Gujarat Board Solutions Class 10 Maths Chapter 6 Triangle

 Gujarat Board Solutions Class 10 Maths Chapter 6 Triangle

Gujarat Board Textbook Solutions Class 10 Maths Chapter 6 Triangle

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. (a) equal (b) proportional.

Question 2.
Give two different examples of pair of
(i) Similar figure
(ii) Non-similar figure.
Solution:
(i) (a) Two equilateral triangle are similar.
(b) Two squares are similar.
(ii) (a) Two triangles i.e. one scalene and other isosceles.
(b) Two rectangles i.e. one square and other rectangle.

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

Solution:
Angles of both figure are not equal so they are not similar.

Ex 6.2

Question 1.
In figure (i) and (ii). DE II BC. Find EC in (i) and AD in (ii).

     

Question 7.
Using theorem 6.1 (i.e. Basic proportionality theorem), prove that 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) (CBSE 2012)
Solution:
Given:
∆ABC in which, D is the midpoint of AB and
DE || BC

Question 8.
Using theorem 6.2 (i.e. Converse of basic proportionality theorem), prove that the line joining the midpoints of any two sides a triangle is a parallel to the third side (Recall that you have done it in class IX)
Solution:
Given: ∆EC in which D and E are the midpoints of sides AB and AC respectively. DE is the line joining D and E.

   

∴ In ∆ADC
OE || CD (by converse of BPT)
But OE || AB (by construction)
therefore AB ||CD
Hence quadrilateral ABCD is a trapezium.

Ex 6.3

Question 1.
State in which pairs of triangles in the figures are similar. Write the similarity criteria used by you for answering the question and also write the pairs of similar triangles in the symbolic form.

  

(vi) In ∆DEF and ∆PQR
∠F = 180° – (70° + 80°)
= 180° – 150° = 30°
∠P = 180° – (80° + 30°)
∠p = 70°
∠D = ∠P (each 70°)
∠E = ∠Q (each 80°)
∠F = ∠R (each 30°)
Hence ∆DEF ~ ∆PQR (by AAA similarity)

Question 2.
In the figure ∆ODC – ∆OBA, ∆BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB

Solution:
We have
∠BOC = 125°
and ∠CDO = 70°
Since ∠DOC + ∠BOC = 1800 (linear pair)
∠DOC + 125° = 1800
∠DOC = 180° – 125°
∠DOC = 55° ……….(1)

In ∆DOC
∠DCO + ∠CDO + ∠DOC = 180°
(by angle sum property)
∠DCO + 70° + 55° = 180°
∠DCO 180° – 125°
∠DCO = 55° ……..(2)
Since ODC – ∠OBA (given)
∠OCD = ∠OAB = 55° ………(3)
(corresponding angle of similar triangles)
Hence, ∠DOC = 55°, ∠DCO = 55°
∠OAB = 55°

 

Question 5.
S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ – ∆RTS. (CBSE 2012)
Solution:
We have T is a point on QR and S is a point on PR.

Now in ∆RPQ and ∆RTS
∠RPQ = ∠RTS (given)
and ∠PRQ = ∠TRS (common)
∴ ∆RPQ ~ ∆RTS (by AA similarity)

Question 6.
In figure, if ∆ABE MCD, show that ∆ADE – ∆ABC.

Question 7.
In the figure, altitude AD and CE of ∆ABC intersect each other at the P. Show that
(i) ∆AEP ~ ∆CDP
(ii) ∆ABD ~ ∆CBE
(iii) ∆AEP ~ ∆ADB
(iv) ∆PDC ~ ∆BEC

Solution:
∆ABC In which AD and CF intersect each other at P.
(i) In ∆AEP and ∆CDP
∠AEP = ∠CDP (each 90°)
∠APE = ∠CPD (vertically opposite angles)
∴ ∆AEP ~ ∆CDP (by AA Similarity)

(ii) In ∆ABD and ∆CBE
∠ADB = ∠CEB (each 90°)
∠ABD =∠CBE (common)
∆RD ~ ACRE (by AA similarity)

(iii) In ∆AEP and ∆ADB
∠AEP = ∠ADP (each 90°)
and ∠EAP = ∠DAB (common)
∴ ∆AEP ~ ∆ADB (by AA similarity)

(iv) In ∆PDC and ∆BEC
∠PDC = ∠BEC (each 90°)
and ∠DCP = ∠ECB (common)
∴ ∆PDC ~ ∆BEC (by AA similarity)

Question 8.
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ~ ∆CFB.
Solution:
A parallelogram ABCD in which AD is produced to E and BE is pointed such that BE intersects CD at F.

