worksheet 1 Triangles

worksheet 1 Triangles

1. Give two different examples of pair of (i) similar figures. (ii) non-similar figures.

Solution:Two square of sides 4 cm and 8 cm each.
A rhombus and a trapezium .

2. State whether the following quadrilaterals are similar or not

Solution:Similar

3. In the figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Solution:(i)

In DABC
DE is parallel to BC
By Basic Proportionality Theorem
------------------- (1)
Given: AD = 1.5 cm, DB = 3 cm, AE = 1 cm
Let EC = ‘x’ cm
Applying in (1)

1.5x = 3
x =
x = 2 cm
EC = 2cm

(ii)

Since DE || BC, using BPT
…………………………. (1)
Given: DB = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm
Let AD be = x
sub. in (1)

x =
=
\ AD = 2.4 cm

4. E and F are points on the sides PQ and PR respectively of a Δ PQR. For each of the following cases, state whether EF || QR
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

Solution:

(i) PE = 3.9 cm, EQ = 3 cm
PF = 3.6 cm, FR = 2.4 cm
=
=
\
EF is not parallel to QR by convene of BDT

(ii) PE= 4cm QE = 4.5cm PF = 8 RF = 9cm
=

\
EF || QR by convene of BPT.

(iii) PQ = 1.28 cm ,  PR = 2.56 cm,  PE = 0.18 cm,  PF = 0.36 cm
EQ = PQ – PE
     = 1.28 – 0.18
     = 1.10
FR = PR – PF
     = 2.56 – 0.36
     = 2.20

Þ
Þ EF is  parallel to QR

5. In the figure, if LM || CB and LN || CD, prove that

Solution:

Given: LM || CB and LN || CD
To prove:
ProofIn D ABC
LM || BC using basic proportionality Theorem
\  ………………………….. (1)
Also in D ADC
LN ||  DC
\  using basic proportionality Theorem …………….. (2)

from (1) and D

Here proved .

6. In the figure, DE || AC and DF || AE. Prove that

Solution:

Given: ABC is a triangle and DE || AC and DF is parallel to AE
To prove:
ProofIn D ABC,
DE || AC (given)
(By BPT) …………………….. (1)

In D AEB,
\ DF || AE
(By BPT) ………………………. (2)
comparing equation (1) and equation (2)

Here proved.

7. In the figure, DE || OQ and DF || OR. Show that
EF || QR.

Solution:

Given: DE || OQ and DF || OR
To prove : EF || QR
Proof
In D POQ
DE || OQ (given)
By using BPT
………………………… (1)
In D POR
………………………….(2)
By comparing equations (1) and (2)

By using inverse of BPT
EF || QR
Here proved.

8. In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR.
Show that BC || QR.

Solution:

Given: A, B and C are points on OP, OQ and OR respectively such that AB || PQ, AC || PR
Proof
In D OPQ
AB || PQ (Given)
\ (By using BPT) ……………………………. (1)
In D OPR
Since AC || PQ
(By using BPT) ………………….. (2)

By comparing (1) and (2)

By using converse of BPT BC || QR
Here proved.

9. Using Theorem , prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

Solution:

Given: ABCD is a trapezium and the diagonals AC and BD intersect at 0.

To prove: The ratio
Construction : Draw OM || AB meeting BC at ‘M’
Proof: In D ACB OM || AB
\ By using BPT ………………………………………. (1)
|||ly In D BDC
OM || CD [ \(OM || AB AND AB || CD Þ OM || CD)]
\ using BPT
Taking the reciprocal
…………………………….(2)
from (1) and (2)

(or)
Here proved.

10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that. Show that ABCD is a trapezium.

