NCERT Solutions for class 10th Triangles - Chapter 6

Exercise 6.2

 

1. In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

 

Answer

 

(i) In △ ABC, DE∥BC (Given)

∴ AD/DB = AE/EC [By using Basic proportionality theorem]

⇒ 1.5/3 = 1/EC

⇒ Σ EC = 3/1.5

EC = 3×10/15 = 2 cm
Hence, EC = 2 cm.

(ii) In △ ABC, DE∥BC (Given)
∴ AD/DB = AE/EC [By using Basic proportionality theorem]
⇒ AD/7.2 = 1.8/5.4
⇒ AD = 1.8×7.2/5.4 = 18/10 × 72/10 × 10/54 = 24/10
⇒ AD = 2.4
Hence, AD = 2.4 cm.

2. E and F are points on the sides PQ and PR respectively of a ΔPQR. 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 = 0.63 cm
Answer
In ΔPQR, E and F are two points on side PQ and PR respectively.

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

PF = 3.6 cm, FR = 2,4 cm (Given)

∴ PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3 [By using Basic proportionality theorem]

And, PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5

So, PE/EQ ≠ PF/FR

Hence, EF is not parallel to QR.

 

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

∴ PE/QE = 4/4.5 = 40/45 = 8/9 [By using Basic proportionality theorem]

And, PF/RF = 8/9

So, PE/QE = PF/RF

Hence, EF is parallel to QR.

 

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

Here, EQ = PQ - PE = 1.28 - 0.18 = 1.10 cm

And, FR = PR - PF = 2.56 - 0.36 = 2.20 cm

So, PE/EQ = 0.18/1.10 = 18/110 = 9/55 … (i)

And, PE/FR = 0.36/2.20 = 36/220 = 9/55 … (ii)

∴ PE/EQ = PF/FR.

Hence, EF is parallel to QR.

 

3. In the fig 6.18, if LM || CB and LN || CD, prove thatAM/MB = AN/AD

Answer

 

In the given figure, LM || CB

By using basic proportionality theorem, we get,

AM/MB = AL/AC … (i)
Similarly, LN || CD
∴ AN/AD = AL/AC … (ii)

From (i) and (ii), we get
AM/MB = AN/AD

4. In the fig 6.19, DE||AC and DF||AE. Prove that

BF/FE = BE/EC

 

Answer

 

In ΔABC, DE || AC (Given)

∴ BD/DA = BE/EC …(i) [By using Basic Proportionality Theorem]

 

In ΔABC, DF || AE (Given)

∴ BD/DA = BF/FE …(ii) [By using Basic Proportionality Theorem]

From equation (i) and (ii), we get
BE/EC = BF/FE

5. In the fig 6.20, DE||OQ and DF||OR, show that EF||QR.

 

Answer

 

In ΔPQO, DE || OQ (Given)

∴ PD/DO = PE/EQ …(i) [By using Basic Proportionality Theorem]

 

In ΔPQO, DE || OQ (Given)

∴ PD/DO = PF/FR …(ii) [By using Basic Proportionality Theorem]

 

From equation (i) and (ii), we get

PE/EQ = PF/FR

 

In ΔPQR, EF || QR. [By converse of Basic Proportionality Theorem]

 

6. In the 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.

 

Answer

 

In ΔOPQ, AB || PQ (Given)

∴ OA/AP = OB/BQ …(i) [By using Basic Proportionality Theorem]

 

In ΔOPR, AC || PR (Given)

∴ OA/AP = OC/CR …(ii) [By using Basic Proportionality Theorem]

 

From equation (i) and (ii), we get

OB/BQ = OC/CR

 

In ΔOQR, BC || QR. [By converse of Basic Proportionality Theorem].

 

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

Given: ΔABC in which D is the mid point of AB such that AD=DB.

A line parallel to BC intersects AC at E as shown in above figure such that DE || BC.

 

To Prove: E is the mid point of AC.

 

Proof: D is the mid-point of AB.

∴ AD=DB

⇒ AD/BD = 1 … (i)

In ΔABC, DE || BC,

Therefore, AD/DB = AE/EC [By using Basic Proportionality Theorem]

⇒1 = AE/EC [From equation (i)]
∴ AE =EC
Hence, E is the mid point of AC.

8. Using Converse of basic proportionality theorem, 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: ΔABC in which D and E are the mid points of AB and AC respectively such that AD=BD and AE=EC.

