RS Aggarwal 2020 2021 Solutions for Class 7 Maths Chapter 15 Properties Of Triangles are provided here with simple step-by-step explanations. These solutions for Properties Of Triangles are extremely popular among class 7 students for Maths Properties Of Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RS Aggarwal 2020 2021 Book of class 7 Maths Chapter 15 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RS Aggarwal 2020 2021 Solutions. All RS Aggarwal 2020 2021 Solutions for class 7 Maths are prepared by experts and are 100% accurate.

Page No 183:

Question 1:

Answer:

Sum of the angles of a triangle is 180°.

A+B+C = 180°72° + 63° +C = 180°C = 45° 

Hence, ∠C measures 45°.

Page No 183:

Question 2:

Sum of the angles of a triangle is 180°.

A+B+C = 180°72° + 63° +C = 180°C = 45° 

Hence, ∠C measures 45°.

Answer:

Sum of the angles of any triangle is 180°.
In ∆DEF:
D+E+F=180°D+105°+40°=180°or D=180°-(105°+40°)or D=35°

Page No 183:

Question 3:

Sum of the angles of any triangle is 180°.
In ∆DEF:
D+E+F=180°D+105°+40°=180°or D=180°-(105°+40°)or D=35°

Answer:

Sum of the angles of any triangle is 180°.
In ∆XYZ:

X+Y+Z=180°90°+Y+48°=180°=>Y=180°-138°=42° 

Page No 183:

Question 4:

Sum of the angles of any triangle is 180°.
In ∆XYZ:

X+Y+Z=180°90°+Y+48°=180°=>Y=180°-138°=42° 

Answer:

Suppose the angles of the triangle are (4x)o, (3x)o and (2x)o.

Sum of the angles of any triangle is 180o.

∴ 4x + 3x + 2x = 180
9x = 180
x = 20
Therefore, the angles of the triangle are (4×20)°, (3×20)° and ( 2×20)°, i.e . 80°, 60° and 40°.

Page No 183:

Question 5:

Suppose the angles of the triangle are (4x)o, (3x)o and (2x)o.

Sum of the angles of any triangle is 180o.

∴ 4x + 3x + 2x = 180
9x = 180
x = 20
Therefore, the angles of the triangle are (4×20)°, (3×20)° and ( 2×20)°, i.e . 80°, 60° and 40°.

Answer:

Sum of the angles of a triangle is 180°.
Suppose the other angle measures x.

It is a right angle triangle. Hence, one of the angle is 90°.

 36° + 90° +x = 180°x= 54°
Hence, the other angle measures 54°.

Page No 183:

Question 6:

Sum of the angles of a triangle is 180°.
Suppose the other angle measures x.

It is a right angle triangle. Hence, one of the angle is 90°.

 36° + 90° +x = 180°x= 54°
Hence, the other angle measures 54°.

Answer:

Suppose the acute angles are (2x)° and (x)°
Sum of the angles of any triangle is 180°

∴ 2x+x+ 90 = 180
(3x) = 180-90
(3x) = 90
x = 30
So, the angles measure (2×30)° and 30°i.e. 60° and 30°

Page No 183:

Question 7:

Suppose the acute angles are (2x)° and (x)°
Sum of the angles of any triangle is 180°

∴ 2x+x+ 90 = 180
(3x) = 180-90
(3x) = 90
x = 30
So, the angles measure (2×30)° and 30°i.e. 60° and 30°

Answer:

The other two angles are equal. Let one of these angles be x°.

Sum of angles of any triangle is 180°.

x + x+ 100 = 180
2x = 80
x = 40

Hence, the equal angles of the triangle are 40° each.



Page No 184:

Question 8:

The other two angles are equal. Let one of these angles be x°.

Sum of angles of any triangle is 180°.

x + x+ 100 = 180
2x = 80
x = 40

Hence, the equal angles of the triangle are 40° each.

Answer:

Suppose the third angle of the isosceles triangle is xo.
Then, the two equal angles are (2x)o and (2x)o.
Sum of the angles of any triangle is 180o.

∴ 2x +2x+ x= 180
5x  = 180
x = 36

Hence, the angles of the triangle are 36°, (2×36)° and (2×36)°, i.e. 36°, 72°and 72°.

