Question 1 Find the point on the x-axis which is equidistant from (2 -5) and (-29)

Suggested MarksSuggested Answers

Part Answer1

Let P(x0) be any point on x-axis and let A(2-5) and B(-29) be given pointsPA = PB (Given)

there4 xminus 2( )2 + 0 + 5( )2radic = x+ 2( )2 + 0 minus 9( )2radic

rArr x2 + 4 minus 4x+ 25radic = x2 + 4 + 4x+ 81radic

Part Answer2

Squaring both sides we getx2 minus 4x+ 29 = x2 + 4x+ 85rArr x2 minus 4xminus x2 minus 4x = 85 minus 29rArr minus 8x = 56 rArr x = minus 7there4 Coordinates of P are (minus70)

Question 2 Find the area of the triangle whose vertices are (2 3) (-1 0) and (2 -4)Suggested Answers

Part Answer1

Let the vertices of Δ ABC be A(23) = (x1 y1)B(-10) = (x2 y2) and C (2 -4) = (x3 y3)

= 12⎡⎣x1 y2 minus y3( ) + x2 y3 minus y1( ) + x3 y1 minus y2( )⎤⎦

MATH MOCK TEST (2010-11) - 5

TEST ID MATHX005Class - X

ANSWER KEY

Question 1

Question 2

Part Answer2

= 12⎡⎣2 0 + 4( ) + minus1( ) minus4 minus 3( ) + 2 3 minus 0( )⎤⎦

= 12⎡⎣8 + 7 + 6⎤⎦

= 12times21 = 21

2there4 Area of Δ ABC = 21

2 sq units

Question 3 In Fig find the length of AP

Suggested MarksSuggested Answers

Part Answer1

AD = BC = 40 cmangDAP = 90deg minus 30deg = 60degthere4 In Δ APDAPAD = sec 60deg = 2there4 AP = 2times AD = 2times40 rArr AP = 80 cm

Question 4 In Fig find the length of CF

Suggested Answers

Part Answer1

BD = AF = 10cmDF = AB = 2 cmIn Δ BDCCDBD = tan 30deg rArr CD

BD = 13radic

rArr CD = 13radic BD rArr CD = 1

3radic times10 = 10 3radic3

Question 3

Question 4

Part Answer2

there4 CF = CD + DF = 10 3radic3 + 2

= 10 3radic +63 cm

Question 5 In Fig Find the coordinate of A

Suggested MarksSuggested Answers

Part Answer1

Centroid

G = x+3minus73 yminus5+4

3

2 minus 1( ) = xminus43 yminus1

3

Part Answer2

there4 xminus43 = 2 and yminus1

3 = minus 1xminus 4 = 6 and yminus 1 = minus 3x = 10 and y = minus 2there4 Point A(10 minus2)

Question 6 Find the coordinates of a point A where AB is the diameter of a circle whose centre is(23) and B is (14)

Suggested Answers

Part Answer1

Let the coordinates of A be (xy) and the given points be C(2 -3) and (xy) B (14)

∵ Centre is the mid point of the diameterthere4 Coordinates of C = x+1

2 y+42

Question 5

Question 6

Part Answer2

Since the coordinates of C are (2-3)

there4 x+12 = 2

y+4

2 = minus 3

x+ 1 = 4x = 4 minus1x = 3

y+ 4 = minus 6y = minus 6 minus 4y = minus 10

there4 Coordinates of A are (3 minus10)

Question 7 Show that points (a b +c) (b c + a) and (c a + b)are collinear

Suggested MarksSuggested Answers

Part Answer1

Let the given points be A (a b + c) = (x1 y1) B (b c + a)= (x2 y2) and C(c a+b) = (x3 y3)there4 Area of Δ ABC= 1

2⎡⎣x1 y2 minus y3( ) + x2 y3 minus y1( ) + x3 y1 minus y2( )⎤⎦

= 12⎡⎣a c+ aminus aminus b( ) + b a+ bminus bminus c( ) + c b+ cminus cminus a( )⎤⎦

Part Answer2

= 12⎡⎣a cminus b( ) + b aminus c( ) + c bminus a( )⎤⎦

= 12⎡⎣acminus ab+ abminus bc+ bcminus ac⎤⎦

= 12times0 = 0

Since the area of the triangle is zerothere4 The points are collinear

Question 8 In fig ABCD is a rectangle in which segments AP and AQ are drawn as shown Findthe length of (AP + AQ)

Suggested MarksSuggested Answers

Question 7

Question 8

Part Answer1

In rt Δ ADQADAQ = sin 30deg rArr AD

AQ = 12

rArr AQ = 2AD rArr AQ = 2times30 = 60 cm hellip i( )

Part Answer2

In rt Δ ABPABAP = cos 60deg rArr AB

AP = 12

rArr AP =2AB rArr AP = 2times60 = 120 cm hellip ii( )there4 AP+AQ = 120+60 = 180 cm

Question 9 The horizontal distance between two trees of different heights is 60 m The angle ofdepression of the first tree when seen from the top of the second tree is 45deg If the height of the

second tree is 80 m find the height of the first treeSuggested Answers

Part Answer1

Let height of first tree AB = x mHeight of second treeCD = 80 M (Given)Difference BD = AF = 60 mIn rt Δ AFCCFAF = tan 45deg