To Prove: DE || BC

Proof: D is the mid point of AB (Given)

∴ AD=DB

⇒ AD/BD = 1 … (i)

 

Also, E is the mid-point of AC (Given)
∴ AE=EC

⇒AE/EC = 1 [From equation (i)]

From equation (i) and (ii), we get
AD/BD = AE/EC
Hence, DE || BC [By converse of Basic Proportionality Theorem]

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: ABCD is a trapezium in which AB || DC in which diagonals AC and BD intersect each other at O.

To Prove: AO/BO = CO/DO

Construction: Through O, draw EO || DC || AB

Proof: In ΔADC, we have
OE || DC (By Construction)

∴ AE/ED = AO/CO …(i) [By using Basic Proportionality Theorem]

In ΔABD, we have
OE || AB (By Construction)

∴ DE/EA = DO/BO …(ii) [By using Basic Proportionality Theorem]

From equation (i) and (ii), we get
AO/CO = BO/DO
⇒ AO/BO = CO/DO

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

Answer
Given: Quadrilateral ABCD in which diagonals AC and BD intersects each other at O such that AO/BO = CO/DO.

 

To Prove: ABCD is a trapezium

 

Construction: Through O, draw line EO, where EO || AB, which meets AD at E.

 

Proof: In ΔDAB, we have

EO || AB

∴ DE/EA = DO/OB …(i) [By using Basic Proportionality Theorem]

Also, AO/BO = CO/DO (Given)

⇒ AO/CO = BO/DO

⇒ CO/AO = BO/DO

⇒ DO/OB = CO/AO …(ii)

 

From equation (i) and (ii), we get

DE/EA = CO/AO

Therefore, By using converse of Basic Proportionality Theorem, EO || DC also EO || AB

⇒ AB || DC.

Hence, quadrilateral ABCD is a trapezium with AB || CD.

 

Exercise 6.3

 

1. 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:

 

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)

Page No: 139

2. In the fig 6.35, ΔODC ∝ ¼ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.

Answer

DOB is a straight line.
Therefore, ∠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°
∴ ∠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/ODAnswer

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]

∴ DO/BO = OC/OA [ Corresponding sides are proportional]

⇒ OA/OC = OB/OD

 

Page No: 140

4. In the fig.6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.

 

Answer

In ΔPQR, ∠PQR = ∠PRQ
∴ PQ = PR …(i)
Given,QR/QS = QT/PR
Using (i), we get
QR/QS = QT/QP …(ii)
In ΔPQS and ΔTQR,
QR/QS = QT/QP [using (ii)]
∠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 fig 6.37, 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 fig 6.38, 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,
∠A = ∠C (Opposite angles of a parallelogram)
∠AEB = ∠CBF (Alternate interior angles as AE || BC)
∴ ΔABE ~ ΔCFB (By AA similarity criterion)

 

9. In the fig 6.39, 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 ΔABC and ΔAMP, we have

∠A = ∠A (common angle)

∠ABC = ∠AMP = 90° (each 90°)

∴ ΔABC ~ ΔAMP (By AA similarity criterion)

 

(ii) As, ΔABC ~ ΔAMP (By AA similarity criterion)

If two triangles are similar then the corresponding sides are equal,

Hence, CA/PA = BC/MP

 

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
(i) It is given that ΔABC ~ ΔFEG.
∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE
∠ACB = ∠FGE
∴ ∠ACD = ∠FGH (Angle bisector)
And, ∠DCB = ∠HGE (Angle bisector)
In ΔACD and ΔFGH,
∠A = ∠F (Proved above)
∠ACD = ∠FGH (Proved above)
∴ ΔACD ~ ΔFGH (By AA similarity criterion)

⇒ CD/GH = AC/FG

 

(ii) In ΔDCB and ΔHGE,
∠DCB = ∠HGE (Proved above)
∠B = ∠E (Proved above)
∴ ΔDCB ~ ΔHGE (By AA similarity criterion)

(iii) In ΔDCA and ΔHGF,
∠ACD = ∠FGH (Proved above)
∠A = ∠F (Proved above)
∴ ΔDCA ~ ΔHGF (By AA similarity criterion)
Page No: 141

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

It is given that ABC is an isosceles triangle.
∴ AB = AC
⇒ ∠ABD = ∠ECF
In ΔABD and ΔECF,
∠ADB = ∠EFC (Each 90°)
∠BAD = ∠CEF (Proved above)
∴ ΔABD ~ ΔECF (By using AA similarity criterion)

 

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 Fig 6.41). Show that ΔABC ~ ΔPQR.

