Exercise 6.1

1. Fill in the blanks using the correct word given in brackets:

(i) All circles are _______________. (congruent, similar)

(ii) All squares are _______________. (similar, congruent)

(iii) All _______________ triangles are similar. (isosceles, equilateral)

(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are _______________ and (b) their corresponding sides are _______________. (equal, proportional)

Answer:

(i) similar

(ii) similar

(iii) equilateral

(iv) equal, proportional

2. Give two different examples of pair of:

(i) similar figures

(ii) non-similar figures

Answer:

(i) Two different examples of pair of similar figures are:

(a) Any two rectangles

(b) Any two squares

(ii) Two different examples of pair of non-similar figures are:

(a) A scalene and an equilateral triangle

(b) An equilateral triangle and a right angled triangle

3. State whether the following quadrilaterals are similar or not:

Answer:

On looking at the given figures of the quadrilaterals, we can say that they are not similar because their angles are not equal.

EXERCISE 6.2

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

Answer:

(i) Since DE || BC,

Let, EC = x cm

It is given that DE || BC

By using basic proportionality theorem, we obtain

x = 2

EC = 2 cm.

(ii) Let AD = x cm.

It is given that DE || BC.

By using basic proportionality theorem, we obtain

x = 2.4

AD = 2.4cm.

2. E and F are points on the sides PQ and PR respectively of a ∆ PQR. For each of the following cases, state whether EF II QR:

(i) PE = 3.9 cm, EQ = 3cm, 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

Answer:

(i)Given: PE = 3.9 cm, EQ = 3cm, PF = 3.6 cm and FR = 2.4 cm

 

 

Hence,

Therefore, EF is not parallel to QR.

(ii)Given: PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

 

Therefore,EF is parallel to QR.

(iii)Given: PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

And ER = PR – PF = 2.56 – 0.36 = 2.20 cm

 

Hence,

Therefore, EF is parallel to QR.

3. In figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD.

Answer:

In figure, LM || CB

By using Basic Proportionality theorem

And in ∆ ACD, LN || CD

  ……i

Similalry, in ∆ ACD, LN || CD

  …….ii

By using Basic Proportionality theorem

From eq. (i) and (ii), we have

4. In figure, DE || AC and DF || AE. Prove that BF/GE = BE/EC .

Answer:

In ∆ BCA, DE || AC

[Basic Proportionality theorem] ……….(i)

And in ∆ BEA, DF || AE

[Basic Proportionality theorem] ……….(ii)

From eq. (i) and (ii), we have

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

Answer:

In ∆ PQO, DE || OQ

[Basic Proportionality theorem] ……….(i)

And in ∆ POR, DF || OR

[Basic Proportionality theorem] ……….(ii)

From eq. (i) and (ii), we have

Therefore, EF || QR [By the converse of Basic Proportionality Theorem]

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

Answer:

And in ∆ POQ, AB || PQ

[Basic Proportionality theorem] ……….(i)

And in ∆ OPR, AC || PR

[Basic Proportionality theorem] ……….(ii)

From eq. (i) and (ii), we have

Therefore, BC || QR (By the converse of Basic Proportionality Theorem)

7. Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Answer:

Consider the given figure in which l  is a line  drawn through the mid-point P of line segment AB meeting AC at Q, such that  PQ || BC

 By using Basic Proportionality theorem, we obtain,

    (P is the midpoint of AB ∴ AP = PB)

⇒ AQ = QC

Or, Q is the mid-point of AC.

8. Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

Answer:

Given: A triangle ABC, in which P and Q are the mid-points of

sides AB and AC respectively.

Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and AC respectively.

i.e., AP = PB and AQ = QC

It can be observed that

and 

 Therefore, 

Hence, by using basic proportionality theorem, we obtain

PQ || BC.

9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO .

Answer:

Given: A trapezium ABCD, in which AB || DC and its diagonals

AC and BD intersect each other at O.

Draw a line EF through point O, such that EF || CD

In ΔADC, EO || CD

By using basic proportionality theorem, we obtain

   …i

In ΔABD, OE || AB

So, by using basic proportionality theorem, we obtain

⇒   …ii

From eq. (i) and (ii), we get

⇒

10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO Such that ABCD is a trapezium.

Answer:

Given: A quadrilateral ABCD, in which its diagonals AC and

BD intersect each other at O such that , i.e.

Quadrilateral ABCD is a trapezium.

Construction: Through O, draw OE || AB meeting AD at E.

In ∆ ADB, we have OE || AB [By construction] By Basic Proportionality theorem

 …..i

However, it is given that

 …..ii

From eq. (i) and (ii), we get

⇒ EO || DC [By the converse of basic proportionality theorem]

⇒ AB || OE || DC

⇒ AB || CD

∴ ABCD is a trapezium.

EXERCISE 6.3

1. State which pairs of triangles in figure, 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:

Answer:

(i) In  ΔABC and ΔPQR, we have
∠A = ∠P = 60° (Given)
∠B = ∠Q = 80° (Given)
∠C = ∠R = 40° (Given)
∴ ΔABC ~ ΔPQR (AAA similarity criterion)

(ii) In  ΔABC and ΔPQR, we have
AB/QR = BC/RP = CA/PQ
∴  ΔABC ~ ΔQRP (SSS similarity criterion)

(iii) In ΔLMP and ΔDEF, we have
LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6
MP/DE = 2/4 = 1/2
PL/DF = 3/6 = 1/2
LM/EF= 2.7/5 = 27/50
Here, MP/DE = PL/DF ≠ LM/EF
Hence, ΔLMP and ΔDEF are not similar.