Question 9

Part Answer2

rArr CF60 = 1 rArr CF = 60

Hence height of the first tree AB = CD - CF = 80 - 60 = 20 m

Question 10 Show that the points (aa) (-a -a) and minus 3radic a 3radic a( ) are the vertices of anequilateral triangle

Suggested MarksSuggested Answers

Part Answer1

Let A (aa) B (-a -a) and C minus 3radic a 3radic a( ) be the vertices of Δ ABC

AB = minusaminus a( )2 + minusaminus a( )2radic

= 4a2 + 4a2radic = 8a2radic = 2 2radic a

Part Answer2

BC = minus 3radic a+ a( )2 + 3radic a+ a( )2radic

= 3a2 + a2 minus 2 3radic a2 + 3a2 + a2 + 2 3radic a2radic

= 8a2radic = 2 2radic a

Part Answer3

AC = minus 3radic aminus a( )2 + 3radic aminus a( )2radic

= 3a2 + a2 + 2 3radic a2 + 3a2 + a2 minus 2 3radic a2radic

= 8a2radic = 2 2radic athere4 AB = BC = AC

Hence the triangle ABC is an equilateral triangle

Question 11 Find the coordinates of the points which divide the line-segment joining the points(57) and (8 10) in three equal parts

Suggested MarksSuggested Answers

Question 10

Question 11

Part Answer1

Let Points A = (5 7) and B = (8 10)Let Points C and D divide line AB in three equal parts

there4 ACCB = 12CDDB = 11 [ ∵ D is the mid point of CB]

Part Answer2

Point C = 1times8+2times51+2 1times10+2times7

1+2⎛⎝

⎞⎠

hellip[Using Section formula]= 8+10

3 10+143

⎛⎝

⎞⎠or 18

3 243

⎛⎝

⎞⎠or 68( )

Part Answer3

Point D = 6+82 8+10

2⎛⎝

⎞⎠or 14

2 182

⎛⎝

⎞⎠or (79)

Question 12 A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaffof height 5 m From a point on the plane the angles of elevation of the bottom and top of the

flagstaff are respectively 30deg and 60deg Find the height of the tower

Suggested MarksSuggested Answers

Question 12

Part Answer2

= 12⎡⎣2 0 + 4( ) + minus1( ) minus4 minus 3( ) + 2 3 minus 0( )⎤⎦

= 12⎡⎣8 + 7 + 6⎤⎦

= 12times21 = 21

2there4 Area of Δ ABC = 21

2 sq units

Question 3 In Fig find the length of AP

Suggested MarksSuggested Answers

Part Answer1

AD = BC = 40 cmangDAP = 90deg minus 30deg = 60degthere4 In Δ APDAPAD = sec 60deg = 2there4 AP = 2times AD = 2times40 rArr AP = 80 cm

Question 4 In Fig find the length of CF

Suggested Answers

Part Answer1

BD = AF = 10cmDF = AB = 2 cmIn Δ BDCCDBD = tan 30deg rArr CD

BD = 13radic

rArr CD = 13radic BD rArr CD = 1

3radic times10 = 10 3radic3

Question 3

Question 4

Part Answer2

there4 CF = CD + DF = 10 3radic3 + 2

= 10 3radic +63 cm

Question 5 In Fig Find the coordinate of A

Suggested MarksSuggested Answers

Part Answer1

Centroid

G = x+3minus73 yminus5+4

3

2 minus 1( ) = xminus43 yminus1

3

Part Answer2

there4 xminus43 = 2 and yminus1

3 = minus 1xminus 4 = 6 and yminus 1 = minus 3x = 10 and y = minus 2there4 Point A(10 minus2)

Question 6 Find the coordinates of a point A where AB is the diameter of a circle whose centre is(23) and B is (14)

Suggested Answers

Part Answer1

Let the coordinates of A be (xy) and the given points be C(2 -3) and (xy) B (14)

∵ Centre is the mid point of the diameterthere4 Coordinates of C = x+1

2 y+42

Question 5

Question 6

Part Answer2

Since the coordinates of C are (2-3)

there4 x+12 = 2

y+4

2 = minus 3

x+ 1 = 4x = 4 minus1x = 3

y+ 4 = minus 6y = minus 6 minus 4y = minus 10

there4 Coordinates of A are (3 minus10)

Question 7 Show that points (a b +c) (b c + a) and (c a + b)are collinear

Suggested MarksSuggested Answers

Part Answer1

Let the given points be A (a b + c) = (x1 y1) B (b c + a)= (x2 y2) and C(c a+b) = (x3 y3)there4 Area of Δ ABC= 1

2⎡⎣x1 y2 minus y3( ) + x2 y3 minus y1( ) + x3 y1 minus y2( )⎤⎦

= 12⎡⎣a c+ aminus aminus b( ) + b a+ bminus bminus c( ) + c b+ cminus cminus a( )⎤⎦