Now, in ∆ABE and ∆CFB
∠BAE = ∠FCB
(opposite angle of || are equal)
∠AEB = ∠CBF ( AE || BC alternate interior angles)
∆ABE ~ ∆CFB (by AA similarity)

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

 

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 ABC and EFG respectively. If ∆ABC ~ ∆FEG show that

 

Question 11.
In figure, E is a point on side CB produced on an isosceles triangle ABC with AB = AC. If AD⊥BC and EF⊥AC, Prove that ∆ABD ~ ∆ECF. (CBSE 2012)

Solution:
Given: ∆ABC is an isosceles triangle, E is a point on BC produced
AB = AC
Also AD ⊥ BC andEF ⊥ AC
To Prove: ∆ABD – ∆ECF
Proofs In ∆ABD and ∆ECF
∠ABD = ∠ECF
(AB = AC, angle opposite to equal sides are equal)
and ∠ADB = ∠EFC (each 90°)
∠ABD – ∠ECF (by AA similarity)

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.

   

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:
Le ∆AB = 6m be the pole and BC = 4m be its shadow whereas the shadow of tower is 28 m

Question 16.
If AD and PM are medians of triangles ∆ABC and ∆PQR, respectively where ∆ABC ~ ∆PQR.
Prove that: = AB/PQ=AD/PM
Solution:
Given: AD and PM are medians of triangle ABC and PQR, where ∆ABC ~ ∆PQR

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:
We have
∆ABC ~ ∆DEF
ar ∆ABC = 64 cm2
ar ∆DEF = 14cm2
and EF = 15.4 cm

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. (CBSE 2012)
Solution:
Trapezium ABCD, in which AB DC and diagonal AC and BD intersect at O.

In ∆AOB and ∆COD
∠AOB = ∠COD (Vertically opposite angle)
∠OAB = ∠OCD (Alternate Interior angles)
∴ ∆AOB ~ ∆COD (by AA similarity)

∴ar ∆AOB: ar ∆COD = 4: 1

Question 3.
In Figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, arAABC AO
show that =  (CBSE 2005, 2012)

Solution:
∆ABC and IDBC are two triangles on the same base BC and diagonals AD and BC Intersect each other at O.
Let us draw AE ⊥ BC and DF ⊥ BC
In ∆AOE and ∆DOF
∠AOE = ∠DOF (Vertically opposite angles)
∠AEO = ∠DFO (each 900)
∆AOE ~ ∆DOF (by AA similarity)
AE/DF = AO/DO ………(1)
(Corresponding sides of two similar triangles are proportional)

From equations (1) and (2)

Question 4.
If the area of two similar triangles are equal, prove that they are congruent. (CBSE 2012)
Solution:
Given: ∆ABC and ∆DEF are two similar triangles such that, ar ∆ABC = ar ∆DEF

Question 5.
D, E and F are respectively the midpoints of sides AB, BC and CA of ∆ABC. Find the ratio of the area of ∆DEF and ∆ABC.
Solution:
D, E and F are respectively the midpoints of the sides of ∆ABC. Jour the points D, E and F to form ∆DEF.

∠ADF = ∠ABC (corresponding angles)
and ∆AFD = ∆ACE ……….(4)
(corresponding angles)
∴ ∆DEF ~ ∆ABC (by AA similarity)

(The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides)

Line drawn from mid points of two sides of a triangle is parallel to third and half of the third.

ar ∆DEF : ar ∆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 and ∆DEF such that
∆ABC ~ ∆DEF
AM and DN are their corresponding medians.

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 its diagonals. (CBSE 2012)

Question 8.
ABC and BDE are two equilateral triangles such that D is the midpoint of BC. The ratio of the areas of triangles ABC and BDE
(a) 2 : 1
(b) 1 : 2
(c) 4 : 1
(d) 1 : 4
Solution:
∆ABC and ∆BDE are both equilateral triangles
∴ ∆ABC ~ ∆BDE (by AAA similarity)

(Ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides)

(D is the mid point of BC and AB = BC = CA)
∴ar ∆ABC : ar ∆BDE = 4 : 1
Thus correct option is (c).

