Bihar Board Class 10th Maths Chapter 6 Triangles Solution

 

BSEB Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.1

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

1. All circles are ________. (congruent, similar)
2. All squares are ________. (similar, congruent)
3. All triangles are similar, (isosceles, equilateral)
4. Two polygons of the same number of sides are similar, if (a) their corresponding angles are and (b) their corresponding sides are ________. (equal, proportional)

Solution:

1. similar
2. similar
3. equilateral
4. equal, proportional.

Question 2.
Give two different examples of pair of
(i) similar figures
(ii) non-similar figures.
Solution:
(i) Two different examples of pair of similar figures are :

• any two circles
• any two squares.

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

• a scalene and an equilateral triangle.
• an equilateral triangle and a right angled triangle.

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

Solution:
On looking at the given figures of the quadrilaterals, we can say that they are not similar. Here, sides are proportional, but corresponding angles are not equal.

BSEB Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.2

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

Solution:
(i) In Fig. (i),
since DE || BC, therefore
ADDB = AEEC or 1.53 = 1EC
or EC = 31.5 = 3×1015 Cm = 2 cm

(ii) In Fig. (ii),
since DE || BC, therefore
ADDB =AEEC or AD7.2 = 1.85.4
or AD = 1854 x 7210 cm = 2.4 cm.

Question 2.
E and F are points on the sides PQ and PR respectively of a A 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) We have :
PE = 3.9 cm, EQ = 3 cm,
PF = 3.6 cm and FR = 2.4 cm

i. e., EF does not divide the sides PQ and PR of A PQR in the same ratio. Therefore, EF is not parallel to QR.

(ii) We have : PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
Now, PEEQ = 44.5 = 4045 = 89
and PFFR = 89
So, PEEQ = PFFR
Thus, EF divides sides PQ and PR of ∆ PQR in the same ratio. Therefore, by the converse of Basic Proportionality Theorem, we have EF || QR.

(iii) We have :PQ= 1.28 cm, PR = 2.56 cm,
PE = 0.18 cm and PF 0.36 cm
∴ EQ = PQ – PE = (1.28 – 0.18) cm = 1.10 cm
and FR = PR – PF = (2.56 – 0.36) = 2.20 cm
Now, PEEQ = 0.181.10 = 18110 = 955
and PFFR = 0.362.20 = 36220 = 955
So, PEEQ = PFFR
Thus, EF divides sides PQ and PR of ∆ PQR in the same ratio. Therefore, by the converse of Basic Proportionality Theorem, we have EF || QR.

Question 3.
In the figure, if LM || CB and LN || CD, prove that AMAB = ANAD

Solution:
In ∆ ABC, we have :
LM || CB
∴By a result based on Basic Proportionality Theorem (a corollary), we have :
AMAB = ALAC … (1)
In ∆ ACD, we have :
LN || CD [Given]
∴By a result based on Basic Proportionality Theorem, we have :
ALAC = ANAD
From (1) and (2), we obtain that
AMAB = ANAD

Question 4.
In the figure, DE || AC and DF || AE. Prove that BFFE = BEEC

Solution:
In ∆ BGA, we have :
DE || ‘AC
∴By Basic Proportionality Theorem, we have :
BEEC = BDDA …. (1)
In ∆ BEA, we have :
DF || AE [Given]
∴By Basic Proportionality Theorem, we have
BFFE = BDDA …. (2)
From (1) and (2), we obtain that
BFFE = BEEC.

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

Solution:
In ∆ PQO, we have :
DE || OQ [Given]
∴By Basic Proportionality Theorem, we have :
PEEQ = PDDO … (1)
In ∆ POR, We have :
DF || OR [Given]
∴By Basic Proportionality Theorem, we have
PDDO = PFFR … (2)
From (1) and (2), we obtain that
PEEQ = PFFR
So, EF || QR [By the converse of BPT]

Question 6.
In the figure, 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 : O is any point within ∆ PQR, AB || PQ and AC || PR
To prove : BC || QR.
Construction : Join BC.
Proof:
In ∆ OPQ, we have :
AB || PQ [Given]
∴By Basic Proportionality Theorem, we have :
OAAP = OBBQ … (1)
In ∆ OPR, We have :
AC || PR [Given]
∴ By Basic Proportionality Theorem, we have
OAAP = OCCR … (2)
From (1) and (2), we obtain that
OBBQ = OCCR
Thus, in ∆ OQR, B and C are points dividing the sides OQ and OR in the same ratio. Therefore, by the converse of Basic Proportionality Theorem, we have :
BC || QR.

Question 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).

Solution:
Given : ∆ ABC, in which D is the mid-point of side AB and the line DE is drawn parallel to BC, meeting AC in E.
To prove : AE = EC
Proof:
Since DE || BC, therefore by Basic Proportionality Theorem, we have :
ADDB = AEEC … (1)
But AD = DB [∵D is the mid-point of AB]
i.e., ADDB = 1
∴ From (1), AEEC = 1 or AE = EC
Hence, E is the mid-point of the third side AC.

Question 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).

Solution:
Given : ∆ ABC, in which D and E are the mid¬points of sides AB and AC respectively.
To prove : DE || BC.
Proof :
Since D and E are the mid-points of AB and AC respectively, therefore
AD = DB and AE = EC
Now, AD = DB
∴ ADDB = 1
and AE = EC
∴ AEEC = 1
Thus, in ∆ ABC, D and E are points dividing the sides AB and AC in the same ratio. Therefore, by the converse of Basic Proportionality Theorem (Theorem 6.2), we have :
DE || BC.