Part Answer2

= 12⎡⎣a cminus b( ) + b aminus c( ) + c bminus a( )⎤⎦

= 12⎡⎣acminus ab+ abminus bc+ bcminus ac⎤⎦

= 12times0 = 0

Since the area of the triangle is zerothere4 The points are collinear

Question 8 In fig ABCD is a rectangle in which segments AP and AQ are drawn as shown Findthe length of (AP + AQ)

Suggested MarksSuggested Answers

Question 7

Question 8

Part Answer1

In rt Δ ADQADAQ = sin 30deg rArr AD

AQ = 12

rArr AQ = 2AD rArr AQ = 2times30 = 60 cm hellip i( )

Part Answer2

In rt Δ ABPABAP = cos 60deg rArr AB

AP = 12

rArr AP =2AB rArr AP = 2times60 = 120 cm hellip ii( )there4 AP+AQ = 120+60 = 180 cm

Question 9 The horizontal distance between two trees of different heights is 60 m The angle ofdepression of the first tree when seen from the top of the second tree is 45deg If the height of the

second tree is 80 m find the height of the first treeSuggested Answers

Part Answer1

Let height of first tree AB = x mHeight of second treeCD = 80 M (Given)Difference BD = AF = 60 mIn rt Δ AFCCFAF = tan 45deg

Question 9

Part Answer2

rArr CF60 = 1 rArr CF = 60

Hence height of the first tree AB = CD - CF = 80 - 60 = 20 m

Question 10 Show that the points (aa) (-a -a) and minus 3radic a 3radic a( ) are the vertices of anequilateral triangle

Suggested MarksSuggested Answers

Part Answer1

Let A (aa) B (-a -a) and C minus 3radic a 3radic a( ) be the vertices of Δ ABC

AB = minusaminus a( )2 + minusaminus a( )2radic

= 4a2 + 4a2radic = 8a2radic = 2 2radic a

Part Answer2

BC = minus 3radic a+ a( )2 + 3radic a+ a( )2radic

= 3a2 + a2 minus 2 3radic a2 + 3a2 + a2 + 2 3radic a2radic

= 8a2radic = 2 2radic a

Part Answer3

AC = minus 3radic aminus a( )2 + 3radic aminus a( )2radic

= 3a2 + a2 + 2 3radic a2 + 3a2 + a2 minus 2 3radic a2radic

= 8a2radic = 2 2radic athere4 AB = BC = AC

Hence the triangle ABC is an equilateral triangle

Question 11 Find the coordinates of the points which divide the line-segment joining the points(57) and (8 10) in three equal parts

Suggested MarksSuggested Answers

Question 10

Question 11

Part Answer1

Let Points A = (5 7) and B = (8 10)Let Points C and D divide line AB in three equal parts

there4 ACCB = 12CDDB = 11 [ ∵ D is the mid point of CB]

Part Answer2

Point C = 1times8+2times51+2 1times10+2times7

1+2⎛⎝

⎞⎠

hellip[Using Section formula]= 8+10

3 10+143

⎛⎝

⎞⎠or 18

3 243

⎛⎝

⎞⎠or 68( )

Part Answer3

Point D = 6+82 8+10

2⎛⎝

⎞⎠or 14

2 182

⎛⎝

⎞⎠or (79)

Question 12 A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaffof height 5 m From a point on the plane the angles of elevation of the bottom and top of the

flagstaff are respectively 30deg and 60deg Find the height of the tower

Suggested MarksSuggested Answers

Question 12

Part Answer2

there4 CF = CD + DF = 10 3radic3 + 2

= 10 3radic +63 cm

Question 5 In Fig Find the coordinate of A

Suggested MarksSuggested Answers

Part Answer1

Centroid

G = x+3minus73 yminus5+4

3

2 minus 1( ) = xminus43 yminus1

3

Part Answer2

there4 xminus43 = 2 and yminus1

3 = minus 1xminus 4 = 6 and yminus 1 = minus 3x = 10 and y = minus 2there4 Point A(10 minus2)

Question 6 Find the coordinates of a point A where AB is the diameter of a circle whose centre is(23) and B is (14)

Suggested Answers

Part Answer1

Let the coordinates of A be (xy) and the given points be C(2 -3) and (xy) B (14)

∵ Centre is the mid point of the diameterthere4 Coordinates of C = x+1

2 y+42

Question 5

Question 6

Part Answer2

Since the coordinates of C are (2-3)

there4 x+12 = 2

y+4

2 = minus 3

x+ 1 = 4x = 4 minus1x = 3

y+ 4 = minus 6y = minus 6 minus 4y = minus 10

there4 Coordinates of A are (3 minus10)

Question 7 Show that points (a b +c) (b c + a) and (c a + b)are collinear

Suggested MarksSuggested Answers

Part Answer1

Let the given points be A (a b + c) = (x1 y1) B (b c + a)= (x2 y2) and C(c a+b) = (x3 y3)there4 Area of Δ ABC= 1

2⎡⎣x1 y2 minus y3( ) + x2 y3 minus y1( ) + x3 y1 minus y2( )⎤⎦

= 12⎡⎣a c+ aminus aminus b( ) + b a+ bminus bminus c( ) + c b+ cminus cminus a( )⎤⎦