(iv) In ΔMNL and ΔQPR, we have
MN/QP = ML/QR = 1/2
∠M = ∠Q = 70°
∴ ΔMNL ~ ΔQPR (SAS similarity criterion)

(v) In ΔABC and ΔDEF, we have
AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80°
Here, AB/DF = 2.5/5 = 1/2
And, BC/EF = 3/6 = 1/2
⇒ ∠B ≠ ∠F
Hence, ΔABC and ΔDEF are not similar.

(vi) In ΔDEF,we have
∠D + ∠E + ∠F = 180° (sum of angles of a triangle)
⇒ 70° + 80° + ∠F = 180°
⇒ ∠F = 180° – 70° – 80°
⇒ ∠F = 30°
In PQR, we have
∠P + ∠Q + ∠R = 180 (Sum of angles of Δ)
⇒ ∠P + 80° + 30° = 180°
⇒ ∠P = 180° – 80° -30°
⇒ ∠P = 70°
In ΔDEF and ΔPQR, we have
∠D = ∠P = 70°
∠F = ∠Q = 80°
∠F = ∠R = 30°
Hence, ΔDEF ~ ΔPQR (AAA similarity criterion)

2. In the figure, ΔODC ∝ ¼ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.

Answer:

DOB is a straight line.

∴ ∠DOC + ∠COB = 180°

⇒ ∠DOC = 180° − 125°

= 55°

In ΔDOC,

∠DCO + ∠CDO + ∠DOC = 180°

(Sum of the measures of the angles of a triangle is 180º.)

⇒ ∠DCO + 70º + 55º = 180°

⇒ ∠DCO = 55°

It is given that ΔODC ∼ ΔOBA.

∴ ∠OAB = ∠ OCD [Corresponding angles are equal in similar triangles.]

⇒ ∠OAB = 55°.

3. 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 AO/OC = OB/OD

Answer:

Given: ABCD is a trapezium in which AB DC.

In ΔDOC and ΔBOA,

∠CDO = ∠ABO [Alternate interior angles as AB || CD]

∠DCO = ∠BAO [Alternate interior angles as AB || CD]

∠DOC = ∠BOA [Vertically opposite angles]

∴ ΔDOC ∼ ΔBOA [AAA similarity criterion]

 ∴ …..  [Corresponding sides are proportional]

Hence, 

4. In the figure, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.

Answer:

We have, 

In ΔPQR, ∠PQR = ∠PRQ

∴ PQ = PR …..(i)

Given,

Using (i), we get,

In ΔPQS & In ΔTQR

∠Q = ∠Q

∴ ΔPQS ~ ΔTQR [SAS similarity criterion]

5. S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.

Answer:

In ΔRPQ and ΔRST,
∠RTS = ∠QPS (Given)
∠R = ∠R (Common angle)
∴ ΔRPQ ~ ΔRTS (By AA similarity criterion)

6. In the figure, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.

Answer:

It is given that ΔABE ≅ ΔACD.
∴ AB = AC [By cpct] …(i)
And, AD = AE [By cpct] …(ii)
In ΔADE and ΔABC,
AD/AB = AE/AC [Dividing equation (ii) by (i)]
∠A = ∠A [Common angle]
∴ ΔADE ~ ΔABC [By SAS similarity criterion]

7. In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
(i) ΔAEP ~ ΔCDP
(ii) ΔABD ~ ΔCBE
(iii) ΔAEP ~ ΔADB
(iv) ΔPDC ~ ΔBEC

Answer:

(i) In ΔAEP and ΔCDP,

∠AEP = ∠CDP (Each 90°)
∠APE = ∠CPD (Vertically opposite angles)
Hence, by using AA similarity criterion,
ΔAEP ~ ΔCDP

(ii) In ΔABD and ΔCBE,

∠ADB = ∠CEB (Each 90°)
∠ABD = ∠CBE (Common)
Hence, by using AA similarity criterion,
ΔABD ~ ΔCBE

(iii) In ΔAEP and ΔADB,

∠AEP = ∠ADB (Each 90°)
∠PAE = ∠DAB (Common)
Hence, by using AA similarity criterion,
ΔAEP ~ ΔADB

(iv) In ΔPDC and ΔBEC,

∠PDC = ∠BEC (Each 90°)
∠PCD = ∠BCE (Common angle)
Hence, by using AA similarity criterion,
ΔPDC ~ ΔBEC

8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.