Solution:

Given: ABCD is a quadrilateral and the diagonals AC and BD intersect of ‘O’ such that

To prove: ABCD is a trapezium
Construction: Draw OM || AB
ProofIn D ACB \ OM || AB (By construction)
………………………. (1)
Given:
(or)
sub = in equation (1)
\ =
or =
In D BDC by converse of BPT
OM || DC. But OM || AB
Þ AB || CD
Þ ABCD is a trapezium.
11. State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :
Solution:(i) Similar
AAA D ABC ~ D PQR
(ii) Similar
SSS, D ABC ~ D QRP
(iii) Not similar
(iv) Not smilar
(v) Similar
(vi) Not similar
12. DODC ~ D OBA, Ð BOC = 125°and Ð CDO = 70°. Find Ð DOC, Ð DCO and Ð OAB.
Solution:

Given: D ODC ~ D OBA
Ð BOC= 125⁰
Ð CDO = 70⁰
To find Ð DOC, Ð DCO and Ð OAB
Ð DOC= 180 – 125 = 55
Ð DCO = 125 – 70 = 55
Ð OAB =55 ( Q D ODC ~ D OBA).
13. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles show that
Solution:

Given. ABCD is a trapezium with AB || CD and the diagonals AC and BD intersect at ‘O’.

To prove: =

Proof
In the figure consider the triangle OAB and OCD

Ð DOC = Ð AOB (Vertically opposite angles are equal)

Ð DCO = Ð OAB (Alternate angles are equal)

\ By AA corollary of similar triangles.
\ D OAB ~ D OCB When the two triangle are similar, the side are proportionally.
Þ =
Here proved.
14. In the figure, = and Ð 1 = Ð 2. Show that Δ PQS ~ Δ TQR.
Solution:Given: =
and Ð 1 = Ð 2
To prove: D PQS ~ D TQR
Proof
In D PQR Ð 1 = Ð 2 Ð PQR = Ð PRQ
Þ PQ = PR
=
=
= ( \ QP = PR proved above )Ð 1 is a common angle
By SAS similarity D PQS ~ D TQR Hence proved.
15. S and T are points on sides PR and QR of D PQR such that Ð P = Ð RTS. Show that D RPQ ~ D RTS.
Solution:
Given: S and T are the points on sides. PR and QR of a. D PQR such that Ð P = Ð RTS
To prove: D RPQ ~ D RTS
Proof:
D S PQR and D T TRs
Ð QRP = Ð TRS (common angle) Ð STR = Ð QPR (given)
\ By similarly
D S RPQ ~ D RTS
Here proved.
16. In the figure, if ΔABE @ Δ ACD, show that Δ ADE ~ Δ ABC.
Solution:Given: D ABE @ D ACD
To prove: Δ ADE ~ Δ ABC
Prove: D ABE @ D ACD
\ AB = AC …………………… (1) (since corresponding parts of D ^s are equal)
and AD = AE …………………… (2)

and ÐA is a common angle \ÐDAE = ÐBAC
By SAS similarity condition
D ADE ~ D ABC .
17. In the Figure, altitudes AD and CE of Δ ABC intersect each other at the point P. Show that:
(i) Δ AEP ~ Δ CDP
(ii) Δ ABD ~ Δ CBE
(iii) Δ AEP ~ Δ ADB
(iv) Δ PDC ~ Δ BEC.
Solution:

Given: D ABC in which the altitudes AD and CE intersect at P.
To prove:
(i) D AEP ~ D COP (ii) D ABD ~ D CBE
(iii) D AEP ~ D AOB (iv) D PDC ~ D BEC
Proof :(i) D AEP and D CDP
Ð APE = Ð CPD                      (Vertically opposite angle and equal)
Ð AEP = Ð CDP                      (Both are equal to 90⁰ since CE and AD are altitudes)
\ BY AA similarity
D AEP ~ D CDP
Here proved.