Part Answer2

= 12⎡⎣a cminus b( ) + b aminus c( ) + c bminus a( )⎤⎦

= 12⎡⎣acminus ab+ abminus bc+ bcminus ac⎤⎦

= 12times0 = 0

Since the area of the triangle is zerothere4 The points are collinear

Question 8 In fig ABCD is a rectangle in which segments AP and AQ are drawn as shown Findthe length of (AP + AQ)

Suggested MarksSuggested Answers

Question 7

Question 8

Part Answer1

In rt Δ ADQADAQ = sin 30deg rArr AD

AQ = 12

rArr AQ = 2AD rArr AQ = 2times30 = 60 cm hellip i( )

Part Answer2

In rt Δ ABPABAP = cos 60deg rArr AB

AP = 12

rArr AP =2AB rArr AP = 2times60 = 120 cm hellip ii( )there4 AP+AQ = 120+60 = 180 cm

Question 9 The horizontal distance between two trees of different heights is 60 m The angle ofdepression of the first tree when seen from the top of the second tree is 45deg If the height of the

second tree is 80 m find the height of the first treeSuggested Answers

Part Answer1

Let height of first tree AB = x mHeight of second treeCD = 80 M (Given)Difference BD = AF = 60 mIn rt Δ AFCCFAF = tan 45deg

Question 9

Part Answer2

rArr CF60 = 1 rArr CF = 60

Hence height of the first tree AB = CD - CF = 80 - 60 = 20 m

Question 10 Show that the points (aa) (-a -a) and minus 3radic a 3radic a( ) are the vertices of anequilateral triangle

Suggested MarksSuggested Answers

Part Answer1

Let A (aa) B (-a -a) and C minus 3radic a 3radic a( ) be the vertices of Δ ABC

AB = minusaminus a( )2 + minusaminus a( )2radic

= 4a2 + 4a2radic = 8a2radic = 2 2radic a

Part Answer2

BC = minus 3radic a+ a( )2 + 3radic a+ a( )2radic

= 3a2 + a2 minus 2 3radic a2 + 3a2 + a2 + 2 3radic a2radic

= 8a2radic = 2 2radic a

Part Answer3

AC = minus 3radic aminus a( )2 + 3radic aminus a( )2radic

= 3a2 + a2 + 2 3radic a2 + 3a2 + a2 minus 2 3radic a2radic

= 8a2radic = 2 2radic athere4 AB = BC = AC

Hence the triangle ABC is an equilateral triangle

Question 11 Find the coordinates of the points which divide the line-segment joining the points(57) and (8 10) in three equal parts

Suggested MarksSuggested Answers

Question 10

Question 11

Part Answer1

Let Points A = (5 7) and B = (8 10)Let Points C and D divide line AB in three equal parts

there4 ACCB = 12CDDB = 11 [ ∵ D is the mid point of CB]

Part Answer2

Point C = 1times8+2times51+2 1times10+2times7

1+2⎛⎝

⎞⎠

hellip[Using Section formula]= 8+10

3 10+143

⎛⎝

⎞⎠or 18

3 243

⎛⎝

⎞⎠or 68( )

Part Answer3

Point D = 6+82 8+10

2⎛⎝

⎞⎠or 14

2 182

⎛⎝

⎞⎠or (79)

Question 12 A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaffof height 5 m From a point on the plane the angles of elevation of the bottom and top of the

flagstaff are respectively 30deg and 60deg Find the height of the tower

Suggested MarksSuggested Answers

Question 12

Part Answer2

Since the coordinates of C are (2-3)

there4 x+12 = 2

y+4

2 = minus 3

x+ 1 = 4x = 4 minus1x = 3

y+ 4 = minus 6y = minus 6 minus 4y = minus 10

there4 Coordinates of A are (3 minus10)

Question 7 Show that points (a b +c) (b c + a) and (c a + b)are collinear

Suggested MarksSuggested Answers

Part Answer1

Let the given points be A (a b + c) = (x1 y1) B (b c + a)= (x2 y2) and C(c a+b) = (x3 y3)there4 Area of Δ ABC= 1

2⎡⎣x1 y2 minus y3( ) + x2 y3 minus y1( ) + x3 y1 minus y2( )⎤⎦

= 12⎡⎣a c+ aminus aminus b( ) + b a+ bminus bminus c( ) + c b+ cminus cminus a( )⎤⎦

Part Answer2

= 12⎡⎣a cminus b( ) + b aminus c( ) + c bminus a( )⎤⎦

= 12⎡⎣acminus ab+ abminus bc+ bcminus ac⎤⎦

= 12times0 = 0

Since the area of the triangle is zerothere4 The points are collinear

Question 8 In fig ABCD is a rectangle in which segments AP and AQ are drawn as shown Findthe length of (AP + AQ)

Suggested MarksSuggested Answers

Question 7

Question 8

Part Answer1

In rt Δ ADQADAQ = sin 30deg rArr AD

AQ = 12

rArr AQ = 2AD rArr AQ = 2times30 = 60 cm hellip i( )