Answer:

In ΔABE and ΔCFB,we have,

∠A = ∠C (Opposite angles of a parallelogram)
∠AEB = ∠CBF (Alternate interior angles as AE || BC)

∴By AA-criterion of similarity, we have

∴ ΔABE ~ ΔCFB (By AA similarity criterion)

9. In the figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:
(i) ΔABC ~ ΔAMP
(ii) CA/PA = BC/MP

Answer:

(i) In Δ ABC and AMP, we have,

∠ABC = ∠AMP =90⁰ [Given]

∠BAC = ∠MAP [Common angles]

∴   Δ ABC ~ Δ AMP  [By AA-criterion of similarity, we have]

⇒ CA/PA = BC/MP ….. (Corresponding sides of similar trianlges are proportional)

10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, Show that:
(i) CD/GH = AC/FG
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF

Answer:

We have, Δ ABC ~ Δ FEG

∴  ∠A = ∠F, ∠B = ∠E, & ∠ACB = ∠FGE

∴ ∠ACD = ∠FGH (Angle Bisector)

And ∠DCB = ∠HGE (Angle Bisector)

In ΔACD & ΔFGH

∠A = ∠F (Proved above)

∠ACD = ∠FGH (Proved above)

∴  Δ ACD ~ Δ FGH [By AA similarity criterion]

⇒ [(CD)/ (GH)] = [(AC) / (FG)]

In ΔDCB & ΔHGE,

∠DCB= ∠HGE (Proved above)

∠B= ∠E (Proved above)

∴  Δ DCB~ Δ HGE [By AA similarity criterion]

In ΔDCA & ΔHGF,

∠ACD = ∠FGH (Proved above)

∠A = ∠F (Proved above)

∴  Δ DCA~ Δ HGF [By AA similarity criterion]

11. In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.

Answer:

Here Δ ABC is isosceles with AB = AC

∠B = ∠C

In Δ ABD and ECF, we have

∠ABD = ∠ECF[Each 90°]

∠ABD = ∠ECF = [Proved above]

By AA-criterion of similarity, we have

Δ ABD ~ Δ ECF

12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see figure). Show that ΔABC ~ ΔPQR.

Answer:

Given: AD is the median of Δ ABC and PM is the median of Δ PQR such that

Median divides the opposite side.

∴   BD = BC / 2 [Given]

And QM = QR / 2 [Given]

Given that,

AB/PQ = BC/QR = AD/PM

⇒  AB/PQ =[( ½BC) / (½QR) ]= AD/PM

 ⇒   AB/PQ = BD/QM = AD/PM

In ΔABD and ΔPQM,

  AB/PQ = BD/QM = AD/PM [ Proved above]

∴ ΔABD ∼ ΔPQM (By SSS similarity criterion)

⇒ ∠ABD = ∠PQM (Corresponding angles of similar triangles)

In ΔABC and ΔPQR,

∠ABD = ∠PQM (Proved above)

AB / PQ = BC / QR

∴ ΔABC ∼ ΔPQR (By SAS similarity criterion)

13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB.CD

Answer:

In triangles ABC and DAC,

∠ADC = ∠BAC (Given)
∠ACD = ∠BCA (Common angle)
∴ ΔADC ~ ΔBAC (By AA similarity criterion)
We know that corresponding sides of similar triangles are in proportion.
∴ CA/CB =CD/CA
⇒ CA² = CB x CD.

14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR.

Answer:

Given: AD is the median of Δ ABC and PM is the median of Δ PQR such that

⇒AB / PQ = AC / PR = AD / PM

Let us extend AD and PM up to point E and L respectively, such that AD = DE and PM = ML. Then, join B to E, C to E, Q to L, and R to L.

We know that medians divide opposite sides.

Therefore, BD = DC and QM = MR

Also, AD = DE (By construction)

And, PM = ML (By construction)

In quadrilateral ABEC, diagonals AE and BC bisect each other at point D.

Therefore, quadrilateral ABEC is a parallelogram.

∴ AC = BE and AB = EC (Opposite sides of a parallelogram are equal)

Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, PQ = LR

It was given that

⇒AB / PQ = AC / PR = AD / PM

⇒AB / PQ = BE / QL = [(2AD) / (2PM)]

⇒AB / PQ = BE / QL = AE / PL

∴ ΔABE ∼ ΔPQL (By SSS similarity criterion)

We know that corresponding angles of similar triangles are equal.

∴ ∠BAE = ∠QPL … (1)

Similarly, it can be proved that ΔAEC ∼ ΔPLR and

∠CAE = ∠RPL … (2)

Adding equation (1) and (2), we obtain

∠BAE + ∠CAE = ∠QPL + ∠RPL

⇒ ∠CAB = ∠RPQ … (3)

In ΔABC and ΔPQR,

AB / PQ = AC / PR

∠CAB = ∠RPQ [Using equation (3)]

∴ ΔABC ∼ ΔPQR (By SAS similarity criterion)

15. 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.

Answer:

Let AB the vertical pole and AC be its shadow. Also, let DE be the vertical tower and DF be its shadow. Joined BC and EF.

Length of the vertical pole = 6m (Given)

Shadow of the pole = 4 m (Given)

Let Height of tower = h m

Length of shadow of the tower = 28 m (Given)

In ΔABC and ΔDEF,

∠C = ∠E (angular elevation of sum)

∠B = ∠F = 90°

∴ ΔABC ~ ΔDEF (By AA similarity criterion)

∴ AB/DF = BC/EF (If two triangles are similar corresponding sides are proportional)

∴ 6/h = 4/28

⇒ h = 6 x 28/4

⇒ h = 6 x 7

⇒ h = 42 m

Hence, the height of the tower is 42 meters.

16. If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR prove that AB/PQ = AD/PM.

Answer:

Given: AD and PM are the medians of triangles

ABC and PQR respectively, where

It is given that Δ ABC ~ Δ PQR

We know that the corresponding sides of similar triangles are in proportion.

∴ AB / PQ = AC / AD and BC / QR ….(1)

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R … (2)

Since AD and PM are medians, they will divide their opposite sides.