Page No 184:

Question 9:

Suppose the third angle of the isosceles triangle is xo.
Then, the two equal angles are (2x)o and (2x)o.
Sum of the angles of any triangle is 180o.

∴ 2x +2x+ x= 180
5x  = 180
x = 36

Hence, the angles of the triangle are 36°, (2×36)° and (2×36)°, i.e. 36°, 72°and 72°.

Answer:


Suppose the angles are A, B and C.Given: A = B +CAlso, A +B+C = 180° A+A= 180°2A = 180° A=90°                  (Sum of the angles of a triangle is 180°)

Hence, the triangle ABC is right angled at A.

Page No 184:

Question 10:


Suppose the angles are A, B and C.Given: A = B +CAlso, A +B+C = 180° A+A= 180°2A = 180° A=90°                  (Sum of the angles of a triangle is 180°)

Hence, the triangle ABC is right angled at A.

Answer:

Suppose: 2A = 3B = 6C = x°
Then, Ax2°
B =x3°and C =x6°


Sum of the angles of any triangle is 180°.
A +B +C = 180°
x2+x3+x6= 180°3x+2x+x6=180°6x6=180°x=180


  A=1802°=90°
B =1803°=60°C =1806°=30°

Page No 184:

Question 11:

Suppose: 2A = 3B = 6C = x°
Then, Ax2°
B =x3°and C =x6°


Sum of the angles of any triangle is 180°.
A +B +C = 180°
x2+x3+x6= 180°3x+2x+x6=180°6x6=180°x=180


  A=1802°=90°
B =1803°=60°C =1806°=30°

Answer:

We know that the angles of an equilateral triangle are equal.
Let the measure of each angle of an equilateral triangle be x°.

x + x + x = 180
x = 60
Hence, the measure of each angle of an equilateral triangle is 60°.

Page No 184:

Question 12:

We know that the angles of an equilateral triangle are equal.
Let the measure of each angle of an equilateral triangle be x°.

x + x + x = 180
x = 60
Hence, the measure of each angle of an equilateral triangle is 60°.

Answer:

(i)
DEBC ABC =ADE = 55°                   
(Corresponding angles)

(ii) Sum of the angles of any triangle is 180°.

A+B+C = 180°C = 180° -(65° +55°) = 60°

 DE || BC
 AED =ACB =60°                (corresponding angles)

(iii)  We have found in point (ii) that C is equal to 60°.

Page No 184:

Question 13:

(i)
DEBC ABC =ADE = 55°                   
(Corresponding angles)

(ii) Sum of the angles of any triangle is 180°.

A+B+C = 180°C = 180° -(65° +55°) = 60°

 DE || BC
 AED =ACB =60°                (corresponding angles)

(iii)  We have found in point (ii) that C is equal to 60°.

Answer:

(i) No. This is because the sum of all the angles is 180°.
(ii) No. This is because a triangle can only have one obtuse angle.
(iii) Yes
(iv) No. This is because the sum of the angles cannot be more than 180°.
(v) No. This is because one angle has to be more than 60° as the sum of all angles is always 180°.
(vi) Yes, it will be an equilateral triangle.

Page No 184:

Question 14:

(i) No. This is because the sum of all the angles is 180°.
(ii) No. This is because a triangle can only have one obtuse angle.
(iii) Yes
(iv) No. This is because the sum of the angles cannot be more than 180°.
(v) No. This is because one angle has to be more than 60° as the sum of all angles is always 180°.
(vi) Yes, it will be an equilateral triangle.

Answer:

(i) Yes, it will be an isosceles right triangle.

(ii) Yes, a right triangle can have all sides of different measures. For example, 3, 4 and 5 are the sides of a scalene right triangle.

(iii) No, it cannot be an equilateral triangle since the hypotenuse square will be the sum of the square of the other two sides.

(iii) Yes, if an obtuse triangle has an obtuse angle of 120° and the other two angles of 30° each, then it will be an isosceles triangle.

Page No 184:

Question 15:

(i) Yes, it will be an isosceles right triangle.

(ii) Yes, a right triangle can have all sides of different measures. For example, 3, 4 and 5 are the sides of a scalene right triangle.