Part Answer2

In rt Δ ABPABAP = cos 60deg rArr AB

AP = 12

rArr AP =2AB rArr AP = 2times60 = 120 cm hellip ii( )there4 AP+AQ = 120+60 = 180 cm

Question 9 The horizontal distance between two trees of different heights is 60 m The angle ofdepression of the first tree when seen from the top of the second tree is 45deg If the height of the

second tree is 80 m find the height of the first treeSuggested Answers

Part Answer1

Let height of first tree AB = x mHeight of second treeCD = 80 M (Given)Difference BD = AF = 60 mIn rt Δ AFCCFAF = tan 45deg

Question 9

Part Answer2

rArr CF60 = 1 rArr CF = 60

Hence height of the first tree AB = CD - CF = 80 - 60 = 20 m

Question 10 Show that the points (aa) (-a -a) and minus 3radic a 3radic a( ) are the vertices of anequilateral triangle

Suggested MarksSuggested Answers

Part Answer1

Let A (aa) B (-a -a) and C minus 3radic a 3radic a( ) be the vertices of Δ ABC

AB = minusaminus a( )2 + minusaminus a( )2radic

= 4a2 + 4a2radic = 8a2radic = 2 2radic a

Part Answer2

BC = minus 3radic a+ a( )2 + 3radic a+ a( )2radic

= 3a2 + a2 minus 2 3radic a2 + 3a2 + a2 + 2 3radic a2radic

= 8a2radic = 2 2radic a

Part Answer3

AC = minus 3radic aminus a( )2 + 3radic aminus a( )2radic

= 3a2 + a2 + 2 3radic a2 + 3a2 + a2 minus 2 3radic a2radic

= 8a2radic = 2 2radic athere4 AB = BC = AC

Hence the triangle ABC is an equilateral triangle

Question 11 Find the coordinates of the points which divide the line-segment joining the points(57) and (8 10) in three equal parts

Suggested MarksSuggested Answers

Question 10

Question 11

Part Answer1

Let Points A = (5 7) and B = (8 10)Let Points C and D divide line AB in three equal parts

there4 ACCB = 12CDDB = 11 [ ∵ D is the mid point of CB]

Part Answer2

Point C = 1times8+2times51+2 1times10+2times7

1+2⎛⎝

⎞⎠

hellip[Using Section formula]= 8+10

3 10+143

⎛⎝

⎞⎠or 18

3 243

⎛⎝

⎞⎠or 68( )

Part Answer3

Point D = 6+82 8+10

2⎛⎝

⎞⎠or 14

2 182

⎛⎝

⎞⎠or (79)

Question 12 A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaffof height 5 m From a point on the plane the angles of elevation of the bottom and top of the

flagstaff are respectively 30deg and 60deg Find the height of the tower

Suggested MarksSuggested Answers

Question 12

Part Answer1

In rt Δ ADQADAQ = sin 30deg rArr AD

AQ = 12

rArr AQ = 2AD rArr AQ = 2times30 = 60 cm hellip i( )

Part Answer2

In rt Δ ABPABAP = cos 60deg rArr AB

AP = 12

rArr AP =2AB rArr AP = 2times60 = 120 cm hellip ii( )there4 AP+AQ = 120+60 = 180 cm

Question 9 The horizontal distance between two trees of different heights is 60 m The angle ofdepression of the first tree when seen from the top of the second tree is 45deg If the height of the

second tree is 80 m find the height of the first treeSuggested Answers

Part Answer1

Let height of first tree AB = x mHeight of second treeCD = 80 M (Given)Difference BD = AF = 60 mIn rt Δ AFCCFAF = tan 45deg

Question 9

Part Answer2

rArr CF60 = 1 rArr CF = 60

Hence height of the first tree AB = CD - CF = 80 - 60 = 20 m

Question 10 Show that the points (aa) (-a -a) and minus 3radic a 3radic a( ) are the vertices of anequilateral triangle

Suggested MarksSuggested Answers

Part Answer1

Let A (aa) B (-a -a) and C minus 3radic a 3radic a( ) be the vertices of Δ ABC

AB = minusaminus a( )2 + minusaminus a( )2radic

= 4a2 + 4a2radic = 8a2radic = 2 2radic a

Part Answer2

BC = minus 3radic a+ a( )2 + 3radic a+ a( )2radic

= 3a2 + a2 minus 2 3radic a2 + 3a2 + a2 + 2 3radic a2radic

= 8a2radic = 2 2radic a

Part Answer3

AC = minus 3radic aminus a( )2 + 3radic aminus a( )2radic

= 3a2 + a2 + 2 3radic a2 + 3a2 + a2 minus 2 3radic a2radic

= 8a2radic = 2 2radic athere4 AB = BC = AC

Hence the triangle ABC is an equilateral triangle

Question 11 Find the coordinates of the points which divide the line-segment joining the points(57) and (8 10) in three equal parts

Suggested MarksSuggested Answers

Question 10

Question 11

Part Answer1

Let Points A = (5 7) and B = (8 10)Let Points C and D divide line AB in three equal parts

there4 ACCB = 12CDDB = 11 [ ∵ D is the mid point of CB]