(ii) To prove D ABD ~ D CBE
In D ABD and D CBE
Ð CBE = Ð ABD                              (Common angle)
  Ð ADB = Ð CEB                             (Both are equal to 90^o)
\ By AA similarity
D ABD ~ CDE

(iii) To prove. D AEP ~ ADB
In D AEP and D ADB
Ð PAE = Ð DAB                             (common angle)
Ð PAE = Ð DAB                      (Both are equal to 90^o)
Þ By AA similarity
D AEP ~ D ADB

(iv) To prove: D PDC ~ D BEC
Proof: In D PDC and D BEC
Ð PCD = Ð ECB                             (common angle)Ð PDC = Ð BLC                             (Both are equal to 90⁰)
D PDC ~ BFC (AA similarity).
18. In the figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
(i) D ABC ~ D AMP
(ii)
Solution:

(i) To prove D ABC ~ D AMP
In D ABC and D AMP
(Both are equal to 90⁰)
(common angle)
\ BY AA corollary
&D ABC ~ AMP

(ii) To prove

Proof
As proved above
D ABC ~ D AMP
Since the corresponding sides of the similar triangles are proportional

Þ
Here proved.
19. CD and GH are respectively the bisectors of Ð ACB and Ð EGF such that D and H lie on sides AB and FE of D ABC and D EFG respectively. If D ABC ~ D FEG, show that:
(i)
(ii) D DCB ~ D HGE
(iii) D DCA ~ D HGF
Solution:

Given: D ABC ~ D FEG and CD and GH are bisectors of the Ð ACB and Ð EGF of the D ACB and D EGF respectively.
To prove:
(i)
(ii) D DCB ~ D HGE
(iii) D DCA ~ D HGF
Proof: (i)
To prove:

consider D ADC and D FGH
Ð BAC = Ð EFG (Q D ABC ~ D FEG) is given)
(or)Ð DAC = Ð HFG
Also
Ð BCA = Ð EGF (Q D ABC ~ D FEG is given)
´ by
Ð BCA = Ð EGF
=> Ð DCA = Ð HGF

\ By AA corollary
D DAC ~ D HFG
\

(ii) To prove: D DCB ~ D HGE
In D DBC and D HEG
Ð ABC = Ð FEG [QD ABC ~ D FEG]
(or) Ð DBC = Ð HEG [ Q D and H lie on the line AB and EF respectively]
Ð BCA = Ð EGF …………………..(1) [Q D ABC ~ D FEG]
multiplying by
Ð BCA = Ð EGF
(or) Ð DCB = Ð HGE …………………………..(2) [Q CD and GH are the bisectors of the Ð ACB and Ð EGF of the D ACB and D EGF respectively]

from (1) and (2)
D DCB ~ D HGE [By AA corollary]

(iii) Already proved in (i)
In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD perpendicular to BC and EF perpendicular to AC prove that D ABD ~ D ECF.

Given: D ABC is isosceles with AB = AC. E is a point as CB such that EF perpendicular to AC. AD is perpendicular to BC.
To prove: D ABD ~ D ECF
Proof :
D ABC is isosceles with AB = AC (Given)
\ ÐABC = ÐACB ………………. (1) Property of an isosceles D
and ÐADB = ÐEFC ……………………(2) Both are equal to 90⁰Now, in D ABD and D ECFÐABD = ÐECF From equation (1)
and ÐADB = ÐEFC From equation (2)
\ D ABD ~ D ECF .
20. In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD perpendicular to BC and EF perpendicular to AC prove that D ABD ~ D ECF.
Solution:

Given:
Replacing BC = 2BD and QR = 2QM
Since AD and PM are the medians from A and P respectively.

Þ
Since the sides of D ABD ~ D PQM ÐABD = ÐPQM
ÐABC = ÐPQR -------------------------- (1)
Now from given data
ÐABC = ÐPQR
Þ D ABC ~ D PQR (By SAS similarity condition).

21.  A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Solution:

Since the shadow is cast of the since from the angle made by the sun’s ray is same
2 D ABC and PQR
ÐACB = ÐPRQ
ÐABC = ÐPQR          (Q both are equal to 90⁰)
Þ D ABC ~ D PQR (By AA corollary)
\ =

or 6 ´  = 42
Height of the tower = 42 m .