Question 9.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AOBO = CODO.
Solution:
Given : A trapezium ABCD, in which AB || DC and its diagonals AC and BD intersect each other at O.
To prove : AOBO = CODO.
Construction : Through O, draw OE || AB, i.e., OE || DC.
Proof:
In ∆ ADC, we have :
OE || DC [Construction]
∴ By Basic Proportionality Theorem, we have :

AEED = AOCO … (1)
Again, in ∆ ABD, we have :
OE || AB [Construction]
∴By Basic Proportionality Theorem, we have :0
EDAE = DOBO or AEED = BODO … (2)
From (1) and (2), we obtain that
AOCO = BODO or AOBO = CODO.

Question 10.
The diagonals of a quadrilateral ABCD intersect each other at the point O such that AOBO = CODO that ABCD is a trapezium.
Solution:
Given : A quadrilateral ABCD in which its diagonals AC and BD intersect each other at the point O such that AOBO = CODO, i.e., AOCO = BODO
To prove : Quadrilateral ABCD is a trapezium.
Construction : Through O, draw OE || AB meeting AD in E.
Proof:
In ∆ ADB, we have :
OE || AB [Construction]
∴ By Basic Proportionality Theorem, we have :
DEEA = ODBO
or EADE = BODO
i.e., EADE = BODO = AOCO [ ∵AOCO = BODO(given)]
or EADE = AOCO
Thus, in ∆ ADC, points E and O are dividing the sides AD and AC in the same ratio. Therefore, by the converse of Basic Proportionality Theorem, we have :
EO || DC
But, EO || AB [construction]
Hence, AB || DC
∴ Quadrilateral ABCD is a trapezium

BSEB Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.3

Question 1.
State which pairs of triangle in the figures, 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) In ∆s ABC and PQR, we observe that
∠A = ∠P = 60°, ∠B = ∠Q = 80° and ∠C = ∠R = 40°. By AAA criterion of similarity,
∆ ABC ~ ∆ PQR.

(ii) In ∆s ABC and PQR, we observe that
\(\frac { AB }{ QR }\) = \(\frac { BC }{ RP }\) = \(\frac { CA }{ PQ }\) = \(\frac { 1 }{ 2 }\)
∴By SSS criterion of similarity, ∆ ABC ~ ∆ QRP.

(iii) In ∆s LMP and DEF, we observe that the ratio of the sides of these triangles are not equal.
[ \(\frac { MP }{ DE }\) = \(\frac { LP }{ DF }\) but it is not equal to \(\frac { LM }{ EF }\) ]
So, the two triangles are not similar.

(iv) In ∆s MNL and QPR, we observe that
∠M = ∠Q = 70°
But, \(\frac { MN }{ PQ }\) ≠ \(\frac { ML }{ QR }\)
These two triangles are not similar as they do not satisfy SAS criterion of similarity.

(v) In ∆s ABC and FDE, we have
∠A = ∠F = 80°
But, \(\frac { AB }{ DE }\) ≠ \(\frac { ML }{ QR }\) [∵ AC is not given]
DE DF
So, by SAS criterion’ of similarity, these two triangles are not similar.

(vi) In ∆s DEF and PQR, we have :
∠D = ∠P = 70° [∵∠P = 180° – 80° – 30° = 70°]
∠E = ∠Q = 80°
So, by AAA criterion of similarity,
∆ DEF ~ ∆ PQR.

Question 2.
In the figure,

∆ ODC ~ ∆ OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
Solution:
Since BD is a line and OC is a ray on it,
∴∠DOC + ∠BOC = 180°
So, ∠DOC + 125° = 180°
or ∠DOC = 180°
In ∆ CDO, we have :
∠CDO + ∠DOC + ∠DCO = 180°
or 70° + 55° + ∠DCO = 180°
or ∠DCO = 180° – 125° = 55°
It is given that ∆ ODC ~ A OBA
∴ ∠ODC = ∠OBA, ∠OCD = ∠OAB
or ∠OBA = 70° and ∠OAB = 55°
Hence, ∠DOC = 55° ∠DCO = 55° and ∠OAB = 55°

Question 3.
Diagonals AC and BD of a trapezium ABCD with | DC intersect each other at the point O. Using asimilarity criterion for two triangles, show that \(\frac { OA }{ OC }\) = \(\frac { OB }{ OD }\)
Solution:
Given : ABCD is a trapezium in which AB || DC.
To prove : \(\frac { OA }{ OC }\) = \(\frac { OB }{ OD }\)
Proof:
In As OAB and OCD, we have :
∠AOB = ∠COD [Vert. opp. ∠s]
∠OAB = ∠OCD [Alternate ∠s]
and ∠OBA = ∠ODC [Alternate ∠s]

∴ By AAA criterion of similarity,
∆ OAB ~ ∆ OCD
Hence \(\frac { OA }{ OC }\) = \(\frac { OB }{ OD }\) [ ∵ In case of two similar triangles the ratio of their corresponding sides are equal]

Question 4.
In the figure, \(\frac { QR }{ QS }\) = \(\frac { QT }{ PR }\) and ∠1 = ∠2. Show that ∆ PQS ~ ∆ TQR

Solution:

Question 5.
S and T are points on sides PR and QR of ∆ PQR such that ∠P = ∠RTS. Show that ∆ RPQ ~ ∆ RTS.
Solution:
In ∆s RPQ and RTS, we have :

∠RPQ = ∠RTS [Given]
∠PRQ = ∠TRS [Common]
∴ By AA criterion of similarity,
∆ RPQ ~ ∆ RTS.