Part Answer2

Point C = 1times8+2times51+2 1times10+2times7

1+2⎛⎝

⎞⎠

hellip[Using Section formula]= 8+10

3 10+143

⎛⎝

⎞⎠or 18

3 243

⎛⎝

⎞⎠or 68( )

Part Answer3

Point D = 6+82 8+10

2⎛⎝

⎞⎠or 14

2 182

⎛⎝

⎞⎠or (79)

Question 12 A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaffof height 5 m From a point on the plane the angles of elevation of the bottom and top of the

flagstaff are respectively 30deg and 60deg Find the height of the tower

Suggested MarksSuggested Answers

Question 12

Part Answer2

rArr CF60 = 1 rArr CF = 60

Hence height of the first tree AB = CD - CF = 80 - 60 = 20 m

Question 10 Show that the points (aa) (-a -a) and minus 3radic a 3radic a( ) are the vertices of anequilateral triangle

Suggested MarksSuggested Answers

Part Answer1

Let A (aa) B (-a -a) and C minus 3radic a 3radic a( ) be the vertices of Δ ABC

AB = minusaminus a( )2 + minusaminus a( )2radic

= 4a2 + 4a2radic = 8a2radic = 2 2radic a

Part Answer2

BC = minus 3radic a+ a( )2 + 3radic a+ a( )2radic

= 3a2 + a2 minus 2 3radic a2 + 3a2 + a2 + 2 3radic a2radic

= 8a2radic = 2 2radic a

Part Answer3

AC = minus 3radic aminus a( )2 + 3radic aminus a( )2radic

= 3a2 + a2 + 2 3radic a2 + 3a2 + a2 minus 2 3radic a2radic

= 8a2radic = 2 2radic athere4 AB = BC = AC

Hence the triangle ABC is an equilateral triangle

Question 11 Find the coordinates of the points which divide the line-segment joining the points(57) and (8 10) in three equal parts

Suggested MarksSuggested Answers

Question 10

Question 11

Part Answer1

Let Points A = (5 7) and B = (8 10)Let Points C and D divide line AB in three equal parts

there4 ACCB = 12CDDB = 11 [ ∵ D is the mid point of CB]

Part Answer2

Point C = 1times8+2times51+2 1times10+2times7

1+2⎛⎝

⎞⎠

hellip[Using Section formula]= 8+10

3 10+143

⎛⎝

⎞⎠or 18

3 243

⎛⎝

⎞⎠or 68( )

Part Answer3

Point D = 6+82 8+10

2⎛⎝

⎞⎠or 14

2 182

⎛⎝

⎞⎠or (79)

Question 12 A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaffof height 5 m From a point on the plane the angles of elevation of the bottom and top of the

flagstaff are respectively 30deg and 60deg Find the height of the tower

Suggested MarksSuggested Answers

Question 12

Part Answer1

Let Points A = (5 7) and B = (8 10)Let Points C and D divide line AB in three equal parts

there4 ACCB = 12CDDB = 11 [ ∵ D is the mid point of CB]

Part Answer2

Point C = 1times8+2times51+2 1times10+2times7

1+2⎛⎝

⎞⎠

hellip[Using Section formula]= 8+10

3 10+143

⎛⎝

⎞⎠or 18

3 243

⎛⎝

⎞⎠or 68( )

Part Answer3

Point D = 6+82 8+10

2⎛⎝

⎞⎠or 14

2 182

⎛⎝

⎞⎠or (79)

Question 12 A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaffof height 5 m From a point on the plane the angles of elevation of the bottom and top of the

flagstaff are respectively 30deg and 60deg Find the height of the tower

Suggested MarksSuggested Answers

Question 12

Part Answer1

Let BC be the tower and CD be the flag staff = 5 mLet AB = x mBC = y mIn rt Δ ABC tan 30deg = BC

AB

rArr 13radic = y

x

rArr x = 3radic y hellip i( )

Part Answer2

In rt Δ ABD tan 60deg = BDAB

rArr 3radic = y+5x rArr y+5 = 3radic x

Part Answer3

rArr y+ 5 = 3radic 3radic y( ) hellip[From i( )]rArr y+ 5 = 3y rArr 2y = 5there4 The height of the lower y = 5

2 = 25 m

Question 13 Three vertices of a parallelogram taken in order are (-1 0) (3 1) and (2 2)respectively

Find the coordinates of fourth vertex

Suggested MarksSuggested Answers

Question 13

Part Answer1

Let A (-1 0) B(3 1) C (2 2) and D (x y) be the vertices of parallelogram ABCD taken in order

Since the diagonals of a parallelogram bisect each otherthere4 Coordinates of the midminuspoint of AC

= Coordinates of the mid-point of BD

Part Answer2

rArr minus1+22 0+2

2⎛⎝

⎞⎠ = 3+x

2 1+y2

12 1⎛⎝

⎞⎠ = 3+x

2 y+12

rArr 3+x2 = 1

2 and y+12 = 1

Part Answer3

rArr 3 + x = 1 and y+ 1 = 2rArr x = 1 minus 3 and y = 2 minus 1rArr x = minus 2 and y = 1

Hence coordinates of the fourth vertex = (-2 1)