Question 9.
Sides of two similar triangles are in the ratio 4: 9. The area of these triangles are in the ratio.
(a) 2 : 3
(b) 4 : 9
(c) 81 : 16
(d) 16 : 81
Solution:
Let ∆ABC and ∆DEF are similar then

ar ∆ABC : ar ∆DEF = 16 : 81
Thus (d) is the correct answer.

Ex 6.5

Question 1.
Sides of triangles are given below. Determine which of these are 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) 7 cm, 24 cm, 25 cm.
Here 72 = 49cm
242 = 576 cm
252 = 625 cm
we observed that
49 + 576 = 625
72 + 242 = 252
The given triangle is right angled triangle.
Hypotenuse = 25 cm

(ii) 3 cm, 8 cm, 6 cm
Here 32 = 9cm
82 = 64 cm
62 = 36 cm
we observed that
32 + 62 + 82
The given triangle is not right angled triangle.

(iii) 50 cm, 80 cm, 100 cm.
Here 502 = 2500 cm
802 = 6400 cm
1002 = 10,000 cm
We observed that
502 + 802 ≠ 1002
The given triangle is not right angled.

(iv) 13 cm, 12 cm, 5 cm.
Here 132 = 169 cm
122 = 144 cm
52 = 25 cm
we observed that
122 + 52 = 132
∴ The given triangle is not right angled.
Hypotenuse = 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:

Question 3.
In the 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 (CBSE 2010)

(ii) In ∆ABC
∠BAC + ∠CBA = 90° ………(1) (by angle sum property)
In ∆ABD
∠BDA + ∠CBA = 90° …….(2)
From eqn (1) and (2)
∠BAC + ∠CBA = ∠BDA + ∠CBA
∠BAC = ∠BDA
In ∆ACB and ∆DCA
∠ACB = ∠DCA (each 900)
∠BAC = ∠BDA
(proved above from eqn (3))
∴ ∆ACB ~ ∆DCA (by AA similarity)
AC/DC=BC/AC
(corresponding sides of two similar triangle are proportional)
⇒ AC2 = BC.DC

(iii) In ∆ADB and ∆CDA
∠DAB = ∠DCA (each 900)
∠ADB = ∠CDA (Common)
∴ ∆ADB ~ ∆CDA (by AA similarity)
Hence = AD/CD=BD/CD
(Corresponding sides of two similar triangles are proportional)
⇒ AD2 = BD.CD

Question 4.
ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2. (CBSE 2012)
Solution:
We have ∆ABC is right angled such that
∠C = 90° and AC = AB

Applying Pythagoras theorem
– AC2 + BC2
= AC2 + AC2 (BC = AC given)
Thus AB2 = 2AC2

Question 5.
ABC is an isosceles triangle with AC = BC, If BC2 = 2AC2, prove that ABC is a right triangle.
Solution:
We have ∆ABC such that AB = AC

Also AB2 = 2AC2 = AC2 + AC2
AB2 = BC2 +AC2(BC =AC given)
∠ACB = 90° (by converse of Pythagoras theorem)
i.e. ∆ABC is a right angle triangle

Question 6.
ABC is an equilateral triangle of side 2a. Find each of its attitudes.
Solution:
It is given that LABC is an equilateral triangle in which

AB = AC = CB = 2a
Let us draw AD ⊥ BC
Now in ∆ABD and ∆ACD
AB = AC (Side of equilateral triangle)
AD = AD (Common)
∠ADB = ∠ADC (each 90°)
∴ ∆ABD ≅ ∆ACD (by RHS)
∴ BD = CD (by CPCT)
i.e. D is a mid point of BC
Hence BD = 2a/2 = a
Now in right ∆ABD
AB2 = BD2 + AD2
(2a)2 = a2 + AD2
4a2 – a2 = AD2
AD2 = 3a2
AD = √3
Similarly BE = CF = √3