∴ BD = BC/2 &  QM = QR/2 …(3)

From equations (1) and (3), we obtain

AB/PQ = BD/QM … (4)

In ΔABD and ΔPQM,

∠B = ∠Q [Using equation (2)]

AB/PQ = BD/QM

∴ ΔABD ∼ ΔPQM (By SAS similarity criterion)

AB/PQ = BD/QM  = AD/PM

EXERCISE 6.4

1. Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm² and 121 cm². If EF = 15.4 cm, find BC.

Answer:

It is given that,
Area of ΔABC = 64 cm²
Area of ΔDEF = 121 cm²
EF = 15.4 cm
and, ΔABC ~ ΔDEF
∴ Area of ΔABC/Area of ΔDEF = AB²/DE²
= AC2/DF² = BC²/EF² …(i)
[If two triangles are similar, ratio of their areas are equal to the square of the ratio of their corresponding sides]
∴ 64/121 = BC²/EF²
⇒ (8/11)² = (BC/15.4)²
⇒ 8/11 = BC/15.4
⇒ BC = 8×15.4/11
⇒ BC = 8 × 1.4
⇒ BC = 11.2 cm

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

Answer:

ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect each other at point O.
In ΔAOB and ΔCOD, we have
∠1 = ∠2 (Alternate angles)
∠3 = ∠4 (Alternate angles)
∠5 = ∠6 (Vertically opposite angle)
∴ ΔAOB ~ ΔCOD [By AAA similarity criterion]
Now, Area of (ΔAOB)/Area of (ΔCOD)
= AB²/CD² [If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides]
= (2CD)²/CD² [∴ AB = CD]
∴ Area of (ΔAOB)/Area of (ΔCOD)
= 4CD²/CD = 4/1
Hence, the required ratio of the area of ΔAOB and ΔCOD = 4:1

3. In figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area (ΔABC)/area (ΔDBC) = AO/DO.

Answer:

Let us draw two perpendiculars AP and DM on line BC.

We know that area of a triangle = 1/2 x Base x Height

 ∴  

In ΔAPO and ΔDMO,

∠APO = ∠DMO (Each = 90°)

∠AOP = ∠DOM (Vertically opposite angles)

∴ ΔAPO ∼ ΔDMO (By AA similarity criterion)

∴  AP/DM = AO/DO

⇒ 

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

Answer:

Let us assume two similar triangle as ΔABC ~ ΔPQR

Given that, ar(ΔABC) = ar(ΔABC)

Putting this value in equation (1) we obtain

 1 = 

 ⇒  AB = PQ, BC = QR and AC = PR

∴  ΔABC ≅ ΔPQR   (By SSS congruence criterion)

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

Answer:

D and E are the mid-points of ΔABC

:.DE || AC and DE = 1/2 AC

In ΔBED and ΔBCA

∠BED = ∠BCA   (Corresponding angle)

∠BDE = ∠BAC   (Corresponding angle)

∠EBD = ∠CBA   (Common angles)

∴ΔBED ~ ΔBCA (AAA similarity criterion)

⇒

Similary

⇒

Also ar(ΔDEF) = ar(ΔABC) –  [ar(ΔBED) + ar(ΔCFE) + ar(ΔADF)]

⇒

⇒

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

Answer:

Let us assume two similar triangles as ΔABC ∼ ΔPQR. Let AD and PS be the medians of these triangles.

∵  ΔABC ∼ ΔPQR

:.(AB)/(PQ) = (BC)/(QR) = (AC)/(PR)…(1)

∠A = ∠P, ∠B = ∠Q, ∠C = ∠R … (2)

Since AD and PS are medians,

∴ BD = DC = BC/2

And, QS = SR = QR/2

Equation (1) becomes

(AB)/(PQ) = (BD)/(QS) = (AC)/(PR)  ….(3)

In ΔABD and ΔPQS,

∠B = ∠Q [Using equation (2)]

and (AB)/(PQ) = (BD)/(QS)    [Using equation (3)]

∴ ΔABD ∼ ΔPQS (SAS similarity criterion)

Therefore, it can be said that

AB/PQ = BD/QS = AD/PS ….(4)

From equations (1) and (4), we may find that

AB/PQ = BC/QR = AC/PR = AD/PS

And hence,

7. 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 the diagonals.
Tick the correct answer and justify:

Answer:

Let ABCD be a square of side a.

Therefore, its diagonal = √2a

Two desired equilateral triangles are formed as ΔABE and ΔDBF.

Side of an equilateral triangle, ΔABE, described on one of its sides = a

Side of an equilateral triangle, ΔDBF, described on one of its diagonals =√2a

We know that equilateral triangles have all its angles as 60º and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.

8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is:

(A) 2: 1

(B) 1: 2

(C) 4: 1

(D) 1: 4

Answer:

We know that equilateral triangles have all its angles as 60º and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.

Let side of ΔABC = x

Therefore, side of  ΔBDE = x/2

Hence, (C) is the correct answer.

9. 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) 81: 16

(D) 16: 81

Answer:

If two triangles are similar to each other, then the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides of these triangles.

It is given that the sides are in the ratio 4:9.

Therefore, ratio between areas of these triangles = (4/9)² = 16/81

Hence,(D) is the correct answer.