Question 14 Draw a quadrilateral in the cartesian plane whose vertices are (-4 5) (07) (5-5) and(-4 -2) Also find its area

Suggested MarksSuggested Answers

Question 14

Part Answer1

Let the given vertices of quadrilateral be A(-4 5) B (07) C(5 -5) and (-4-2) Join ACthere4 Area of the quadrilateral ABCD

= Area of Δ ABC + Area of Δ ACD

Part Answer2

Area of Δ ABC = 12⎡⎣x1 y2 minus y3( ) + x2 y3 minus y1( ) + x3 y1 minus y2( )⎤⎦

= 12⎡⎣minus 4 7 + 5( ) + 0 minus5 minus 5( ) + 5 5 minus 7( )⎤⎦ = 1

2⎡⎣minus 48 + 0 minus 10⎤⎦

= 12times minus 58 = minus 29

∵ Area is a measure which cannot be negativethere4 Area of Δ ABC = 29 sq units

Part Answer3

Now Area of Δ ACD

= 12⎡⎣minus 4 minus5 + 2( ) + 5 minus2 minus 5( ) + minus4( ) 5 + 5( )⎤⎦

= 12⎡⎣12 minus 35 minus 40⎤⎦

= 12⎡⎣12 minus 75⎤⎦ = 1

2times minus 63 = minus632 = minus 315

Part Answer4

∵ Area is a measure which cannot be negativethere4 Area of Δ ACD = 315 sq unitsthere4 Area of quad ABCD = 29+315 = 605 sq units

Question 15 Two pillars of equal height stand on either side of a roadway which is 150 m wideFrom a point on the roadway between the pillars the elevations of the top of the pillars are 60deg and

30deg Find the height of the pillars and the position of the pointSuggested Answers

Part Answer1

Let AB and DE be two equal pillars and C be the point on BD (roadway)Let BC = x mThen CD = (150-x) mLet AB = DE = y m

Part Answer2

In rt Δ ABC tan 60deg = ABBC

rArr 3radic1 = y

x rArr y = 3radic x hellip i( )

Part Answer3

In rt Δ CDE tan 30deg = DECD

rArr 13radic = y

150minusxrArr 3radic y = 150 minus xrArr 3radic 3radic x( ) = 150 minus x hellip[From i( )]

Question 15

Part Answer4

rArr 3x = 150 minus x rArr 4x = 150rArr x = 150

4 = 375m = BCCD = 150 - 375 = 1125 mthere4 Pt C is 375 m from 1st end and 1125 m from 2nd end

Part Answer5

From i( ) y = 3radic x rArr y = 173( ) 375( )rArr y = 64875 mthere4 Height of the pillars = 6488 m

Part Answer1

Let A (-1 0) B(3 1) C (2 2) and D (x y) be the vertices of parallelogram ABCD taken in order

Since the diagonals of a parallelogram bisect each otherthere4 Coordinates of the midminuspoint of AC

= Coordinates of the mid-point of BD

Part Answer2

rArr minus1+22 0+2

2⎛⎝

⎞⎠ = 3+x

2 1+y2

12 1⎛⎝

⎞⎠ = 3+x

2 y+12

rArr 3+x2 = 1

2 and y+12 = 1

Part Answer3

rArr 3 + x = 1 and y+ 1 = 2rArr x = 1 minus 3 and y = 2 minus 1rArr x = minus 2 and y = 1

Hence coordinates of the fourth vertex = (-2 1)

Question 14 Draw a quadrilateral in the cartesian plane whose vertices are (-4 5) (07) (5-5) and(-4 -2) Also find its area

Suggested MarksSuggested Answers

Question 14

Part Answer1

Let the given vertices of quadrilateral be A(-4 5) B (07) C(5 -5) and (-4-2) Join ACthere4 Area of the quadrilateral ABCD

= Area of Δ ABC + Area of Δ ACD

Part Answer2

Area of Δ ABC = 12⎡⎣x1 y2 minus y3( ) + x2 y3 minus y1( ) + x3 y1 minus y2( )⎤⎦

= 12⎡⎣minus 4 7 + 5( ) + 0 minus5 minus 5( ) + 5 5 minus 7( )⎤⎦ = 1

2⎡⎣minus 48 + 0 minus 10⎤⎦

= 12times minus 58 = minus 29

∵ Area is a measure which cannot be negativethere4 Area of Δ ABC = 29 sq units

Part Answer3

Now Area of Δ ACD

= 12⎡⎣minus 4 minus5 + 2( ) + 5 minus2 minus 5( ) + minus4( ) 5 + 5( )⎤⎦

= 12⎡⎣12 minus 35 minus 40⎤⎦

= 12⎡⎣12 minus 75⎤⎦ = 1

2times minus 63 = minus632 = minus 315

Part Answer4

∵ Area is a measure which cannot be negativethere4 Area of Δ ACD = 315 sq unitsthere4 Area of quad ABCD = 29+315 = 605 sq units

Question 15 Two pillars of equal height stand on either side of a roadway which is 150 m wideFrom a point on the roadway between the pillars the elevations of the top of the pillars are 60deg and

30deg Find the height of the pillars and the position of the pointSuggested Answers