Answer

 

Given: ΔABC and ΔPQR, AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR

i.e., AB/PQ = BC/QR = AD/PM

 

To Prove: ΔABC ~ ΔPQR

 

Proof: AB/PQ = BC/QR = AD/PM

⇒ AB/PQ = BC/QR = AD/PM (D is the mid-point of BC. M is the mid point of QR)

⇒ ΔABD ~ ΔPQM [SSS similarity criterion]

∴ ∠ABD = ∠PQM [Corresponding angles of two similar triangles are equal]

⇒ ∠ABC = ∠PQR

In ΔABC and ΔPQR

AB/PQ = BC/QR …(i)

∠ABC = ∠PQR …(ii)

From equation (i) and (ii), we get

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.CDAnswer

In ΔADC and ΔBAC,
∠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.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: Two triangles ΔABC and ΔPQR in which AD and PM are medians such that AB/PQ = AC/PR = AD/PM

 

To Prove: ΔABC ~ ΔPQR

 

Construction: Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN.

 

Proof: In ΔABD and ΔCDE, we have

AD = DE [By Construction]

BD = DC [∴ AP is the median]

and, ∠ADB = ∠CDE [Vertically opp. angles]

∴ ΔABD ≅ ΔCDE [By SAS criterion of congruence]

⇒ AB = CE [CPCT] …(i)

Also, in ΔPQM and ΔMNR, we have

PM = MN [By Construction]

QM = MR [∴ PM is the median]

and, ∠PMQ = ∠NMR [Vertically opposite angles]

∴ ΔPQM = ΔMNR [By SAS criterion of congruence]

⇒ PQ = RN [CPCT] …(ii)

Now, AB/PQ = AC/PR = AD/PM

⇒ CE/RN = AC/PR = AD/PM …[From (i) and (ii)]

⇒ CE/RN = AC/PR = 2AD/2PM

⇒ CE/RN = AC/PR = AE/PN [∴ 2AD = AE and 2PM = PN]

∴ ΔACE ~ ΔPRN [By SSS similarity criterion]

Therefore, ∠2 = ∠4

Similarly, ∠1 = ∠3

∴ ∠1 + ∠2 = ∠3 + ∠4

⇒ ∠A = ∠P …(iii)

Now, In ΔABC and ΔPQR, we have

AB/PQ = AC/PR (Given)

∠A = ∠P [From (iii)]

∴ ΔABC ~ ΔPQR [By SAS similarity criterion]

 

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

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 = 6×28/4

⇒ h = 6 × 7

⇒ h = 42 m

Hence, the height of the tower is 42 m.

 

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

 

Answer

 

It is given that ΔABC ~ ΔPQR
We know that the corresponding sides of similar triangles are in proportion.∴ AB/PQ = AC/PR = BC/QR …(i)
Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R …(ii)
Since AD and PM are medians, they will divide their opposite sides.∴ BD = BC/2 and QM = QR/2 …(iii)
From equations (i) and (iii), we get
AB/PQ = BD/QM …(iv)
In ΔABD and ΔPQM,
∠B = ∠Q [Using equation (ii)]
AB/PQ = BD/QM [Using equation (iv)]
∴ ΔABD ~ ΔPQM (By SAS similarity criterion)⇒ AB/PQ = BD/QM = AD/PM.

 

Exercise 6.3

1. Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm²and 121 cm². If EF = 15.4 cm, find BC.AnswerIt 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 the fig 6.53, 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
Given: ABC and DBC are triangles on the same base BC. Ad intersects BC at O.

 

To Prove: area (ΔABC)/area (ΔDBC) = AO/DO.

 

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

 

Proof: We know that area of a triangle = 1/2 × Base × Height

In ΔAPO and ΔDMO,
∠APO = ∠DMO (Each equals to 90°)
∠AOP = ∠DOM (Vertically opposite angles)
∴ ΔAPO ~ ΔDMO (By AA similarity criterion)∴ AP/DM = AO/DO
⇒ area (ΔABC)/area (ΔDBC) = AO/DO.

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

Answer

Given: ΔABC and ΔPQR are similar and equal in area.