Question 7.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. (CBSE 2012) of its diagonals. (CBSE 2012)
Solution:
Given: ABCD is a rhombus i.e. AB = BC = CD =
DA whose diagonals AC and BD intersect at O.
To prove: AC2 + BD2
= AB2 + BC2 + CD2 + DA2

Proof: We know that diagonals of a rhombus bisect each other at right angles.
∴ OA = OC and OB = OD
Now in right ∆AOB
AB2 = OA2 + OB2 (by Pythagoras theorem)
In right ∆BOC
BC2 = OB2 + OC2……..(2)
Similarly in ∆COD
CD2 = OC2 + OD2
and in right ∆AOD
AD2 – OA2 + OD2
Now adding eqn (1), (2), (3) and (4) we get
AB2 + BC2 + CD2 + DA2
= OA2 + OB2 + OC2 + OB2 + OD2 + OC2 + OD2 + OA2
= 20A2 + 20B2 + 20C2 + 20D2
= 20A2 + 20B2 + 20A2 + 20B2
= 40A2 + 40B2
= (20A)2 + (20B)2 + AB2 + BC2 + CD2 + DA2
= AC2 + BD2
(20A = AC and 20B = BD)

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 – DE2 – OF2 = AF2 + BD2 + CE2
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2

Solution:
In the given figure, 0 is in the interior of ∆ABC,
OD ⊥ BC, OF ⊥AB, OE ⊥AC
To prove:
(i) OA2 + OB2 + OC2 + 0D2 + 0E2 + OF2 = AF2 + BD2 + CE2
(ii) AF2 + BD2 + CE2 = 2 + CD2 + BF2
Construction: Join OA, OB and OC
Proof: In right ∆AOF
OA2 = AF2 + OF2 (by Pythagoras theorem)
AF2 = OA2 + OF2 ……….(1)

In right ∆OBD
OB2 + BD2 + OD2
BD2 = OB2 + OD2 ………..(2)

In right ∆OCE
OC2 = CE2 + OE2
CE2 = OC2 + OE2
Adding eqn (1), (2) and (3) we get
AF2 + BD2 + CE2 = OA2 + OB2 + OC2 + OD2 + OE2 – OF2 ………..(3)

(ii) In right ∆OBD
OB2 = OD2 + BD2
(by Pythagoras theorem)
⇒ BD2 = OB2 – OD2 ……….(1)

In right triangle ∆OCD
OC2 = OD2 + CD2
(by Pythagoras theorem)
⇒ CD2 = OC2 – OD2 ……….(2)

Subtracting eqn (2) from eqn (1)
BD2 – CD2 = OB2 – OD2 – OC2 – OD2
BD2 – CD2 = OB2 – OC2 ……….(3)
Similarly in right triangle OCE and OAE we get
CE2 = OC2 – OE2 ………(4)
(by Pythagoras theorem)
AE2 = OA2 – OE2 ………(5)

Subtracting eqn (5) from eqn (4)
CE2 – AE2 = OC2 – OE2 – OA2 + OE2
⇒ CE2 – AE2 = OC2 – OA2 ……….(6)

And in right triangle OAF and OBF
AF2 = OA2 – OF2 ……….(7)
BF2 = OB2 – OF2 ………..(8)

Subtracting eqn (8) form (7)
AF2 – BF2 = AO2 – OF2 – OB2 + OF2
2 – BF2 = OA2 – OB2 ………..(9)

Adding eqn (3), (6) and eqn (9) we get
BD2 + CE2 + AF2 – CD2 – AE2 – BF2
⇒ OB2 + OC2 + OA2 – OC2 – OA2 – 0B2
BD2 + CE2 + AF2 – CD2 – AE2 – BF2 = O
BD2 + CE2 + AF2 = CD2 + AE2 + BF2

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 base of the wall.
Solution:
Let AB is a ladder and A is the foot of the ladder BC is the wall.

Now in right triangle ABC
AB2 = AC2 + BC2
(by Pythagoras theorem)
(10)2 + AC2 + 82
⇒ 100 – 64 = AC2
⇒ AC2 = 36
⇒ AC = 6m
Hence, the distance of the foot of the ladder from the base of wall is 6 m.

Question 10.
A guy wire attached to a vertical pole of height 18 m is 24 m long and has a state 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 is the wire and BC is the vertical pole. The point A is the stake.