(iii) No, it cannot be an equilateral triangle since the hypotenuse square will be the sum of the square of the other two sides.

(iii) Yes, if an obtuse triangle has an obtuse angle of 120° and the other two angles of 30° each, then it will be an isosceles triangle.

Answer:

(i) obtuse (since the sum of the other two angles of the right triangle is 90o)
(ii) equal to the sum of 90o
(iii) 45o (since their sum is equal to 90o)
(iv) 60o
(v) a hypotenuse
(vi) perimeter



Page No 186:

Question 1:

(i) obtuse (since the sum of the other two angles of the right triangle is 90o)
(ii) equal to the sum of 90o
(iii) 45o (since their sum is equal to 90o)
(iv) 60o
(v) a hypotenuse
(vi) perimeter

Answer:

We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

ACD= CAB+CBAACD=75°+45°=120°

Page No 186:

Question 2:

We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

ACD= CAB+CBAACD=75°+45°=120°

Answer:

We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

 BAC +ABC =ACDx +68 = 130x = 62
Sum of the angles in any triangle is 180o.

 BAC +ABC+ACB =180°62 + 68 + y = 180y = 50

Page No 186:

Question 3:

We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

 BAC +ABC =ACDx +68 = 130x = 62
Sum of the angles in any triangle is 180o.

 BAC +ABC+ACB =180°62 + 68 + y = 180y = 50

Answer:

We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

 BAC + CBA = ACD32 + x = 65x = 33

Also, sum of the angles in any triangle is 180°.

 BAC+CBA +ACB= 180°32 +33+ y = 180y = 115

x= 33
    y =115

Page No 186:

Question 4:

We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

 BAC + CBA = ACD32 + x = 65x = 33

Also, sum of the angles in any triangle is 180°.

 BAC+CBA +ACB= 180°32 +33+ y = 180y = 115

x= 33
    y =115

Answer:

Suppose the interior opposite angles are (2x)° and (3x)°.
We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

∴ 3x +2x= 110
x = 22

The interior opposite angles are (2×22)° and (3×22)°, i.e. 44° and 66°.
Suppose the third angle of the triangle is y°.
Now, sum of the angles in any triangle is 180°.

∴ 44 + 66 + y = 180
y = 70

Hence, the angles of the triangle are 44°, 66° and 70°.

Page No 186:

Question 5:

Suppose the interior opposite angles are (2x)° and (3x)°.
We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

∴ 3x +2x= 110
x = 22

The interior opposite angles are (2×22)° and (3×22)°, i.e. 44° and 66°.
Suppose the third angle of the triangle is y°.
Now, sum of the angles in any triangle is 180°.

∴ 44 + 66 + y = 180
y = 70

Hence, the angles of the triangle are 44°, 66° and 70°.

Answer:

Suppose the interior opposite angles of an exterior angle 100oare xo and xo.
We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

x + x = 100
2x= 100
x= 50

Also, sum of the angles of any triangle is  180°.
Let the measure of the third angle be y°.

x + x + y = 180
50  + 50 + y= 180
y  = 80

Hence, the angles are of the measures 50°, 50° and 80°.

Page No 186:

Question 6:

Suppose the interior opposite angles of an exterior angle 100oare xo and xo.
We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

x + x = 100
2x= 100
x= 50

Also, sum of the angles of any triangle is  180°.
Let the measure of the third angle be y°.

x + x + y = 180
50  + 50 + y= 180
y  = 80

Hence, the angles are of the measures 50°, 50° and 80°.

Answer:

We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

In ABC:
ACD=BAC+ABC=25°+45°ACD=70°(ii) In ECD:AED=ECD+EDC=70°+40°=>AED=110°

Page No 186:

Question 7:

We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

In ABC:
ACD=BAC+ABC=25°+45°ACD=70°(ii) In ECD:AED=ECD+EDC=70°+40°=>AED=110°

Answer:

Sum of the angles of a triangle is 180°.

InABC:BAC+CBA+ACB=180°BAC=180°-(40°+100°)=>BAC=40° 


We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

 ACD=BAC+CBA=40°+40°=80°(i) ACD=80°(ii) In  ACD:CAD+ACD+ADC=180°=>ADC=180°-(50°+80°)=>ADC=50° ADC= 50°(iii) DAB+DAE=180°        (since BE is a straight line) DAE=180°-(DAC+CAB)DAE=180°-(50°+40°)DAE=90°

Page No 186:

Question 8:

Sum of the angles of a triangle is 180°.