To Prove: ΔABC ≅ ΔPQR

Proof: Since, ΔABC ~ ΔPQR
∴ Area of (ΔABC)/Area of (ΔPQR) = BC²/QR²
⇒ BC²/QR² =1 [Since, Area(ΔABC) = (ΔPQR)
⇒ BC²/QR²
⇒ BC = QR
Similarly, we can prove that
AB = PQ and AC = PR
Thus, ΔABC ≅ ΔPQR [BY SSS criterion of congruence]

5. D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC.

 

Answer

Given: D, E and F are the mid-points of the sides AB, BC and CA respectively of the ΔABC.

To Find: area(ΔDEF) and area(ΔABC)

Solution: In ΔABC, we have
F is the mid point of AB (Given)
E is the mid-point of AC (Given)
So, by the mid-point theorem, we have
FE || BC and FE = 1/2BC
⇒ FE || BC and FE || BD [BD = 1/2BC]
∴ BDEF is parallelogram [Opposite sides of parallelogram are equal and parallel]
Similarly in ΔFBD and ΔDEF, we have
FB = DE (Opposite sides of parallelogram BDEF)
FD = FD (Common)
BD = FE (Opposite sides of parallelogram BDEF)
∴ ΔFBD ≅ ΔDEF
Similarly, we can prove that
ΔAFE ≅ ΔDEF
ΔEDC ≅ ΔDEF
If triangles are congruent,then they are equal in area.
So, area(ΔFBD) = area(ΔDEF) …(i)
area(ΔAFE) = area(ΔDEF) …(ii)
and, area(ΔEDC) = area(ΔDEF) …(iii)
Now, area(ΔABC) = area(ΔFBD) + area(ΔDEF) + area(ΔAFE) + area(ΔEDC) …(iv)
area(ΔABC) = area(ΔDEF) + area(ΔDEF) + area(ΔDEF) + area(ΔDEF)
⇒ area(ΔDEF) = 1/4area(ΔABC) [From (i), (ii) and (iii)]
⇒ area(ΔDEF)/area(ΔABC) = 1/4
Hence, area(ΔDEF):area(ΔABC) = 1:4

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

Given: AM and DN are the medians of triangles ABC and DEF respectively and ΔABC ~ ΔDEF.

 

To Prove: area(ΔABC)/area(ΔDEF) = AM²/DN²
Proof: ΔABC ~ ΔDEF (Given)
∴ area(ΔABC)/area(ΔDEF) = (AB²/DE²) …(i)
and, AB/DE = BC/EF = CA/FD …(ii)

In ΔABM and ΔDEN, we have
∠B = ∠E [Since ΔABC ~ ΔDEF]
AB/DE = BM/EN [Prove in (i)]
∴ ΔABC ~ ΔDEF [By SAS similarity criterion]
⇒ AB/DE = AM/DN …(iii)
∴ ΔABM ~ ΔDEN
As the areas of two similar triangles are proportional to the squares of the corresponding sides.
∴ area(ΔABC)/area(ΔDEF) = AB²/DE² = AM²/DN²

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 its diagonals.

Answer
Given: ABCD is a square whose one diagonal is AC. ΔAPC and ΔBQC are two equilateral triangles described on the diagonals AC and side BC of the square ABCD.

To Prove: area(ΔBQC) = 1/2area(ΔAPC)

Proof: ΔAPC and ΔBQC are both equilateral triangles (Given)
∴ ΔAPC ~ ΔBQC [AAA similarity criterion]
∴ area(ΔAPC)/area(ΔBQC) = (AC²/BC²) = AC²/BC²

⇒ area(ΔAPC) = 2 × area(ΔBQC)
⇒ area(ΔBQC) = 1/2area(ΔAPC)
Tick the correct answer and justify:

8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is
   (A) 2 : 1
   (B) 1 : 2
   (C) 4 : 1
   (D) 1 : 4Answer

ΔABC and ΔBDE are two equilateral triangle. D is the mid point of BC.

∴ BD = DC = 1/2BC

Let each side of triangle is 2a.

As, ΔABC ~ ΔBDE

∴ area(ΔABC)/area(ΔBDE) = AB²/BD² = (2a)²/(a)² = 4a²/a² = 4/1 = 4:1

Hence, the correct option is (C).

 

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 : 81Answer

Let ABC and DEF are two similarity triangles ΔABC ~ ΔDEF (Given)

and, AB/DE = AC/DF = BC/EF = 4/9 (Given)

∴ area(ΔABC)/area(ΔDEF) = AB²/DE² [the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides]

∴ area(ΔABC)/area(ΔDEF) = (4/9)² = 16/81 = 16:81

Hence, the correct option is (D).