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 plane after 1 1/2 hours?
Solution:
Let the point A represent the airport. First plane flies due North with speed of 1000 km/h.
Distance of plane after 1 1/2 hour from airport = speed x time

Question 12.
Two poles of height 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 and CD be the two poles. Distance between their feet is 12 m. Height of pole 1st
AB = 11 m and height of pole lind CD = 6 m

AB = 11m
CD = 6m
BB = AB – AE (AE = CD)
⇒BE =11 – 6
⇒BE = 5m

In ∆BED
BD2 = DE2 + BE2
= 122 + 52 = 144 + 25
⇒ BD2 = 169
⇒ BD = 13m
Thus the distance between tops is 13 metres.

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:
Given: ∆ABC in which ∠C = 90°, D and E are the mid prints on the sides AC and BC.

To prove: AB2 + BD2 = AB2 + DE2
Proof: In right angle AACB
AB2 = AC2 + BC2 (by Pythagoras theorem)
In DCE
DE2 = CD2 + CE2 ………(2)
(by Pythagoras theorem)
Adding eqn (1) and (2)
We get AB2 + DE2 = AC2 + BC2 + CD2 + CE2
AB2 + DE2 = (AC2 + CE2) +(BC2 + CD2) …….(3)

Now in ∆ACE
AE2 = AC2 + CE2 ……. (4)
In ∆BCD
BC2 + CD2 = BD2 ……(5)
from eqn (3), (4) and (5)
AB2 + DE2 = AE2 + BD2

Question 14.
The perpendicular from A on side BC at D of a ¿ABC intersects BC at D such that DB = 3CD (see figure). Prove that 2AB2 = 2AC2 + BC2.
Solution:
Given: ∆ABC such that AD ⊥ BC. The position of D is such that BD = 3 CD.

Question 15.
In an equilateral triangle ABC, D is a point on side BC such that BD = BC. Prove that
Solution:
Given: AABC is an equilateral triangle.

BD = 1/3 BC
To prove: 9AD2 = 7AB2
Construction: Draw AP⊥BC
Proof: In right AAPB, we have
AB2 =AP2 + BP2 …….(1) (by Pythagoras theorem))
Now, in right triangle APD, we have
AD2 = AP2 + DP2
(by Pythagoras theorem)
AP2 =AD2 – DP2
From eqn (1) and (2) we get
AB2 =AD2 – DP2 + BP2
AB2 =AD2 – DP2 + (BP = BC)
AD2 = AB2 + DP2 – 4
Note that, DP = BP – 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 altitude.
Solution:
Given: ABC is an equilateral triangle,
To Prove: 3AB2 = 4AD2

Ex 6.6

Question 1.
In the figure, PS is the bisector of ∠QPR of

Question 2.
In the given figure, D is a point on hypotenuse AC of ∆ABC, such that BD ⊥ AC, DM ⊥BC and DN ⊥ AB. Prove that:
(i) DM2 = DN.
(ii) DM2 = DM . AN

Solution:
In the given figure ∆ABC in which
AB ⊥ BC and DM ⊥ BC,
AB || DM
Similarly CB ⊥ AB and DN ⊥ AB
CB || DN
Hence quadrilateral BMDN is a rectangle
BM = ND
(i) In ∆BMD
∠1 + ∠BMD + ∠2 = 180°
(by Angle sum property)
∠1 + 90° +∠2 = 180°
∠1 + ∠2 = 90° ……………(1)
Similarly, in ∆DMC we have
∠3 + ∠4 = 90° ……………. (2)
Since BD ⊥ AC
Therefore ∠2 + ∠3 = 90° ………….(3)
From eqn (1) and (3)
∠1 + ∠2 = ∠2 + ∠3
∠1 = ∠3
Also from eqn (2) and (3)
∠3 + ∠4 = ∠2 + ∠3
∠2 = ∠4
Thus is ∆DMB and ∆DMC
We have
∠1 + ∠3 and ∠2 + ∠4
∆B/ND ~ ∆D/NA
BN/DN = NDDA
DM/DN = DN/MC
DN2 = DM . MC