Exercise 6.5

1. Sides of triangles are given below. Determine which of them are right triangles? In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm

Answer:
(i) Given that the sides of the triangle are 7 cm, 24 cm, and 25 cm.
Squaring the lengths of these sides, we will get 49, 576, and 625.
49 + 576 = 625
(7)² + (24)² = (25)²
The sides of the given triangle are satisfying Pythagoras theorem.Hence, it is right angled triangle.
Length of Hypotenuse = 25 cm

(ii) Given that the sides of the triangle are 3 cm, 8 cm, and 6 cm.
Squaring the lengths of these sides, we will get 9, 64, and 36.
However, 9 + 36 ≠ 64
Or, 3² + 6² ≠ 8²
Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
Therefore, the given triangle is not satisfying Pythagoras theorem.

(iii) Given that sides are 50 cm, 80 cm, and 100 cm.
Squaring the lengths of these sides, we will get 2500, 6400, and 10000.
However, 2500 + 6400 ≠ 10000
Or, 50² + 80² ≠ 100²
Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
Therefore, the given triangle is not satisfying Pythagoras theorem.
Hence, it is not a right triangle.

(iv) Given that sides are 13 cm, 12 cm, and 5 cm.
Squaring the lengths of these sides, we will get 169, 144, and 25.
Clearly, 144 +25 = 169
Or, 12² + 5² = 13²
The sides of the given triangle are satisfying Pythagoras theorem.
Therefore, it is a right triangle.
Length of the hypotenuse of this triangle is 13 cm.

2. PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM × MR.

Answer:

Given: ΔPQR is right angled at P is a point on QR such that PM ⊥QR.
To prove:. PM² = QM × MR
Proof: In ΔPQM, we have
PQ² = PM² + QM² [By Pythagoras theorem]
Or, PM² = PQ² – QM² …(i)
In ΔPMR, we have
PR² = PM² + MR² [By Pythagoras theorem]
Or, PM² = PR² – MR² …(ii)
Adding (i) and (ii), we get
2PM² = (PQ² + PM2) – (QM² + MR²)
= QR² – QM² – MR²        [∴ QR² = PQ² + PR²]
= (QM + MR)² – QM² – MR²
= 2QM × MR
∴ PM² = QM × MR

3. In figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that
(i) AB² = BC × BD
(ii) AC² = BC × DC
(iii) AD² = BD × CD

Answer:

Given: ABD is a triangle right angled at A and AC ⊥ BD.

(i) In ΔADB and ΔCAB, we have

∠DAB = ∠ACB (Each equals to 90°)

∠ABD = ∠CBA (Common angle)

∴ ΔADB ~ ΔCAB [AA similarity criterion]

⇒  AB/CB = BD/AB

⇒  AB² = CB × BD

(ii) Let ∠CAB = x

In ΔCBA,

∠CBA = 180° – 90° – x

∠CBA = 90° – x

Similarly, in ΔCAD

∠CAD = 90° – ∠CBA

= 90° – x

∠CDA = 180° – 90° – (90° – x)

∠CDA = x

In ΔCBA and ΔCAD, we have

∠CBA = ∠CAD

∠CAB = ∠CDA

∠ACB = ∠DCA (Each equals to 90°)

∴ ΔCBA ~ ΔCAD [By AAA similarity criterion]

⇒ AC/DC = BC/AC

⇒  AC² = DC × BC

(iii) In ΔDCA and ΔDAB, we have

∠DCA = ∠DAB (Each equals to 90°)

∠CDA = ∠ADB (common angle)

∴ ΔDCA ~ ΔDAB [By AA similarity criterion]

⇒ DC/DA = DA/DA

 ⇒ AD² = BD × CD

4. ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².

Answer:

Since ABC is an isosceles right triangle, right angled at C.

∴ AC = CB

Applying Pythagoras theorem in ΔABC (i.e., right-angled at point C), we obtain

AC² + CB² = AB²

⇒ AC² +  AC² = AB² ( AC = CB)

⇒ 2 AC² = AB²

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

Answer:

Given that ΔABC is an isosceles triangle having AC = BC and AB² = 2AC²
In ΔACB,
AC = BC (Given)
AB² = 2AC² (Given)
AB2 = AC² + AC²
= AC² + BC² [Since, AC = BC]
Hence, By Pythagoras theorem ΔABC is right angle triangle.

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

Answer:

 ABC is an equilateral triangle of side 2a.
Draw, AD ⊥ BC
In ΔADB and ΔADC, we have
AB = AC [Given]
AD = AD [Given]
∠ADB = ∠ADC [equal to 90°]
Therefore, ΔADB ≅ ΔADC by RHS congruence.
Hence, BD = DC [by CPCT]
In right angled ΔADB,
AB² = AD² + BD²
(2a)² = AD² + a²
⇒ AD²  = 4a² – a²
⇒ AD²  = 3a²
⇒ AD  = √3a

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

Answer:

Let the diagonals AC and BD of rhombus ABCD intersect each other at O. Since the diagonals of a rhombus bisect each other at right angles.

In ΔAOB, ΔBOC, ΔCOD, ΔAOD,

Applying Pythagoras theorem, we obtain

AB² = AO² + OB² ….(1)

BC² = BO² + OC² …. (2)

CD² = CO² + OD² …..(3)

AD² = AO² + OD² ……(4)

Adding all  these equation we obtain

AB² + BC² + CD² + AD² = 2(AO² + OB² + OC² + OD²)

=

(Diagonals bisect each other)

 =

 = (AC)² + (BD)²

8. In Fig. O is a point in the interior of a triangle
ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE² ,
(ii) AF² + BD² + CE² = AE² + CD² + BF².