22.  If AD and PM are medians of triangles ABC and PQR, respectively where D ABC ~ D PQR, prove that

Solution:
Given: D ABC ~ D PQR
To prove :
Proof:
In D ABC ~ D PQR
\ = ………………………………(1)
and Ð B = Ð Q ……………………………..(2)
replacing BC = 2BD and QR = 2QM in equation (1)
……………………………..(3)
from (2) and (3)
D ABD ~ D PQM By SAS similarly
\
Here proved .

23.  Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of D PQR. Show that DABC ~ D PQR.

Solution:
Given: D ABC and AD is the median from A to BC and D PQR, PM is the median from P to, QR. Also

=

To prove: D ABC ~ D PQR

Construction : Draw DE || AC and DF || AB in D ABC .

Draw MN || PR and MO || PQ in D PQR .

Proof:

In D ABC, since, AD in the median D is the midpoint of BC and DF is parallel to AC by construction E is the midpoint of AB(using midpoint theorem)

Similarly N is the midpoint of PQ

In D ABC and PQR, given that
=
=  ………………………………… (1)
In D AEO AND PNM, the sides are proportional (from (1)
D AED ~ D PNM
Þ ÐDAE = ÐMPN ……………………….. (2)
|||ly we can prove that D AFD ~ D POM are similar. Hence
ÐDAF = ÐMPO …………………….(3)

Adding equation (2) + Equation (3)

ÐDAE + ÐDAF = ÐMPN + ÐMPO

ÐEAF = ÐNPO

ÐBAC = ÐQPR   …………………….(4)

As  and from equation (4)

\ By SAS similarity criterion

D ABC ~ D QPR

Here the result.

24.  Let D ABC ~ D DEF and their areas be, respectively, 64 cm² and 121 cm2. If EF = 15.4cm, find BC.

Solution:Since D ABC ~ D DEF
=

(or)  =
=
BC =  ´ 15.4
BC = 11.2 cm.

25.  Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.

Solution:
In D AOB and COD
ÐCOD = ÐAOB  (vertically opposite angles)ÐCDO = ÐOBA (Alternate interior angles)
By AA similarity
D COD ~ D AOB
=  =
\  =

26.  In the figure ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that

Solution:

Construction: From A and D draw perpendiculars to BC meeting BC at E and F respectively.
Proof:In D AFO and D DEO

ÐAFO = ÐDEO = 90⁰ ÐAOE = ÐDOE   (vertically opposite angle)
Þ D AOE ~ D DOF (By AA similarity)
\  =  =  ………………………. (1)
\ =
Here proved.

27.  If the areas of two similar triangles are equal, prove that they are congruent.

Solution:Let ABC and DEF be two similar triangles where areas are equal.
(ie) ar(D ABC) = ar(D DEF)
=  =  =
since ar. D ABC = ar.D DEF
=  = 1
\  = 1 or AB² = DE²Þ AB – DF
|||ly BC = EF
and AC = DF
By SSS congruency
D ABC @ D DEF
Here the areas of two similar D ^s are equal, those two triangles are congruent.

28.  D, E and F are respectively the mid-points of sides AB, BC and CA of D ABC. Find the ratio of the areas of D DEF and D ABC.

Solution:
Since D and F are the midpoints of the AB and AC, DF || AC or DF || BE
Similarly EF || AB or EF || DB
AEFD is a parallelogram as both pairs of opposite sides are parallel.
\ By the property of parallelogram

ÐDBE = ÐDFE = (In an parallelogram opposite angle are equal)
or ÐDFE = ÐABC   -------------------(1)
Similarly ÐFEB = ÐACB   …………………………..(2)
In D DEF and D ABC, from
Equation (1) and (2)
D DEF ~ D CAB
\  =  =
\ ar D DEF: ar D CAB = 1 : 4 .

29.  Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Solution:
Given: AM and DN are medians of the D ABC and D DEF and D ABC ~ D DEF
To prove:
=
Proof : since D ABC ~ D DEF

or
and  ( QD ABC ~ D DEF)
\ By SAS similarity
D ABM ~ D DEN
or  ………………………... (1)
=
=  (From (1)
Hence proved.