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

Solution:
Given : ∆ABC in which < ABC > 90° and AD ⊥ CB produced.
To prove: In right triangle ADC
∠D = 90°
AC2 = AD2 + DC2
(by Pythagoras theorem)
AC2 = AD2 + (BD + BC)2
(DC = BD + BC)
AC2 = AD2 + BD2 + BC2 + 2BC . BD ………….(1)
Now in right ∆ADB
AC2 = AD2 + BD2 ………………….(2)
(by Pythagoras theorem)
From eqn (1) and (2)
AC2 = AB2 + BC2 + 2BC . BD

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

Proof: In right ∆ADC
∠D = 90°
AC2 = AD2 + (BC – BD)2
(DC = BC – BD)
AC2 = AD2 + BC2 + BD2 – 2BC . BD
AC2 = AD2 + BD2 + BC2 – 2BC . BD ……………(1)
Now in right ∆ADB
AB2 = AD2 + BD2 …………(2)
(by Pythagoras theorem)
From eqn (1) and (2) we have
AC2 = AB2 + BC2 – 2BC2 . BD

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

  

Question 6.
Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
Solution:
Given:
ABCD is a ||gm whose diagonals are AC and BD.

To prove:
AB2 + BC2 + CD2 + CA2 = AC2 + BD2
Construction:
Draw AM ⊥ DC and BN ⊥ DC (produced)
Proof: In right ∆AMD and ∆BNC
AD = BC (opposite sides of ||gm)
AM = BN (altitudes of same ||gm on the same base)
∆AMD ≅ ∆BNC (by RHS congruence Rule)
MD = CN …………..(1) (by CPCT)
Now in right triangle BND
∠N = 90°
BD2 = BN2 + DN2
(by Pythagoras theorem)
BD2 = BN2 + (DC + CN)2(DN = DC + CN)
BD2 = BN2 + DC2 CN2 + 2DC . CN
BD2 = BN2 + CN2 + DC2 + 2DC . CN ………….(2)
Now in right ∆BNC
BD2 = BN2 + CN2 ……..(3)
From eqn (2) and (3)
BD2 = BC2 + DC2 + 2DC . CN ………………(4)
Now in right ∆ADM
∠M = 90°
AC2 = AM2 + MC2 ………….
(by Pythagoras theorem)
AC2 = AM2 + (DC – DM)2
(MC = DC – DM)
AC2 = AM2 + DC2 + DM2 – 2DC . DM
AC2 = AM2 + DM2 + DC2 – 2DC . DM …………..(5)
Now in right ∆ADM
AD2 = AM2 + DM2 ……………(6)
(by Pythagors theorem)
From eqn (5) and (6) and we have
AC2 = AD2 + DC2 – 2DC . DM
AC2 = AD2 + AB2 – 2DC . CN …………..(7)
(DC = AB opposite sides of ||gm and DM = CN from eqn (1))
Adding eqn (4) and (7)
AC2 + BD2 = AD2 + AB2 + BC2 + DC2
AC2 + BD2 = AB2 + BC2 + CD2 + DA2

Question 7.
In figure, two chords AB and CD intersect each other at the point P prove that
(i) ∆APC ~ ∆DPB
(ii) AP . PB = CP . DP

Question 8.
In figure, two chords AB and CD of a circle intersect each other at the point P (When produce) outside the circle. Prove that
(i) ∆PAC ~ ∆PDB
(ii) PA . PB = PC . PD

 

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

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:
Let us find the length of the string that the has out
In right ∆ABC
AC2 = AB2 + BC2
(by Pythagoras theorem)
AC2 = (2.4)2 + (1.8)2
AC2 = 5.76 + 3.24 = 9
AC = 3m
i.e length of string she has out = 3m since the string is pulled out at the rate of 5 cm/sec.
Length of the string pulled out in
12 seconds = 5 cm x 12 cm = 60 cm
= 60/100 m = 0.60 metre
Remaining string left out = (3 – 0.60) m
= 2.4m
To find horizontal distance
Inright ∆PBC, let PB the required horizontal distance of fly.
Since PC2 = PB2 + BC2
PB2 = PC2 + BC2
PB2 = (2.4)2 – (1.8)2
PB2 = 5.76 – 3.24
PB2 = 2.52
PB = 1.59 (approx)

Thus, the horizontal distance of the fly from nazima after 12 seconds
= 1.59 + 1.2 = 2.79 m (approx)