Answer:

 Applying Pythagoras theorem in ΔAOF, we obtain

OA² = OF² + AF²

Similarly, in ΔBOD

OB² = OD² + BD²

Similarly, in ΔCOE

OC² = OE² + EC²

Adding these equations,

OA² + OB² + OC² = OF² + AF² + OD² + BD² + OE² + EC²

OA² + OB² + OC² – OF² – OD² – OE² = AF² + BD² + EC²

From above result

AF² + BD² + EC² = (OA²- OE² ) + (OC² – OD²) + (OB² – OF² )

∴ AF² + BD² + EC² = AE² + CD² + BF².

9. 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.

Answer:

Let OA be the wall and AB be the ladder.

Therefore, by Pythagoras theorem,

AB² = OA² + BO²

(10 m)² =( 8 m)² + OB²

100m² = 64m² + OB²

OB² = 36m²

OB = 6m

Therefore the distance of the foot of the ladder from the base of the wall is 6 m.

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

Answer:

Let AB (= 24m) be a guy wire attached to a vertical pole. BC of height 18 m. To keep the wire taut, let it be fixed to a stake at A. Then, ABC is a right triangle, right angled at O.

Let OB be the pole and AB be the wire.

By Pythagoras theorem,

AB² = OB² + OA²

(24 m)² = ( 18 m)² + OA²

576m² = 324m² + OA²

OA² = (576 – 324)m² = 25m²

OA² = √252 m = √6 x 6 x 7 = 6√7 m.

OA =6√7 m.

Hence, the stake may be placed at distance of 6√7 m from the base of the pole.

11. An aeroplane leaves an airport and flies due north at a speed of 1000 km pwe 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 hours?

Answer:

Let the first aeroplane starts from O and goes upto A towards north where

 

Distance travelled by the plane flying towards north in  hrs

Similarly, distance travelled by the plane flying towards west inhrs

Let these distances be represented by OA and OB respectively.

Applying Pythagoras theorem,

Distance between these planes after

 = √[(1500)² + (1800)²]

= √(5490000)

Therefore, the distance between these planes will be 300√61 km.

12. 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.

Answer:

Let CD and AB be the poles of height 11 m and 6 m.

Therefore, CP = 11 − 6 = 5 m.

From the figure, it can be observed that AP = 12m

Applying Pythagoras theorem for ΔAPC, we obtain

AP² + PC² = AC²

(12m)² + (5m)² = AC²

AC² = (144 + 25)m² = 169m²

AC = 13m

Therefore, the distance between their tops is 13 m.

13. 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.

Answer:

 Applying Pythagoras theorem in ΔACE, we get
AC² + CE² = AE² ….(i)
Applying Pythagoras theorem in ΔBCD, we get
BC² + CD² = BD² ….(ii)
Using equations (i) and (ii), we get
AC² + CE² + BC² + CD² = AE² + BD² …(iii)
Applying Pythagoras theorem in ΔCDE, we get
DE² = CD² + CE²
Applying Pythagoras theorem in ΔABC, we get
AB² = AC² + CB²
Putting these values in equation (iii), we get
DE² + AB² = AE² + BD².

14. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Fig.). Prove that 2AB² = 2AC² + BC².

Answer:

Given that in ΔABC, we have
AD ⊥BC and BD = 3CD
In right angle triangles ADB and ADC, we have
AB² = AD² + BD² …(i)
AC² = AD² + DC² …(ii) [By Pythagoras theorem]
Subtracting equation (ii) from equation (i), we get
AB² – AC² = BD² – DC²
= 9CD² – CD²  [∴ BD = 3CD]
= 9CD² = 8(BC/4)² [Since, BC = DB + CD = 3CD + CD = 4CD]
Therefore, AB² – AC² = BC²/2
⇒ 2(AB² – AC²) = BC²
⇒ 2AB² – 2AC² = BC²
∴ 2AB² = 2AC² + BC².

15. In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD² = 7AB².

Answer:

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.

∴ BE = EC = BC/2 = a/2

and AE= a√3/2

Given that, BD = 1/3 BC

∴ BD = a/3

DE = BE – BD = a/2 – a/3 = a/6

Applying Pythagoras theorem in ΔADE, we get

AD² = AE² + DE²

AD² = (a√3/2)² + (a/6)²

AD² = (3a²/4) + (a²/36)

 ⇒ AD² = (28a²/36)

 ⇒ AD² = 7/9AB²

⇒ 9AD² = 7AB²

16. 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.

Answer:

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.

∴ BE = EC = BC/2 = a/2

Applying Pythagoras theorem in ΔABE, we obtain

AB² = AE² + BE²

a² = AE² + (a/2)²

AE² = a² – a²/4

 AE²  = 3a²/4

4 AE² = 3a²

⇒ 4 × (Square of altitude) = 3 × (Square of one side).

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

Answer:

Given that, AB = 6√3 cm, AC = 12 cm, and BC = 6 cm
We can observe that
AB² = 108
AC² = 144
And, BC² = 36
AB² + BC² = AC²
The given triangle, ΔABC, is satisfying Pythagoras theorem.
Therefore, the triangle is a right triangle, right-angled at B.
∴ ∠B = 90°
Hence, the correct option is (C).

