC is an isosceles triangle.
Q.5 ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B=∠C
Sol.
In Δs ABP and ACP, we have
AB = AC [Given]
AP = AP [Common]
and ∠APB=∠APC [Since each = 90º]
Therefore , by R.H.S. criterion of congruence we have
ΔABP≅ΔACP
⇒ ∠B=∠C
[Since corresponding parts of congruent triangles are equal]
Exercise 7.4
Q.1 Show that in a right angled triangle, the hypotenuse is the longest side.
Sol.
Let ABC be a right angled Δ in which ∠ABC= 90º
But ∠ABC+∠BCA+∠CAB= 180º [By anglesum property]
⇒ 90º + ∠BCA+∠CAB= 180º
⇒ ∠BCA+∠CAB= 90º
⇒ ∠BCAand∠CAB are acute angles
⇒ ∠BCA<90oand∠CAB<90o
⇒ ∠BCA<∠ABCand∠CAB<∠ABC
⇒ AC > AB and AC > BC [Since side opp. to greater angle is larger]
Hence , in a right triangle , the hypotenuse is the longest side.
Q.2 In figure, sides AB and AC of Δ ABC are extended to points P and Q respectively. Also ∠PBC<∠QCB. Show that AC > AB.
Since ∠PBC<∠QCB (given)
⇒ −∠PBC>−∠QCB (Both sides multiply by – )
⇒ 180º −∠PBC>180o−∠QCB (Adding 180º on both sides)
[Since ∠PBC and ∠ABC as well as, ∠QCB and ∠ACB are linear pair]
⇒ ∠ABC>∠ACB
⇒ AC > AB [Since side opp. to greater angle is larger]
Q.3 In figure , ∠B<∠Aand∠C<∠D. Show that AD < BC.
Since ∠B<∠Aand∠C<∠D (Given)
Therefore AO < BO ………………(1)
and OD < OC…………………..(2)
[Since side opp. to greater angle is larger]
Adding (1) and (2) , we get
AO + OD < BO + OC
⇒ AD < BC.
Q.4 AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that ∠A>∠Cand∠B>∠D
ABCD is a quadrilateral such that AB is its smallest side and CD is its largest side.
Join AC and BD. Since AB is the smallest side of quadrilateral ABCD.
Therefore , in Δ ABC, we have
BC > AB
⇒ ∠8>∠3 … (1)
[Since angle opp. to longer side is greater]
Since CD is the longest side of quadrilateral ABCD.
Now , in Δ ACD, we have
CD > AD
⇒ ∠7>∠4 … (2)
[Since angle opp. to longer side is greater]
Adding (1) and (2), we get
∠8+∠7>∠3+∠4
⇒ ∠A>∠C
Again , in Δ ABD , we have
AD > AB [Since AB is the shortest side]
⇒ ∠1>∠6 … (3)
In Δ BCD, we have
CD > BC [Since CD is the longest side]
⇒ ∠2>∠5 … (4)
Adding (3) and (4), we get
∠1+∠2>∠5+∠6
⇒ ∠B>∠D
Thus , ∠A>∠Cand∠B>∠D
Q.5 In figure PR > PQ and PS bisects ∠QPR. Prove that ∠PSR>∠PSQ
In Δ PQR , we have
PR > PQ [Given]
⇒ ∠PQR>∠PRQ [Since angle opp. to larger side is greater]
⇒ ∠PQR+∠1>∠PRQ+∠1 [Adding ∠1 on both sides]
⇒ ∠PQR+∠1>∠PRQ+∠2 [Since PS is the bisector of ∠P since ∠1=∠2]
Now , in Δs PQS and PSR ,we have
∠PQS+∠1+∠PSQ= 180º
and ∠PRS+∠2+∠PSR= 180º
⇒ ∠PQS+∠1=180o−∠PSQ
and ∠PRS+∠2=180o−∠PSR
Therefore 180º – ∠PSQ>180o−∠PSR [From (1)]
⇒ −∠PSQ>−∠PSR
⇒ ∠PSQ<∠PSR i.e. ∠PSR>∠PSQ
Q.6 Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Sol.
Let P be any point not on the straight line l. PM ⊥ l and N is any point on l other than M.
In Δ PMN , we have
∠M=90o
⇒ ∠N<90o
[Since ∠M=90o ⇒∠MPN+∠PNM=90o⇒ ∠P+∠N=90o⇒∠N<90o]
⇒ ∠N<∠M
⇒ PM < PN [Since side opp. to greater angle is larger]
Hence, PM is the shortest of all line segments from P to AB.
Exercise 7.5
Q.1 ABC is a triangle. Locate a point in the interior of Δ ABC which is equidistant from all the vertices of Δ ABC.
Sol.
Let OD and OE be the perpendicular bisectors of sides BC and CA of Δ ABC.
Therefore O is equidistant from two ends B and C of line segment BC as O lies on the perpendicular bisector of BC. Similarly, O is equidistant from C and A.
Thus , the point of intersection O of the perpendicular bisectors of sides BC, CA and AB is the required point which is equidistant from vertices A,B,C of Δ ABC.
Q.2 In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
Sol.
Let BE and CF be the bisectors of ∠ABC and ∠ACB respectively intersecting AC and AB at E and F respectively.
Since O lies on BE, the bisector of ∠ABC, hence O will be equidistant from AB and BC. Again O lies on the bisector CF of ∠ACB.
Hence , O will be equidistant from BC and AC. Thus, O will be equidistant from AB, BC and CA.
Q.3 In a huge park, people are concentrated at three points (see figure) :
A : where there are different slides and swings for children,
B : near which a man made lake is situated.
C : which is near to a large parking and exit.
Where should an icecream parlour be set up so that maximum number of persons can approach it?
Sol.
The parlour should be equidistant from A, B and C, for which the point of intersection of perpendicular bisector should be located.
Thus O is the required point which is equidistant from A, B and C.
Q.4 Complete the hexagonal and star shaped Rangolies [see figure (i) and (ii)] by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?
Sol.
On filling each figure with equilateral triangles of side 1 cm, we find in figure
(i) number of such triangles is 150, and in figure (ii) number of such triangle is 300.
Figure (ii) has more triangles.