InABC:BAC+CBA+ACB=180°BAC=180°-(40°+100°)=>BAC=40° 


We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.

 ACD=BAC+CBA=40°+40°=80°(i) ACD=80°(ii) In  ACD:CAD+ACD+ADC=180°=>ADC=180°-(50°+80°)=>ADC=50° ADC= 50°(iii) DAB+DAE=180°        (since BE is a straight line) DAE=180°-(DAC+CAB)DAE=180°-(50°+40°)DAE=90°

Answer:

xy=233x = 2yx=23y

We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.


A + B =ACD

x°+ y° = 130°

2y3+y =1305y =130×3 5y= 390y = 78x =23×78x=52

Also, sum of the angles in any triangle is 180°
∴  x+ y + z = 180
z= 180- 78 - 52
z= 50



Page No 188:

Question 1:

xy=233x = 2yx=23y

We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.


A + B =ACD

x°+ y° = 130°

2y3+y =1305y =130×3 5y= 390y = 78x =23×78x=52

Also, sum of the angles in any triangle is 180°
∴  x+ y + z = 180
z= 180- 78 - 52
z= 50

Answer:

(i) Consider numbers 1, 1 and 1.
Clearly, 1 + 1 >1
             1 + 1 >1
             1 + 1 >1

Thus, the sum of any two sides is greater than the third side.
Hence, it is possible to draw a triangle having  sides 1 cm, 1 cm and 1 cm.

(ii)
Clearly, 2 + 3 >4
             3 + 4 >2
             2+ 4 >3
Thus, the sum of any two sides is greater than the third side.
Hence, it is possible to a draw triangle having sides 2 cm, 3 cm and 4 cm.

(iii)
Clearly, 7 + 8 = 15

Thus, the sum of these two numbers is not greater than the third number.
Hence, it is not possible to draw a triangle having sides 7 cm, 8 cm and 15 cm.

(iv) Consider the numbers 3.4, 2.1 and 5.3.

Clearly: 3.4 + 2.1 >5.3
             5.3 + 2.1 > 3.4
             5.3 + 3.4 > 2.1

Thus, the sum of any two sides is greater than the third side.
Hence, it is possible to draw a triangle having sides 3.4 cm, 2.1 cm and 5.3 cm.

(v) Consider the numbers 6, 7 and 14.
Clearly, 6+7 14

Thus, the sum of these two numbers is not greater than the third number.
Hence, it is not possible to draw a triangle having sides 6 cm, 7 cm and 14 cm.

Page No 188:

Question 2:

(i) Consider numbers 1, 1 and 1.
Clearly, 1 + 1 >1
             1 + 1 >1
             1 + 1 >1

Thus, the sum of any two sides is greater than the third side.
Hence, it is possible to draw a triangle having  sides 1 cm, 1 cm and 1 cm.

(ii)
Clearly, 2 + 3 >4
             3 + 4 >2
             2+ 4 >3
Thus, the sum of any two sides is greater than the third side.
Hence, it is possible to a draw triangle having sides 2 cm, 3 cm and 4 cm.

(iii)
Clearly, 7 + 8 = 15

Thus, the sum of these two numbers is not greater than the third number.
Hence, it is not possible to draw a triangle having sides 7 cm, 8 cm and 15 cm.

(iv) Consider the numbers 3.4, 2.1 and 5.3.

Clearly: 3.4 + 2.1 >5.3
             5.3 + 2.1 > 3.4
             5.3 + 3.4 > 2.1

Thus, the sum of any two sides is greater than the third side.
Hence, it is possible to draw a triangle having sides 3.4 cm, 2.1 cm and 5.3 cm.

(v) Consider the numbers 6, 7 and 14.
Clearly, 6+7 14

Thus, the sum of these two numbers is not greater than the third number.
Hence, it is not possible to draw a triangle having sides 6 cm, 7 cm and 14 cm.

Answer:

Let the length of the third side be x cm.

Sum of any two sides of a triangle is greater than the third side.