EXERCISE 6.6

1. In figure, PS is the bisector of ∠QPR of Δ PQR. Prove that QS/SR = PQ/PR .

Answer:

Given:

Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.

Given that, PS is the angle bisector of ∠QPR.

∠QPS = ∠SPR … (1)

By construction,

∠SPR = ∠PRT (As PS || TR) … (2)

∠QPS = ∠QTR (As PS || TR) … (3)

Using these equations, we obtain

∠PRT = ∠QTR

∴ PT = PR

By construction,

PS || TR

By using basic proportionality theorem for ΔQTR,

⇒ QS/SR = QP/PT

⇒ QS/SR = PQ/PR (∴PT = PR)

2. In figure, D is a point on hypotenuse AC of Δ ABC, BD ⊥ AC, DM ⊥ BC and DN ⊥AB. Prove that:

(i) DM²= DN.MC

(ii) DN²= DM.AN

Answer: (i) Let us join DB

We have, DN || CB, DM || AB, and ∠B = 90°

∴ DMBN is a rectangle.

∴ DN = MB and DM = NB

The condition to be proved is the case when D is the foot of the perpendicular drawn from B to AC.

∴ ∠CDB = 90°

⇒ ∠2 + ∠3 = 90° … (1)

In ΔCDM,

∠1 + ∠2 + ∠DMC = 180°

⇒ ∠1 + ∠2 = 90° … (2)

In ΔDMB,

∠3 + ∠DMB + ∠4 = 180°

⇒ ∠3 + ∠4 = 90° … (3)

From equation (1) and (2), we obtain

∠1 = ∠3

From equation (1) and (3), we obtain

∠2 = ∠4

In ΔDCM and ΔBDM,

∠1 = ∠3 (Proved above)

∠2 = ∠4 (Proved above)

∴ ΔDCM ∼ ΔBDM (AA similarity criterion)

⇒ (BM)/(DM) = (DM)/(MC)

⇒ (DN)/(DM) = (DM)/(MC)         ∴ (BM = DN)

⇒ DM² = DN × MC

(ii) In right triangle DBN,

∠5 + ∠7 = 90° … (4)

In right triangle DAN,

∠6 + ∠8 = 90° … (5)

D is the foot of the perpendicular drawn from B to AC.

∴ ∠ADB = 90°

⇒ ∠5 + ∠6 = 90° … (6)

From equation (4) and (6), we obtain

∠6 = ∠7

From equation (5) and (6), we obtain

∠8 = ∠5

In ΔDNA and ΔBND,

∠6 = ∠7 (Proved above)

∠8 = ∠5 (Proved above)

∴ ΔDNA ∼ ΔBND (AA similarity criterion)

=> AN/DN = DN/NB

⇒ DN² = AN × NB

⇒DN² = AN × DM (As NB = DM)

3. In figure, ABC is a triangle in which ∠ABC > 90⁰ and AD ⊥ CB produced. Prove that:

AC² = AB² + BC.BD

Answer

Applying Pythagoras theorem in ΔADB, we obtain

AB² = AD² + DB² …(1)

Applying Pythagoras theorem in ΔACD, we obtain

AC² = AD² + DC²

AC² = AD² + (DB + BC)²

AC² = AD² + DB² + BC² + 2DB x BC

AC² = AB² + BC² + 2DB x BC  [Using equation (1)]

4. In figure, ABC is a triangle in which ∠ABC < 90⁰ and AD ⊥ BC produced. Prove that: AC² =AB² + BC² – 2BC.BD

Answer:

Applying Pythagoras theorem in ΔADB, we obtain

AD² + DB² = AB²

⇒ AD² =  AB² – DB² …(i)

Applying Pythagoras theorem in ΔADC, we obtain

AD² + DC² = AC²

AB² – BD² + DC² = AC² …. Using eqn.(i)

AB² – BD² + (BC – BD)² = AC²

⇒ AB² – BD²  + BC² + BD² – 2BC x BD

 = AB² + BC² – 2BC x BD

5. In figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:

(i) AC² = AD² + BC.DM + (BC/2)²

(ii) AC² = AD² – BC.DM + (BC/2)²

(iii) AC² + AB² = 2AD² + 1/2(BC)²

Answer:

(i) Applying Pythagoras theorem in ΔAMD, we obtain

AM² + MD² = AD² … (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM² + MC² = AC²

AM² + (MD + DC)² = AC²

(AM² + MD²) + DC² + 2MD.DC = AC²

AD² + DC² + 2MD.DC = AC² (Using equation (1))

Using the result, DC = BC/2, we obtain

AD² + (BC/2)² + 2MD.(BC/2) = AC²

AD² + (BC/2)² + MC.BC = AC²

(ii) Applying Pythagoras theorem in ΔABM, we obtain

AB² = AM² + MB²

 = (AD² – DM²) + MB²

 = (AD² – DM²) + (BD – MD)²

 = AD² – DM² + BD² + MD² – 2BD.MD

= AD² + BD² – 2BD x MD

= AD² + (BC/2)² – 2(BC/2) x MD

= AD² + (BC/2)² – BC x MD

(iii)Applying Pythagoras theorem in ΔABM, we obtain

AM² + MB² = AB²  …(1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM² + MC² = AC² …(2)

Adding equations (1) and (2), we obtain

2AM² + MB² + MC² = AB² + AC²

2AM² + (BD-DM)² + (MD+DC)² = AB² + AC²

2AM² + BD² + DM² – 2BD.DM + MD² + DC² + 2MD.DC = AB² + AC²

2AM² + 2MD² + BD² + DC² + 2MD (- BD + DC) = AB² + AC²

2(AM² + MD²)  + (BC/2)² + (BC/2)² + 2MD(-BC/2 + BC/2) = AB² + AC²

2AD² + BC²/2 = AB² + AC²

6. Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

Answer:

Let ABCD be a parallelogram.