Question 6.
In the figure, if ∆ ABE ≅ ∆ ACD, show that ∆ ADE ~ ∆ ABC.

Solution:
It is given that ∆ ABE ≅ ∆ ACD
AB = AC [∵ Corresponding parts of congruent triangles are equal]
and AE = AD
So, \(\frac { AB }{ AD }\) = \(\frac { AC }{ AE }\) = or \(\frac { AB }{ AC }\) = \(\frac { AD }{ AE }\)
∴ In As ADE and ABC, we have :
\(\frac { AB }{ AC }\) = \(\frac { AD }{ AE }\) [ ∵of (1) ]
and ∠BAC = ∠DAE [common]
Thus, by SAS criterion of similarity,
∆ ADE ~ ∆ ABC.

Question 7.
In the figure, altitudes AD and CE of ∆ ABC intersect each other at the point P.
Show that:

(i) ∆AEP ~ ∆ CDP
(ii) ∆ ABD ~ ∆ CBE
(iii) ∆ AEP ~ ∆ ADB
(iv) ∆ PDC ~ ∆ BEC
Solution:
(i) In ∆s AEP and CDP, we have :
∠AEP = ∠CDP = 90° [∵ CE ⊥ AB and AD ⊥ BC]
∠APE = ∠CPD [Vert. opp. ∠s]
∴By AA criterion of similarity, we have :
∆ AEP ~ ∆ CDP.

(ii) In ∆s ∆BD and CBE, we have :
∠ABD = ∠CBE [Common angle]
∠ADB = ∠CEB = 90°
∴By AA criterion of similarity, we have :
∆ ABD – ∆ CBE.

(iii) In ∆s AEP and ADB, we have :
∠AEP = ∠ADB = 90° [∵AD ⊥ BC and CE ⊥ AB]
∠PAE = ∠DAB [Common angle]
∴By AA criterion of similarity, we have :
∆ AEP ~ ∆ ADB.

(iv) In ∆s PDC and BEC, we have :
∠PDC = ∠BEC = 90° [∵ AD ⊥ BC and CE ⊥ AB]
∠PCD = ∠ECB [Common angle]
∴ By AA criterion of similarity, we have :
∆ PDC – ∆ BEC.

Question 8.
E is a point on the side ALD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ ABE ~ ∆ CFB.

Solution:
In ∆s ABE and CFB, we have:
∠AEB = ∠CBF [Alt. ∠s]
∠A = ∠C [Opp. ∠s of a gm]
∴By AA criterion of similarity, B we have :
∆ ABE – ∆ CFB.

Question 9.
In the figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that :

(i) ∆ ABC ~ ∆ AMP
(ii) \(\frac { CA }{ PA }\) = \(\frac { BC }{ MP }\)
Solution:
(i) In ∆s ABC and AMP, we have :
∠ABC = ∠AMP = 90° [Given]
and ∠BAC = ∠MAP [Common ∠s]
∴By AA criterion of similarity, we have :
∆ ABC – ∆ AMP.

(ii) We have :
∆ ABC ~ ∆ AMP [As proved above]
So, \(\frac { CA }{ PA }\) = \(\frac { BC }{ MP }\) [In similar triangles, correspondings sides are proportional.

Question 10.
CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of A ABC and ∆ EFG respectively. If ∆ ABC ~ ∆ FEG, show that :
(i) \(\frac { CD }{ GH }\) = \(\frac { AC }{ FG }\)
(ii) ∆ DCB ~ ∆ HGE
(iii) ∆ DCA ~ ∆ HGF
Solution:

We have : ∆ ABC ~ ∆ FEG
So, ∠A = ∠F
and ∠C = ∠G
∴\(\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 2 }\) ∠G
or ∠1 = ∠3 and ∠2 = ∠4 … (2)
[∵ CD and GH are bisectors of ∠C and ∠G respectively]
∴ In As DCA and HGF, we have :
∠A = ∠F
∠2 = ∠4 [From (2)]
∴By AA criterion of similarity, we have :
∆ DCA – ∆ HGF
which proves the (iii) part.
We have : ∆ DCA ~ ∆ HGF [As proved above]
So, \(\frac { AC }{ FG }\) = \(\frac { CD }{ GH }\) or \(\frac { CD }{ GH }\) = \(\frac { AC }{ FG }\)
which proves the (i) part.
In As DCB and HGE, we have :
∠1 = ∠3 [From (2)]
∠B = ∠E [∵ ∆ ABC ~ ∆ FEG]
∴By AA criterion of similarity, we have :
∆ DCB ~ ∆ HGE
which proves the (ii) part.

Question 11.
In the 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 ∆ ABD ~ ∆ ECF.

Solution:
Here, ∆ ABC is isosceles with AB = AC.
∴ ∠B = ∠C
In ∆s ABD and ECF, we have :
∠ABD = ∠ECF [∵ ∠B = ∠C]
∠ADB = ∠EFC = 90° [∵ AD ⊥ BC and EF ⊥ AC]
∴ By AA criterion of similarity,
∆ ABD ~ ∆ ECF.