∴ 5 + 9 >x
x<14

Hence, the length of the third side must be less than 14 cm.

Page No 188:

Question 3:

Let the length of the third side be x cm.

Sum of any two sides of a triangle is greater than the third side.

∴ 5 + 9 >x
x<14

Hence, the length of the third side must be less than 14 cm.

Answer:

(i) >
(ii) >
(iii) <

The reason for the above three is that the sum of any two sides of a triangle is greater than the third side.

Page No 188:

Question 4:

(i) >
(ii) >
(iii) <

The reason for the above three is that the sum of any two sides of a triangle is greater than the third side.

Answer:

Sum of any two sides of a triangle is greater than the third side.

In AMB:
AB + BM >AM........(i)

In AMC:
AC + CM >AM.........(ii)

Adding the above two equation:
AB + BM + AC + CM >AM + AM
AB + BC + AC > 2AM

Hence, proved.



Page No 189:

Question 5:

Sum of any two sides of a triangle is greater than the third side.

In AMB:
AB + BM >AM........(i)

In AMC:
AC + CM >AM.........(ii)

Adding the above two equation:
AB + BM + AC + CM >AM + AM
AB + BC + AC > 2AM

Hence, proved.

Answer:

Sum of any two sides of a triangle is greater than the third side.

 In APB:AB + BP >APIn APC: AC + PC >APAdding the correspondong sides:AB + BP + AC + PC> AP + APAB + AC+ BC >2AP

Hence, proved.

Page No 189:

Question 6:

Sum of any two sides of a triangle is greater than the third side.

 In APB:AB + BP >APIn APC: AC + PC >APAdding the correspondong sides:AB + BP + AC + PC> AP + APAB + AC+ BC >2AP

Hence, proved.

Answer:

Sum of any two sides of a triangle is greater than the third side.

In ABC:
AB + BC > AC

In ADC:
CD + DA > AC

Adding the above two:

AB + BC + CD + DA  > 2 AC               ... (i)

In ADB:
AD + AB > BD

In BDC:
CD + BC > BD

Adding the above two:

AB + BC + CD + DA  >2 BD             ... (ii)

Adding equation (i) and (ii):

AB + BC + CD + DA+AB + BC + CD + DA> 2(AC+BD)
=> 2(AB + BC + CD + DA)>2(AC+BD)
=> AB + BC + CD + DA > AC+BD

Page No 189:

Question 7:

Sum of any two sides of a triangle is greater than the third side.

In ABC:
AB + BC > AC

In ADC:
CD + DA > AC

Adding the above two:

AB + BC + CD + DA  > 2 AC               ... (i)

In ADB:
AD + AB > BD

In BDC:
CD + BC > BD

Adding the above two:

AB + BC + CD + DA  >2 BD             ... (ii)

Adding equation (i) and (ii):

AB + BC + CD + DA+AB + BC + CD + DA> 2(AC+BD)
=> 2(AB + BC + CD + DA)>2(AC+BD)
=> AB + BC + CD + DA > AC+BD

Answer:

We know that the sum of any two sides of a triangle is greater than the third side.

In AOB:
OA + OB > AB...........(1)

In BOC:
OB + OC > BC........................... (2)

In AOC:
OA + OC > CA.............................(3)

Adding (1), (2) and (3):

OA + OB + OB + OC  + OA + OC >  AB + BC + CA

2( OA + OB  + OC) >  AB +BC + CA

Hence, proved.



Page No 193:

Question 1:

We know that the sum of any two sides of a triangle is greater than the third side.

In AOB:
OA + OB > AB...........(1)

In BOC:
OB + OC > BC........................... (2)

In AOC:
OA + OC > CA.............................(3)

Adding (1), (2) and (3):

OA + OB + OB + OC  + OA + OC >  AB + BC + CA

2( OA + OB  + OC) >  AB +BC + CA

Hence, proved.

Answer:

Suppose the length of the hypotenuse is a cm.

Then, by Pythagoras theorem:

a2 = 92 + 122
=> a2 = 81 + 144
=> a2 =225
=>  a = 225
=> a= 15

Hence, the length of the hypotenuse is 15 cm.

Page No 193:

Question 2:

Suppose the length of the hypotenuse is a cm.