Part Answer1

Let AB and DE be two equal pillars and C be the point on BD (roadway)Let BC = x mThen CD = (150-x) mLet AB = DE = y m

Part Answer2

In rt Δ ABC tan 60deg = ABBC

rArr 3radic1 = y

x rArr y = 3radic x hellip i( )

Part Answer3

In rt Δ CDE tan 30deg = DECD

rArr 13radic = y

150minusxrArr 3radic y = 150 minus xrArr 3radic 3radic x( ) = 150 minus x hellip[From i( )]

Question 15

Part Answer4

rArr 3x = 150 minus x rArr 4x = 150rArr x = 150

4 = 375m = BCCD = 150 - 375 = 1125 mthere4 Pt C is 375 m from 1st end and 1125 m from 2nd end

Part Answer5

From i( ) y = 3radic x rArr y = 173( ) 375( )rArr y = 64875 mthere4 Height of the pillars = 6488 m

Part Answer1

Let the given vertices of quadrilateral be A(-4 5) B (07) C(5 -5) and (-4-2) Join ACthere4 Area of the quadrilateral ABCD

= Area of Δ ABC + Area of Δ ACD

Part Answer2

Area of Δ ABC = 12⎡⎣x1 y2 minus y3( ) + x2 y3 minus y1( ) + x3 y1 minus y2( )⎤⎦

= 12⎡⎣minus 4 7 + 5( ) + 0 minus5 minus 5( ) + 5 5 minus 7( )⎤⎦ = 1

2⎡⎣minus 48 + 0 minus 10⎤⎦

= 12times minus 58 = minus 29

∵ Area is a measure which cannot be negativethere4 Area of Δ ABC = 29 sq units

Part Answer3

Now Area of Δ ACD

= 12⎡⎣minus 4 minus5 + 2( ) + 5 minus2 minus 5( ) + minus4( ) 5 + 5( )⎤⎦

= 12⎡⎣12 minus 35 minus 40⎤⎦

= 12⎡⎣12 minus 75⎤⎦ = 1

2times minus 63 = minus632 = minus 315

Part Answer4

∵ Area is a measure which cannot be negativethere4 Area of Δ ACD = 315 sq unitsthere4 Area of quad ABCD = 29+315 = 605 sq units

Question 15 Two pillars of equal height stand on either side of a roadway which is 150 m wideFrom a point on the roadway between the pillars the elevations of the top of the pillars are 60deg and

30deg Find the height of the pillars and the position of the pointSuggested Answers

Part Answer1

Let AB and DE be two equal pillars and C be the point on BD (roadway)Let BC = x mThen CD = (150-x) mLet AB = DE = y m

Part Answer2

In rt Δ ABC tan 60deg = ABBC

rArr 3radic1 = y

x rArr y = 3radic x hellip i( )

Part Answer3

In rt Δ CDE tan 30deg = DECD

rArr 13radic = y

150minusxrArr 3radic y = 150 minus xrArr 3radic 3radic x( ) = 150 minus x hellip[From i( )]

Question 15

Part Answer4

rArr 3x = 150 minus x rArr 4x = 150rArr x = 150

4 = 375m = BCCD = 150 - 375 = 1125 mthere4 Pt C is 375 m from 1st end and 1125 m from 2nd end

Part Answer5

From i( ) y = 3radic x rArr y = 173( ) 375( )rArr y = 64875 mthere4 Height of the pillars = 6488 m

Part Answer4

∵ Area is a measure which cannot be negativethere4 Area of Δ ACD = 315 sq unitsthere4 Area of quad ABCD = 29+315 = 605 sq units

Question 15 Two pillars of equal height stand on either side of a roadway which is 150 m wideFrom a point on the roadway between the pillars the elevations of the top of the pillars are 60deg and

30deg Find the height of the pillars and the position of the pointSuggested Answers

Part Answer1

Let AB and DE be two equal pillars and C be the point on BD (roadway)Let BC = x mThen CD = (150-x) mLet AB = DE = y m

Part Answer2

In rt Δ ABC tan 60deg = ABBC

rArr 3radic1 = y

x rArr y = 3radic x hellip i( )

Part Answer3

In rt Δ CDE tan 30deg = DECD

rArr 13radic = y

150minusxrArr 3radic y = 150 minus xrArr 3radic 3radic x( ) = 150 minus x hellip[From i( )]

Question 15

Part Answer4

rArr 3x = 150 minus x rArr 4x = 150rArr x = 150

4 = 375m = BCCD = 150 - 375 = 1125 mthere4 Pt C is 375 m from 1st end and 1125 m from 2nd end

Part Answer5

From i( ) y = 3radic x rArr y = 173( ) 375( )rArr y = 64875 mthere4 Height of the pillars = 6488 m

Part Answer4

rArr 3x = 150 minus x rArr 4x = 150rArr x = 150

4 = 375m = BCCD = 150 - 375 = 1125 mthere4 Pt C is 375 m from 1st end and 1125 m from 2nd end

Part Answer5

From i( ) y = 3radic x rArr y = 173( ) 375( )rArr y = 64875 mthere4 Height of the pillars = 6488 m