30.  Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Solution:

Given: ABCD is a square and two equilateral triangle are drawn with BC and DB as are units side respectively.
To Prove: ar D PBC =
Proof:
Consider D ^s PBC and D BQD
Since both these triangles are equilateral triangles, by the property of equilateral D s,
both the D ^s are equiangular with each angle equal to 60^o.
Þ D PBC ~ D QBD
\  =  ………………………… (1)
Consider D BCP
Since it is a right angled triangle
BC² + DC² = BD² ( By using Pythagoras theorem)
But BC = DC
QBC² = BD²
Substituting in (1)
=
\ ar D PBC =  or D BDQ
Here the results.

35.  ABC is an isosceles triangle with AC = BC. If AB² = 2 AC², prove that ABC is a right triangle.

Solution:
Given AB² = 2AC² and AC = BC
To prove: ABC is a right triangle
Proof:
Given: AB² = 2AC²
(or) AB² = AC² +AC² ………………(1)
since AC = BC given, equation (1) becomes
AB² = AC² + BC²
\ By the converse of Pythagoras theorem D ABC is a right triangle with  .

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

Solution:
Given: ABC is an equilateral triangle with each side equal to 2a
Construction: Draw AD ^ r to BC
To find: Length of AD
Construction D ADB and ADC
ÐADB = ÐADC = 90⁰
AC = AB (QABC is isosceles)
AD = AD (common side)
\ D ADB @ D ADB (By RHS property)
Þ BD = DC (CPCT)
BC = BD + DC
= BD + BD
2a = 2BD
BD = a
2D ABD
AB² = AD² + BD²
(2a)² = AD² + a²
4a² = AD² + a²
a² – a² = AB²
3a² = AD²
AD = a
Similarly the other altitudes will also be a only.

37.  Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Solution:
Given: ABCD is a rhombus and the diagonals AC and BD intersect at 0
To prove: AB² + BC² + DC² + AD² = AC² + BD²
Proof: The diagonals of C rhombus intersect at right angle.
\ = 90⁰
In D AOB
AO² + OB² = AB²
+ = AB²
Hence AC² + BD² = 4AB²
Since 4AB² can be taken as sum of the square of the 4 side3 of the rhombus as each side of the rhombus equal, the required result in proved.

38.  In the figure O is a point in the interior of a triangleABC, OD perpendicular to BC, OE perpendicular to AC and OF perpendicular to AB. Show that
(i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²,(ii) AF² + BD² + CE² = AE² + CD² + BF².

Solution:
(i) Given : In D ABC OD ^ BC, OE ^ AC and OF ^ AB
Construction: Join OC, OA and OB.
In D OBD, OB² = OD² + BD²
In D OFA, OA² = OF² + AF²In D OCE, OC² = OE² + CE²\ OB² + OA² + OC² = OD² + OF² + OE² + BD² + AF² + CE²
(or) OB² + OA² + OC² – OD² – OF² – OE²
= BD² + AF² + CF²(ii) From the result above
AF² + BD² + C² = OB² + OA² + OC – OD² – OF² – OE²
                      = (OA² – OE²) + (OC² – OD²) +(OB² – OF²)
                      = AE² + CD² + BF²

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

Solution:
In the figure AC = ladder = 10 m
AB = Height of the window = 8m
BC = Distance from the foot of the wall to the foot of the ladder = x m
ABC is a right angle D
AB² + BC² = AC²
(or) 8² + x² = 10²
x² + 10² - 8²x² = (10 + 8) (10 – 8)
    = (18)(2)
    = 36 x =  = 6 m
\ The distance of the foot of the ladder from the wall = 6 m

40.  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 driven so that the wire will be taut?

Solution:
In D ACB, AC = length of the wire = 24m and the largest of the tower = 18 m
18² + x² = 24² (By using Pythagoras theorem)
x² + 24² = 18²
x²= 576 – 324
   = 252
x =
   = 6m
Therefore, the station should be driven to a distance of 6m so that the wire will be taunt.