Then, by Pythagoras theorem:

a2 = 92 + 122
=> a2 = 81 + 144
=> a2 =225
=>  a = 225
=> a= 15

Hence, the length of the hypotenuse is 15 cm.

Answer:

 Suppose the length of the other side is a cm.

Then, by Pythagoras theorem:

262 = 102 + a2
a2  = 676 - 100 a2 =576 a =576 a=24

Hence, the length of the other side is 24 cm.

Page No 193:

Question 3:

 Suppose the length of the other side is a cm.

Then, by Pythagoras theorem:

262 = 102 + a2
a2  = 676 - 100 a2 =576 a =576 a=24

Hence, the length of the other side is 24 cm.

Answer:

Suppose the length of the other side is a cm.

Then, by Pythagoras theorem:

4.52 + a2 = 7.52a2  = 56.25 - 20.25 a2  = 36a  = 36a  =6

Hence, the length of the other side of the triangle is  6 cm.

Page No 193:

Question 4:

Suppose the length of the other side is a cm.

Then, by Pythagoras theorem:

4.52 + a2 = 7.52a2  = 56.25 - 20.25 a2  = 36a  = 36a  =6

Hence, the length of the other side of the triangle is  6 cm.

Answer:

Suppose the length of the two legs of the right triangle are a cm and a cm.
Then, by Pythagoras theorem:
a2 +a2 = 502a2 =50a2 =25a =25a =5

Hence, the length of each leg is 5 cm.

Page No 193:

Question 5:

Suppose the length of the two legs of the right triangle are a cm and a cm.
Then, by Pythagoras theorem:
a2 +a2 = 502a2 =50a2 =25a =25a =5

Hence, the length of each leg is 5 cm.

Answer:

The largest side of the triangle is 39 cm.

152 + 362=225 + 1296 =1521

Also, 392 = 1521
152 + 362 = 392

Sum of the square of the two sides is equal to the square of the third side.

Hence, the triangle is right angled.

Page No 193:

Question 6:

The largest side of the triangle is 39 cm.

152 + 362=225 + 1296 =1521

Also, 392 = 1521
152 + 362 = 392

Sum of the square of the two sides is equal to the square of the third side.

Hence, the triangle is right angled.

Answer:

Suppose the length of the hypotenuse is c cm.
Then, by Pythagoras theorem:
a2 + b2 = c2c2  = 62 + 4.52 c2 =36 + 20.25 c2 = 56.25c=56.25c=7.5 

Hence, the length of its hypotenuse is 7.5 cm.

Page No 193:

Question 7:

Suppose the length of the hypotenuse is c cm.
Then, by Pythagoras theorem:
a2 + b2 = c2c2  = 62 + 4.52 c2 =36 + 20.25 c2 = 56.25c=56.25c=7.5 

Hence, the length of its hypotenuse is 7.5 cm.

Answer:

(i) Largest side, c = 25 cm

We have:
a2 + b2 = 225 + 400 = 625

Also, c2 = 625

a2 + b2 = c2
Hence, the given triangle is right angled using the Pythagoras theorem.

(ii)  Largest side, c = 16 cm
We have:
a2 + b2  = 81 + 144 = 225

Also,  c2 = 256

Here, a2 + b2c2

Therefore, the given triangle is not right angled.

(iii)  Largest side, c = 26 cm

We have:
a2 + b2  = 100 + 576= 676

Also, c2 = 676

a2 + b2 = c2
Hence, the given triangle is right angled using the Pythagoras theorem.

Page No 193:

Question 8:

(i) Largest side, c = 25 cm

We have:
a2 + b2 = 225 + 400 = 625

Also, c2 = 625

a2 + b2 = c2
Hence, the given triangle is right angled using the Pythagoras theorem.

(ii)  Largest side, c = 16 cm
We have:
a2 + b2  = 81 + 144 = 225

Also,  c2 = 256

Here, a2 + b2c2

Therefore, the given triangle is not right angled.

(iii)  Largest side, c = 26 cm

We have:
a2 + b2  = 100 + 576= 676

Also, c2 = 676

a2 + b2 = c2
Hence, the given triangle is right angled using the Pythagoras theorem.