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

Solution:
Given : AD is the median of ∆ ABC and PM is the median of ∆ PQR such that
\(\frac { AB }{ PQ }\) = \(\frac { BC }{ QR }\) = \(\frac { AD }{ PM }\)
To prove : ∆ ABC ~ ∆ PQR

∴ By SSS criterion of similarity, we have :
∆ ABD ~ ∆ PQM
So, ∠B = ∠Q [Similar ∆s have corresponding ∠s equal]
Also, [Given]
\(\frac { AB }{ PQ }\) = \(\frac { BC }{ QR }\) [Given]
∴ By SAS criterion of similarity, we have :
∆ ABC ~ ∆ PQR.

Question 13.
D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB.CD.
Solution:
In ∆s ABC and DAC, we have :
∠BAC = ∠ADC and ∠C = ∠C
∴ By AA criterion of similarity, we have :

∆ ABC – ∆ DAC
So, \(\frac { AB }{ DA }\) = \(\frac { BC }{ AC }\) = \(\frac { AC }{ DC }\)
Now, from \(\frac { BC }{ AC }\) = \(\frac { AC }{ DC }\)
\(\frac { CB }{ CA }\) = \(\frac { CA }{ CD }\)
or CA² = CB x CD.

Question 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 ∆ ABC ~ ∆ PQR.
Solution:
Produce AD to a point E such that AD = DE and PM to a point N such that PM = MN (see figure). Join CE and RN.

So, from (1) and (3), we have :

Question 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.
Solution:

Let AB be the vertical pole B and AC be its shadow. Also, let DE be the vertical tower and DF be its shadow. 6 m Join BC and EF. Let DE = x metres.
We have : AB = 6 m, AC = 4 m and DF = 28 m.
In As ABC and DEF, we have :

∠A = ∠D = 90°,
and ∠C = ∠F
[∵ Each is the angular elevation of the sun]
∴ By AA criterion of similarity,
∆ ABC ~ ∆ DEF
or \(\frac { AB }{ DE }\) = \(\frac { AC }{ DF }\)
So, \(\frac { 6 }{ x }\) = \(\frac { 4 }{ 28 }\) or \(\frac { 6 }{ x }\) = \(\frac { 1 }{ 7 }\)
or x = 6 x 7 = 42.
Hence, the height of the tower is 42 metres.

Question 16.
If AD and PM are medians of triangles ABC and PQR respectively, where ∆ ABC ~ ∆ PQR, prove \(\frac { AB }{ PQ }\) = \(\frac { AD }{ PM }\)
Solution:
Given : AD and PM are the medians of As ABC and PQR respectively, where
∆ ABC ~ ∆ PQR
To Prove : \(\frac { AB }{ PQ }\) = \(\frac { AD }{ PM }\)
Proof: In As ABD and PQM, we have :

BSEB Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.4

Question 1.
Let ∆ ABC ~ ∆ DEF and their areas be, respectively, 64 cm² and 121 cm². If EF = 15.4 cm, find BC.
Solution:

Question 2.
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 ∆s AOB and COD, we have :
∠AOB = ∠COD [Vert.opp.∠s]
and ∠OAB = ∠OCD [Alternate ∠s]
∴ By AA criterion of similarity, we have:

Hence, area (∆ AOB) : area (∆ COD) = 4 : 1

Question 3.
In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that

= \(\frac { AO }{ DO }\).

Solution:
Given : Two As ABC and DBC which stand on the same base BC but on the opposite sides of BC.
To prove :

= \(\frac { AO }{ DO }\).
Construction : Draw AE ⊥ BC and DF ⊥ BC.

Proof:
In ∆s AOE and DOF, we have :
∠AEO = ∠DFO = 90°
∠AOE = ∠DOF [Vertically opp. ∠s]
∴ By AA criterion of similarity, we have :

Question 4.
If the areas of two similar triangles are equal, prove that they are congruent.
Solution:
Given : Two ∆s ABC and DEF such that ∆ ABC – ∆ DEF
and Area (∆ ABC) = Area (∆ DEF)
To prove : ∆ ABC ≅ ∆ DEF

Proof:
∆ ABC ~ ∆ DEF
So, ∠A = ∠D, ∠B = ∠E, ∠C = ∠F
and \(\frac { AB }{ DE }\) = \(\frac { BC }{ EF }\) = \(\frac { AC }{ DF }\)
To establish ∆ ABC ≅ ∆ DEF, it is sufficient to prove that

Hence ∆ ABC ≅ ∆ DEF [By SSS]

Question 5.
D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ ALBC. Find the ratio of the areas of ∆ DEF and ∆ ABC.
Solution:
Since D and E are respectively, the mid-points of the sides AB and BC of ∆ ABC, therefore
DE [| AC, or DE || FC … (1)
Since D and F are respectively the mid-points of the sides AB and AC
∆ ABC, therefore

DF || BC, or DF || EC … (2)
From (1) and (2), we can say that DECF is a parallelogram.
Similarly, ADEF is a parallelogram.
Now, in ∆s DEF and ABC, we have :
∠DEF = ∠A [Opp. Zs of ||gm ADEF]
and ∠EDF = ∠C [Opp. Zs of ||gm DECF]
∴ By AA criterion of similarity, we have :
∆ DE ≅ ∆ AC

Hence, Area (∆ DEF) : Area (∆ ABC) =1:4.