41.  An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1 hours?

Solution:
Distance traveled by the aeroplane in the North directions is
AB = 1000 ´  = 1500km. (Distance = speed ´ time)
And the Distance traveled by the Distance traveled by the aeroplane in the was direction is AB = 1200 ´  = 1800 km
BC is the distance between the two 11/2 hours.
AB² + AC² = BC²
(1500)² + (1800)² = BC²
2250000 + 3240000 = BC²
5490000 = BC²
BC = 100km.

42.  Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.

Solution:Let AD be a pole of height 11m and BC be a pole of height 6m
Construction: Join CD and draw ^ r from C to AD meeting AD at E.
Proof: ABCD is rectangle
\ AB = CE = 12m
and BC = AE = 6m
\ DE = AD – AE
11 – 6 = 5m
In D PEC
DE² + EC² = DC²
or 5² + 12² = DC²
25 + 144 = DC²
DC =  = 13 m
the distance between the two poles = 13m.

43.  D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.Prove that AE² + BD² = AB² + DE².

Solution:
To prove: AC² + BD² = AB² + DE²
Proof:
In D AEC
AE² = EC² + AC² ………………………….. (1)
In D BDC
BD² = BC² + CD² ………………………….. (2)
(1) + (2)
A ²+ BD² = EC² + BC² + AC² + CO²
               = (EC² + CO²) + (AC² + BC²)
               = DE² + AB²
Here proved.

44.  The perpendicular from A on side BC of a D ABC intersects BC at D such that DB = 3 CD. Prove that 2 AB² = 2 AC² + BC².

Solution:
From DABD

From DADC

From (1) and (2)

BD = 3DC (given)\ BC = BD + DC = 4DC

Now

45.  In an equilateral triangle ABC, D is a point on side BC such that BD =9 AD² = 7 AB².

Solution:
Given: D ABC is an equilateral D . ‘D’ is a point on BC such that BD =  BC.
To prove: 9AD² = 7AB²
construction: Draw AE ^ r to BC meeting C and E
Proof:
In an equilateral triangle the altitude and median are same line regret. Therefore A is the median of the D ABC.E is the midpoint of BC
BC + EC = BC
2EC = BC
BE = EC =  BC ………………………….. (1)
BD = BC (Given)……………………(2)
DC = BC – BD
     = BC - BC .  BC ………………… (3)
DE = BE – B
     = BC -
     = BC
     =  = BC
In D ADE
D ADE
AD² = AE² + ED² …………………(1)
But in D AEC
AC² = AE² + ED²(or) AE² = AC² - EC²substituting in (1)
AD² = AC² – EC² + ED²
      = AC² – DE² - CE²
      = AC² + (DC+CE)(DE – CE)
      = AC² + (DC)(DE – CE)
      = AC² + DC.DE – DC.CE
      = AC² +
      = AC² +
      = AC² +
      = AC² +
      = AC² -
      =
AD² = AB²

46.  In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Solution:
Let the side of the equilateral D ABC be ‘a’
Let AE be the altitude from A to BC meeting
BC at E.
E is the midpoint of is C
Therefore BE =  BC = a
In D ABF AB² = AB² + BE²
AE² = AB² – BE²
      = a² -
      = a² -
      =  =
AE =  ………………………….. (1)
Now,
Three times, square one of its side = 3a²Four times the square of y
Its one of the altitude = 4 = 3a²
Therefore it is proved that in an equilateral triangle three times the square of one side is equal to four times the square of its altitude.

47.  Tick the correct answer and justify : In Ä ABC, AB = 6 3 cm, AC = 12 cm and BC = 6 cm. The angle B is :
(A) 120° (B) 60°
(C) 90° (D) 45°

Solution:In D ABC AB = 6cm AC = 12 cm
BC = 6 cm

AB² = (6)² = 36 ´ 3 = 108
BC² = 6² = 36
AC² = 12² = 144
AB² + BC² = 108 + 36 = 144 = AC²
AB² + BC² = AC²
Þ = 90Answer = (c) = 90^0.