Answer:

We have:
B = 35° and ∠C = 55°

∴ ∠B = 180 - 35 -55 = 90°    (since sum of the angles of any triangle is 180°)

We know that the side opposite to the right angle is the hypotenuse.

By Pythagoras theorem:
BC2 = AB2 + AC2

Hence, (iii) is true.

Page No 193:

Question 9:

We have:
B = 35° and ∠C = 55°

∴ ∠B = 180 - 35 -55 = 90°    (since sum of the angles of any triangle is 180°)

We know that the side opposite to the right angle is the hypotenuse.

By Pythagoras theorem:
BC2 = AB2 + AC2

Hence, (iii) is true.

Answer:

By Pythagoras theorem in ABC:
AB2 = AC2 + BC2 
152 = x2 +122x2 = 225 - 144 x2= 81x2=92x=9
x = 9 cm

Hence, the distance of the foot of the ladder from the wall is 9 cm.

Page No 193:

Question 10:

By Pythagoras theorem in ABC:
AB2 = AC2 + BC2 
152 = x2 +122x2 = 225 - 144 x2= 81x2=92x=9
x = 9 cm

Hence, the distance of the foot of the ladder from the wall is 9 cm.

Answer:

Suppose the foot of the ladder is x m far from the wall.

Let the ladder is represented by AB, the height at which it reaches the wall be AC and the distance between the foot of ladder and wall be BC.

Then, by Pythagoras theorem:
AB2 =AC2 + BC252 =4.82 + x2x2 =25 -23.04 x2 =1.96x2 =(1.4)2x=1.4


Hence, the foot of the ladder is 1.4 m far from the wall.

Page No 193:

Question 11:

Suppose the foot of the ladder is x m far from the wall.

Let the ladder is represented by AB, the height at which it reaches the wall be AC and the distance between the foot of ladder and wall be BC.

Then, by Pythagoras theorem:
AB2 =AC2 + BC252 =4.82 + x2x2 =25 -23.04 x2 =1.96x2 =(1.4)2x=1.4


Hence, the foot of the ladder is 1.4 m far from the wall.

Answer:

Let BD be the height of the tree broken at point C and suppose CD take the position CA




Now as per given conditions we have AB = 9 m , BC = 12 m

By Pythagoras theorem:
AC2 =AB2  + BC2

AC2 = 122 + 92 AC2= 144 + 81 AC2= 225AC2=152AC=15

Length of the tree before it broke = AC + AB
                                                       =  15 + 9
                                                        = 24 m



Page No 194:

Question 12:

Let BD be the height of the tree broken at point C and suppose CD take the position CA




Now as per given conditions we have AB = 9 m , BC = 12 m

By Pythagoras theorem:
AC2 =AB2  + BC2

AC2 = 122 + 92 AC2= 144 + 81 AC2= 225AC2=152AC=15

Length of the tree before it broke = AC + AB
                                                       =  15 + 9
                                                        = 24 m

Answer:

Suppose, the two poles are AB and CD, having the length of 18 m and 13 m, respectively.
Distance between them, BD, is equal to 12 m.
We need to find AC.

From C, draw CEAB.

AE=AB-EB
= AB-CD   (CD = EB)
= 18-13
= 5 cm
EC = BD = 12 m

Now, by Pythagoras theorem in AEC:
AC2 = AE2 + EC2AC2 = 52 + 122 AC2 = 25+ 144 AC2 = 169AC2=132AC=13

Hence, the distance between their tops is 13 m.

Page No 194:

Question 13:

Suppose, the two poles are AB and CD, having the length of 18 m and 13 m, respectively.
Distance between them, BD, is equal to 12 m.
We need to find AC.

From C, draw CEAB.

AE=AB-EB
= AB-CD   (CD = EB)
= 18-13
= 5 cm
EC = BD = 12 m

Now, by Pythagoras theorem in AEC:
AC2 = AE2 + EC2AC2 = 52 + 122 AC2 = 25+ 144 AC2 = 169AC2=132AC=13

Hence, the distance between their tops is 13 m.

Answer:

Suppose the man starts at point A and goes 35 m towards west, say AB. He then goes 12 m north, say BC.



We need to find AC.

By Pythagoras theorem:
AC2 = BC2 + AB2
AC2 = 352 + 122 AC2 =1225 + 144 AC2 = 1369AC2 =372AC=37 m

Hence, the man is 37 m far from the starting point.