Question 6.
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 : ∆ ABC ~ ∆ PQR, AD and PM are the medians of ∆s ABC and PQR respectively.
To Prove : \(\frac{\text { Area }(\Delta \mathrm{ABC})}{\text { Area }(\Delta \mathrm{PQR})}=\frac{\mathrm{AD}^{2}}{\mathrm{PM}^{2}}\)
Proof : Since ∆ ABC ~ ∆ PQR, therefore

Question 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.
Solution:
Given : A square ABCD. Equilateral ∆s BCE and ACF have been drawn on side BC and the diagonal AC respectively.
To prove : Area (∆ BCE) = \(\frac { 1 }{ 2 }\)(Area ∆ ACF)

Proof: ∆ BCE ~ ∆ ACF [All equilateral triangles are similar]

Question 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
Solution:
Since ∆ ABC and ∆ BDE are equilateral triangles, they are equiangular and hence ∆ ABC – ∆ BDE

∴ (C) is the correct answer.

Question 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
Solution:
Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides, therefore
ratio of areas = (4)² : (9)² = 16 : 81
∴ (D) is the correct answer.

BSEB Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5

Question 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
Solution:
(i) Let a = 7 cm, b = 24 cm and c = 25 cm.
Here the largest side is c = 25 cm.
We have : a² + b² = 7² + 24² = 49 + 576 = 625 = c²
So, the triangle with the given sides is a right triangle. Its hypotenuse is 25 cm.

(ii) Let a – 3 cm, 6 = 8 cm and c = 6 cm.
Here, the largest side is 6= 8 cm.
We have : a² + c² = 3² + 6² = 9 + 36 = 45 ≠ b²
So, the triangle with the given sides is not a right triangle.

(iii) Let a = 50 cm, b – 80 cm and c = 100 cm.
Here, the largest side is c = 100 cm.
We have : a² + b² = 50² + 80² = 2500 + 6400
= 8900 ≠ c²
So, the triangle with the given sides is not a right triangle.

(iv) Let a = 13 cm, b – 12 cm and c = 5 cm.
Here, the largest side is a = 13 cm.
We have : b² + c² = 12² + 5² = 144 + 25 = 169 = a²
So, the triangle with the given sides is a right triangle. Its hypotenuse is 13 cm.

Question 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.
Solution:
Given : PQR is a triangle right angled at P and PM ⊥ QR.
To Prove : PM² = QM.MR

Proof: Since PM ⊥ QR, therefore
∆ PQM ~ ∆ RPM
or \(\frac { PM }{ QM }\) = \(\frac { MR }{ PM }\)
So, PM² = QM.MR.

Question 3.
In the figure, ABD is triangle right angled at and AC ⊥ BD. Show that
(i) Ab² = BC.BD
(ii) Ac² = BC.DC
(iii) AD² = BD.CD

Solution:
Given : ABD is a triangle right angled at A and AC ⊥ BD.
To Prove :
(i) Ab² = BC.BD
(ii) Ac² = BC.DC
(iii) AD² = BD.CD
Proof:
(i) Since AC ⊥ BD, therefore
∆ ABC – ∆ DAC and each triangle is similar to the whole ∆ ABD.
∵ ∆ ABC – ∆ DBA
∴ \(\frac { AB }{ DB }\) = \(\frac { BC }{ AB }\) or AB² = BC.BD AB

(ii) Since ∆ ABC ~ ∆ DAC, therefore
\(\frac { AC }{ BC }\) = \(\frac { DC }{ AC }\) or Ac² = BC.DC AC

(iii) Since ∆ ACD ~ ∆ BAD, therefore
∴ \(\frac { AD }{ CD }\) = \(\frac { BD }{ AD }\) or AD² = BD. CD.

Question 4.
ABC is an isosceles triangle right angled at C. Prove that Ab² = 2Ac².
Solution:
Since ABC is an isosceles A right triangle, right angled at C, therefore

Ab² = Ac² + Bc²
or Ab² = Ac² + Ac² [∵ BC = AC, given]
So, Ab² = 2 Ac²

Question 5.
ABC is an isosceles triangle with AC = BC. If Ab² = 2Ac², prove that ABC is a right triangle.
Solution:
Since ABC is an isosceles triangle with AC = BC and Ab² = 2Ac², therefore
Ab² = Ac² + Ac²
or Ab² = Ac² + Bc² [∵ AC = BC, given]
∴ ∆ ABC is right angled at C.

Question 6.
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution:
Let ABC be an equilateral triangle of side 2a units. Draw AD ⊥ BC. Then, D is the mid-point of BC

So, BD = \(\frac { 1 }{ 2 }\) BC
= \(\frac { 1 }{ 2 }\) x 2a = a
Since ABD is a right triangle, right angled at D, therefore
Ab² = AD² + BD²
or (2a)² = AD² + (a)²
or AD² = 4a² – a² = 3a² or AD = \(\sqrt{3}\)a
∴ Each of its altitudes is V3a.

Question 7.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum o diagonals.
Solution:
Let the diagonals AC and BD of a rhombus ABCD intersect each other at O. Since the diagonals of a rhombus bisect each other at right angles.