Let us draw perpendicular DE on extended side AB, and AF on side DC.

Applying Pythagoras theorem in ΔDEA, we obtain

DE² + EA² = DA²…..(i)

Applying Pythagoras theorem in ΔDEB, we obtain

DE² + EB² = DB²

DE² + (EA + AB)² = DB²

(DE² + EA²) + AB² + 2EA x AB = DB²

DA² + AB² + 2EA x AB = DB²  ….(ii)

Applying Pythagoras theorem in ΔADF, we obtain

AD² = AF² + FD²

Applying Pythagoras theorem in ΔAFC, we obtain

AC² = AF² + FC²

AC² = AF² + (DC – FD)²

AC² =AF² + DC² + FD² – 2DC x FD

AC² =(AF² + FD²) + DC² – 2DC x FD

AC² =AD² + DC² – 2DC x FD….(iii)

Since ABCD is a parallelogram,

AB = CD … (iv)

AB = CD … (v)

In ΔDEA and ΔADF,

∠DEA = ∠AFD (Both 90°)

∠EAD = ∠ADF (EA || DF)

AD = AD (Common)

∴ ΔEADΔFDA (AAS congruence criterion)

⇒ EA = DF … (vi)

Adding equations (i) & (iii), we obtain

DA² + AB² + 2EA x AB + AD² + DC² – 2DC x FD = DB² + AC²

DA² + AB² + AD² + DC² + 2EA x AB – 2DC x FD = DB² + AC²

BC² + AB² + AD² + DC² + 2EA x AB – 2AB x EA = DB² + AC²

[Using equations (iv) & (vi)

AB² + BC² + CD² + DA² = AC² + BD²

7. In figure, two chords AB and CD intersect each other at the point P. Prove that:

(i) ΔAPC ~ ΔDPB

(ii) AP.PB = CP.DP

Answer:

(i) In ΔAPC and ΔDPB,

∠APC = ∠DPB (Vertically opposite angles)

∠CAP = ∠BDP (Angles in the same segment for chord CB)

ΔAPC ~ ΔDPB (By AA similarity criterion)

(ii) We have already proved that

ΔAPC ∼ ΔDPB

We know that the corresponding sides of similar triangles are proportional.

∴AP/DP = PC/PB = CA/BD

⇒AP/DP = PC/PB

∴ AP. PB = PC. DP

8. In figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that:

(i) ΔPAC ~ ΔPDB

(ii) PA.PB = PC.PD

Answer:

(i) In ΔPAC and ΔPDB,

∠P = ∠P (Common)

∠PAC = ∠PDB (Exterior angle of a cyclic quadrilateral is ∠PCA = ∠PBD equal to the opposite interior angle)

∴ ΔPAC ∼ ΔPDB

(ii)We know that the corresponding sides of similar triangles are proportional.

∴ PA/PD = AC/DB = PC/PB

⇒PA/PD = PC/PB

∴ PA.PB = PC.PD

9. In figure, D is appointing on side BC of ΔABC such that BD/CD = AB/AC Prove that AD is the bisector of ∠BAC.

Answer:

Let us extend BA to P such that AP = AC. Join PC.

It is given that,

(BD)/(CD) = (AB)/(AC)

⇒ BD/CD = AP/AC

By using the converse of basic proportionality theorem, we obtain

AD || PC

⇒ ∠BAD = ∠APC (Corresponding angles) … (1)

And, ∠DAC = ∠ACP (Alternate interior angles) … (2)

By construction, we have

AP = AC

⇒ ∠APC = ∠ACP … (3)

On comparing equations (1), (2), and (3), we obtain

∠BAD = ∠APC

⇒ AD is the bisector of the angle BAC.

10. Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taur, how much string does she have out (see figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

Answer:

Let AB be the height of the tip of the fishing rod from the water surface. Let BC be the horizontal distance of the fly from the tip of the fishing rod.

Then, AC is the length of the string.

AC can be found by applying Pythagoras theorem in ΔABC.

AC² = AB² + BC²

AB² = (1.8 m)² + (2.4 m)²

AB² = (3.24 + 5.76)m²

AB² = 9 m²

⇒AB = √9 m = 3 m.

Thus, the length of the string out is 3 m.

She pulls the string at the rate of 5 cm per second.

Therefore, string pulled in 12 seconds = 12 × 5 = 60 cm = 0.6 m

Let the fly be at point D after 12 seconds.

Length of string out after 12 seconds is AD.

AD = AC − String pulled by Nazima in 12 seconds

= (3.00 − 0.6) m

= 2.4 m

In ΔADB,

AB² + BD² = AD²

(1.8 m)² + BD² = (2.4 m)²

BD² = (5.76 – 3.24) m² = 2.52 m²

BD = 1.587 m

Horizontal distance of fly = BD + 1.2 m

= (1.587 + 1.2) m

= 2.787 m

= 2.79 m.