Page No 194:

Question 14:

Suppose the man starts at point A and goes 35 m towards west, say AB. He then goes 12 m north, say BC.



We need to find AC.

By Pythagoras theorem:
AC2 = BC2 + AB2
AC2 = 352 + 122 AC2 =1225 + 144 AC2 = 1369AC2 =372AC=37 m

Hence, the man is 37 m far from the starting point.

Answer:

Suppose the man starts from A and goes 3 km north and reaches B.
He then goes 4 km towards east and reaches C.

∴ AB = 3 km
    BC = 4 km
We have to find AC.

By Pythagoras theorem:

AC2 = AB2 + BC2AC2 = 32 + 42 AC2 =25AC2 =52AC=5 km

Hence, he is 5 km far from the initial position.

Page No 194:

Question 15:

Suppose the man starts from A and goes 3 km north and reaches B.
He then goes 4 km towards east and reaches C.

∴ AB = 3 km
    BC = 4 km
We have to find AC.

By Pythagoras theorem:

AC2 = AB2 + BC2AC2 = 32 + 42 AC2 =25AC2 =52AC=5 km

Hence, he is 5 km far from the initial position.

Answer:

Suppose the sides are x and y of lengths 16 cm and 12 cm, respectively.

Let the diagonal be z cm.

Clearly, the diagonal is the hypotenuse of the right triangle with legs x and y.

By Pythagoras theorem:
z2 = x2 + y2z2 = 162 + 122 z2 =256 + 144 z2 =400z2 =202z=20 

Hence, the length of the diagonal is 20 cm.

Page No 194:

Question 16:

Suppose the sides are x and y of lengths 16 cm and 12 cm, respectively.

Let the diagonal be z cm.

Clearly, the diagonal is the hypotenuse of the right triangle with legs x and y.

By Pythagoras theorem:
z2 = x2 + y2z2 = 162 + 122 z2 =256 + 144 z2 =400z2 =202z=20 

Hence, the length of the diagonal is 20 cm.

Answer:


AB = 40 cm
Diagonal, AC = 41 cm

Then, by Pythagoras theorem in right ABC:
AC2 =AB2 + BC2BC2 =412 -402 BC2=1681 - 1600 BC2=81BC2=92BC=9 cm


∴ Length = 40 cm
    Breadth = 9 cm

∴ Perimeter of the rectangle = 2(length + breadth)
                                                = 2(40+9)
                                                = 98 cm

Page No 194:

Question 17:


AB = 40 cm
Diagonal, AC = 41 cm

Then, by Pythagoras theorem in right ABC:
AC2 =AB2 + BC2BC2 =412 -402 BC2=1681 - 1600 BC2=81BC2=92BC=9 cm


∴ Length = 40 cm
    Breadth = 9 cm

∴ Perimeter of the rectangle = 2(length + breadth)
                                                = 2(40+9)
                                                = 98 cm

Answer:

We know that the diagonals of a rhombus bisect each other at right angles.
Therefore, in right triangle AOB, we have:
AO = 8 cm
BO = 15 cm

By Pythagoras theorem in AOB:

AB2=AO2 + BO2AB2 =82 + 152 AB2 =64 +225AB2 =289AB2 =172AB=17 cm

Now, as we know that all sides of a rhombus are equal.
∴ Perimeter of the rhombus = 4(side)
                                               = 4(17)
                                               = 68 cm

Page No 194:

Question 18:

We know that the diagonals of a rhombus bisect each other at right angles.
Therefore, in right triangle AOB, we have:
AO = 8 cm
BO = 15 cm

By Pythagoras theorem in AOB:

AB2=AO2 + BO2AB2 =82 + 152 AB2 =64 +225AB2 =289AB2 =172AB=17 cm

Now, as we know that all sides of a rhombus are equal.
∴ Perimeter of the rhombus = 4(side)
                                               = 4(17)
                                               = 68 cm

Answer:

(i) In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
(ii) If the square of one side of a triangle s equal to the sum of the squares of the other two sides then the triangle is right angled.
(iii) Of all the line segments that can be drawn to a given line from a given point outside it, the perpendicular is the shortest.



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