∠AOB = ∠BOC – ∠COD
= ∠DOA = 90°
and AO = CO, BO = OD
Since AOB is a right triangle right
AB² = Oa² + Ob²
or AB² = (\(\frac { 1 }{ 2 }\)AC)² +(\(\frac { 1 }{ 2 }\)BD)² [∵ OA = OC and OB = OD]
or 4AB² = Ac² + BD² … (1)
Similarly, we have :
4BC² = Ac² + BD² …(2)
4CD² = Ac² + BD² …(3)
and 4AD² = Ac² + BD²
Adding all these results, we get
4(AB² + BC² + CD² + AD)² = 4(AC² + BD²)
or AB² + BC² + CD² + DA² = AC² + BD².

Question 8.
In the figure, 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 B = AF² + BD² + CE²
(ii) AF² + BD² + CE² = AE² + CD² + BF².

Solution:
Join AO, BO and CO.

(i) In right As OFA, ODB and OEC, we have :
OA² = AF² + OF²
Ob² = BD² + OD²
and OC² = CE² + OE²
Adding all these, we get
OA² + OB² + OC² = AF² + BD² + CE² + OF² + OD² + OE² or OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE².

(ii) In-right As ODB and ODC, we have :
OB² = OD² + BD²
and OC² = OD² + CD²
or OB² – OC² – BD² – CD² … (1)
Similarly, we have :
OC² – OA² = CE² – AE² … (2)
and OA² – OB² = AF² – BF² … (3)
Adding equations (1), (2) and (3), we get
(OB² – OC²) + (OC² – OA²) + (OA² – OB²)
= (BD² – CD²) + (CE² – AE²) + (AF² – BF²) or (BD² + CE² + AF²) – (AE² + CD² + BF²) = 0
or AF² + BD² + CE²
= AE² + BF2 + CD².

Aliter for (ii) :
Taking triangles OFB, ODC and OAE, we have :
OA² + OB² + OC²- OD²- OE² – OF² = AE² + BF² + CD²
So, using result (i) above, we have :
AF² + BD² + CE² = AE² + BF² + CD²

Question 9.
A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.
Solution:
Let AB be the ladder, B be the B window and CB be the wall. Then, ABC is a right triangle, right angled at C.

∴ AB² = AC² + Bc²
So, 10² = AC² + 8²
or AC² = 100 – 64
or AC²= 36
or AC = 6 .A C
Hence, the foot of the ladder is at a distance of 6 m from the base of the wall.

Question 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 end. How far from the base of the pole should the stake be driven so that the wire will be taut ?
Solution:
Let AB (= 24 m) 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 C.

∴ AB² = AC²+ Bc²
So, 24² = AC²+ 182
or AC² = 576 – 324
or AC² = 552
or AC = \(\sqrt{252}\) = 6\(\sqrt{7}\)
Hence, the stake may be placed at a distance of 6\(\sqrt{7}\) m from the base of the pole.

Question 11.
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 \(\frac { 1 }{ 2 }\) hours?
Solution:
Let the first aeroplane starts from O and goes upto A towards north, where OA = (1000 x \(\frac { 3 }{ 2 }\)) km = 1500 km.

Let the second aeroplane starts from O at the same time and goes upto B towards west, where OB = (1200 x \(\frac { 3 }{ 2 }\)) km = 1800 km.
According to the problem, the required distance = BA.
∆In right angled ∆ ABC, by Pythagoras theorem, we have :
AB² = OA² + OB²
= (1500)² + (1800)² = 2250000 + 3240000
= 5490000 = 9 x 61 x 100 x 100
or AB = 3 x 100 \(\sqrt{61}\) km = 300 \(\sqrt{61}\) km.

Question 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.
Solution:

Let AB = 11 m and CD = 6 m be the two poles such that BD = 12 m.
Draw CE ⊥ AB and join AC.
∴CE = DB = 12 m,
AE = AB – BE
= AB – CD
= (11 – 6) m = 5 m.
In right angled ∆ ACE, by Pythagoras theorem, we have :
AC² = CE² + AE²
= (12)² + (5)² = 144 + 25 = 169
So, AC = \(\sqrt{169}\) = 13
Hence, the distance between the tops of the two poles is 13 m.

Question 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.
Solution:
In right angled ∆s ACE and DCB, we have :

AE² = AC²+ CE²
and BD² = DC²+ Bc²
So, AE² + BD² = (AC²+ CE²) + (DC²+ Bc²)
AE² + BD² = (AC²+ Bc²) + (DC²+ CE²)
AE² + BD² = AB² + DE² [By Pythagoras theorem, AC²+ BC²= AB² and DC²+ CE² = DE²]

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

Solution:
We have : DB = 3CD
Now, BC = DB + CD
i.e., BC = 3CD + CD [∵ BD = 3CD]
or BC = 4CD
∴ CD = \(\frac { 1 }{ 4 }\) BC and DB = 3CD = \(\frac { 3 }{ 4 }\) BC … (1)
Since ∆ ABD is a right triangle, right angled at D, therefore by Pythagoras theorem, we have :
AB² = AB² + DB²
Similarly, from ∆ACD, we have :
AC² = AD² + CD²
(1) – (2) gives

Question 15.
In an equilateral triangle ABC, D is a point on side BC such that BD = \(\frac { 1 }{ 3 }\) BC. Prove that 9AD² = 7AB².
Solution:
Let ABC be an equilateral triangle and let D be a point on BC such that
BD = \(\frac { 1 }{ 3 }\)BC.
Draw AE ⊥ BC. Join AD.

In ∆s AEB and AEC, we have :
AB = AC [∴∆ABC is equilateral]
∠AEB = ∠AEC [∵ Each = 90°]
and AE = AE
∴ By SAS criterion of congruence, we have :
∆ AEB = ∆ AEC
So, BE = EC
Now, we have :
BD = \(\frac { 1 }{ 3 }\)BC, DC = \(\frac { 2 }{ 3 }\)BC
BE = EC = \(\frac { 1 }{ 2 }\)BC … (1)
Since ∠C = 60°, therefore triangle.
∆ ADC is an acute triangle.

So, 9AD² = 7 AB², which is the required result.
Aliter:
DE = BE – BD

Question 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.
Solution:
Let ABC be an equilateral triangle and let AD ⊥ BC. In ∆ ADB and ∆ ADC, we have :
AB = AC [Given]
AD = AD [Common]
and ∠ADB = ∠ADC [Each = 90°]
By RHS criterion of congruence, we have :
∆ ADB ≅ ∆ ADC
So, BD = DC or BD = DC = \(\frac { 1 }{ 2 }\)BC … (1)
Since ∆ ADB is a right triangle, right angled at D, by Pythagoras theorem, we have :

Hence, the result.

Question 17.
Tick the correct answer and justify :
In ∆ ABC, AB = 6\(\sqrt{3}\)cm, AC = 12 cm and BC = 6 cm. The angle B is :
(A) 120°
(B) 60°
(C) 90°
(D) 45°
Solution:
In ∆ ABC, we have :

AB = 6\(\sqrt{3}\) cm, AC = 12 cm
and BC = 6 cm … (1)
Now, AB² + BC²= (6\(\sqrt{3}\))² + (6)² [From (1)]
= 36 x 3 + 36
= 108 + 36 = 144
= (AC)²
∴ Thus, ∆ ABC is a right triangle, right angled at B.
∠B = 90°
∴ (C) is the correct choice.

BSEB Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.6

Question 1.
In the figure, PS is the bisector of ∠QPR of ∆ PQR. Prove that \(\frac { QS }{ SR }\) = \(\frac { PQ }{ PR }\)

Solution:
Given : PQR is a triangle and PS is the bisector of ∠QPR meeting QR at S.
∴ ∠QPS = ∠SPR
To Prove : \(\frac { QS }{ SR }\) = \(\frac { PQ }{ PR }\)
Construction : Draw RT || SP to cut QP produced at T.
Proof :

Since PS || TR and PR cuts them, hence we have :
∠SPR = ∠PRT … (1) [Alternate ∠S]
and ∠QPS = ∠PTR … (2) [Corresponding ∠S]
But, ∠QPS = ∠SPR [Given]
∴ ∠PRT = ∠PTR [From (1) and (2)]
So, PT = PR … (3) [∵ Sides opp. to equal ∠S are equal]
Now, in ∆ QRT, we have
SP || RT [By Construction]
∴ \(\frac { QS }{ SR }\) = \(\frac { PQ }{ PT }\) [By Basic Proportionality Theorem]
or \(\frac { QS }{ SR }\) = \(\frac { PQ }{ PR }\) [From (3)]

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

(i) DM² = DN.MC
(ii) DN² = DM.AN
Solution:
We have : AB ⊥ BC and DM ⊥ BC

So, AB || DM
Similarly, we have :
BC ⊥ AB and DN ⊥ AB.
So, CB || DN
Hence, quadrilateral BMDN is a rectangle.
BM = ND

(ii) In ∆ BMD, we have :
∠1 + ∠BMD + ∠2 = 180°
or ∠1 + 90° + ∠2 = 180° or ∠1 + ∠2 = 90°
Similarly, in ∆ DMC, we have :
∠3 +∠4 = 90°
Since BD ⊥ AC, therefore
∠2 + ∠3 = 90°
Now, ∠1 + ∠2 = 90° and ∠2 + ∠3 = 90°
∴ ∠1 + ∠2 = ∠2 + ∠3
So, ∠1 = ∠3
Also, ∠3 +∠4 = 90° and ∠2 + ∠3 = 90°
∴ ∠3 +∠4 = ∠2 + ∠3 So, ∠2 =∠4
Thus, in As BMD and DMC, we have :
∠1 = ∠3 and ∠2 =∠4
∴ By AA criterion of similarity, we have :
∆ BMD ~ ∆ DMC
So, \(\frac { BM }{ DM }\) = \(\frac { MD }{ MC }\)
or \(\frac { DN }{ DM }\) = \(\frac { DM }{ MC }\) [∵BM = ND]
so, DM² = DN.MC

(ii) Proceeding as in (i), we can prove that
∆ BMD ~ ∆ DMC
So, \(\frac { BN }{ DN }\) = \(\frac { ND }{ NA }\)
or \(\frac { DM }{ DN }\) = \(\frac { DN }{ AN }\) [∵BN = DM]
so, DN² = DM.AN

Question 3.
In the figure, ABC is a triangle in which ∠ABC >90° and AD ⊥ CB produced. Prove that AC² = AB² + BC² + 2BC.BD.

Solution:
Given : ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced.
To Prove : AC² = AB² + BC² + 2BC.BD.

Proof : Since ∆ ADB is a right triangle, right angled at D, therefore, by Pythagoras theorem, we have :
AB² = AD² + DB² … (1)
Again, ∆ ADC is a right triangle, right angled at D.
Therefore, by Pythagoras theorem, we have
AC² = AD² + DC²
or AC² =AD² + (DB + BC)2
or AC² =AD² + DB²+ BC² + 2DB.BC
or AC² = (AD² + DB²) + BC² + 2BC.BD
or AC² = AB² +BC² + 2BC.BD [Using (1)]
which proves the required result.

Question 4.
In the figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that
AC² = AB² + BC² – 2BC.BD.

Solution: Given : ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC.
To Prove : AC² = AB² + BC² – 2BC . BD.
Proof : Since ∆ ADB is a right triangle, right angled at D, therefore, by Pythagoras theorem, we have :
AB² = AD² + BD² … (1)
Again, ∆ ADC is a right triangle, right angled at D. Therefore, by Pythagoras theorem, we have :
AC² = AD² + DC²
or AC² = AD² + (BC – BD)²
or AC² = AD² + (BC² + BD² – 2BC.BD)
or AC² = (AD² + BD²) + BC² – 2BC.BD
or AC² = AB² + BC² – 2BC.BD [Using (1)]
which proves the required result.

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

(i) AC² = AD² + BC . DM + (\(\frac { BC }{ 2 }\))²
(ii) AB² = AD² – BC . DM + (\(\frac { BC }{ 2 }\))²
(iii) AC² + AB² = 2AD² + \(\frac { 1 }{ 2 }\)BC².
Solution:
Since ∠AMD = 90°, therefore ∠ADM < 90° and ∠ADC > 90°.
Thus, ∠ADB is acute and ∠ADC is obtuse.

(i) In ∆ ADB, ∠ADC is an obtuse angle.

(ii) In ∆ ABD, ∠ADB is an acute angle.

(iii) From (1) and (2), we get
AB² + AC² = 2AD² + \(\frac { 1 }{ 2 }\)BC².

Question 6.
Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.
Solution:
We know that if AD is a median of ∆ ABC, then AB² + AC² = 2AD² + \(\frac { 1 }{ 2 }\)BC²
Since the diagonals of a parallelogram bisect each other, therefore BO and DO are medians of ∆s ABC and ADC respectively.

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

(i) ∆ APC ~ ∆ DPB
(ii) AP.PB = CP.DP
Solution:
(i) In ∆s APC and DPB, we have
∠APC = ∠DPB
∠CAP = ∠BDP [Angles in the same segment of a circle are equal]
∴ By AA criterion of similarity, we have :
∆ APC ~ ∆ DPB.

(ii) Since ∆ APC ~ ∆ DPB, therefore
∴ \(\frac { AP }{ DP }\) = \(\frac { CP }{ PB }\) or AP.PB = CP.DP

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

(i) ∆ PAC ~ ∆ PDB
(ii) PA.PB = PC.PD
Solution:
(i) In ∆s PAC and PDB, we have :
∠APC = ∠BPD [Common]
∠PAC = ∠PDB [∵∠BAC = 180° – ∠PAC and ∠PDB = ∠CDB = 180° – ∠BAC = 180° – (180° – ∠PAC) = ∠PAC]
∴ By AA criterion of similarity, we have
∆ PAC ~ ∆ PDB

(ii) Since ∆ PAC ~ ∆ PDB, therefore
\(\frac { PA }{ PD }\) = \(\frac { PC }{ PB }\) OR PA.PB = PC.PD

Question 9.
In the figure, D is a point on side BC of ∆ ABC such that \(\frac { BD }{ CD }\) = \(\frac { AB }{ AC }\). Prove that AD is the bisector of ∠BAC.

Solution:
Given : ABC is a triangle and D is a point on BC such that
\(\frac { BD }{ CD }\) = \(\frac { AB }{ AC }\)
To Prove : AD is the bisector of ∠BAC.
Construction : Produce BA to E such that AE = AC. Join CE.

Proof : In ∆ AEC, since AE = AC, hence
∠AEC = ∠ACE …(1) [∵ Angles opp. to equal sides of a A are equal]
Now, \(\frac { BD }{ CD }\) = \(\frac { AB }{ AC }\) [Given]
So, \(\frac { BD }{ CD }\) = \(\frac { AB }{ AE }\) [∵ AE = AC, construction]
∴ By converse of Basic Proportionality Theorem, we have :
DA || CE
Now, since CA is a transversal, we have :
∠BAD = ∠AEC … (2) [Corresponding ∠S]
and ∠DAC = ∠ACE … (3) [.Alternate angles]
Also, ∠AEC – ∠ACE [From (1)]
Hence, ∠BAD= ∠DAC [From (2) and (3)]
Thus, AD bisects ∠B AC.

Question 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 taut, 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 ?

Solution:
In fact, we want to find. AC.

By Pythagoras Theorem, we have :
AC² – (2.4)² + (1.8)²
or AC² = 5.76 + 3.24
= 9.00
or AC = 3 m
∴ Length of string she have out = 3m.
Length of the string pulled at the rate of 5 cm/sec in 12 seconds
= (5 x 12) cm
= 60 cm = 0.60 m
Remaining string left out
= (3 – 0.6) m
= 2.4 m

In 2nd case, let us find PB.
PB² = PC² – BC²
= (2.4)² – (1.8)²
= 5.76 – 3.24 = 2.52
or PB = \(\sqrt{2.52}\) = 1.59 (nearly)
Hence, the horizontal distance of the fly from Nazima after 12 seconds
= (1.59 + 1.2) m
= 2.79 m (nearly).