- RD Chapter 8- Quadratic Equations Ex-8.1
- RD Chapter 8- Quadratic Equations Ex-8.2
- RD Chapter 8- Quadratic Equations Ex-8.3
- RD Chapter 8- Quadratic Equations Ex-8.4
- RD Chapter 8- Quadratic Equations Ex-8.5
- RD Chapter 8- Quadratic Equations Ex-8.6
- RD Chapter 8- Quadratic Equations Ex-8.7
- RD Chapter 8- Quadratic Equations Ex-8.8
- RD Chapter 8- Quadratic Equations Ex-8.9
- RD Chapter 8- Quadratic Equations Ex-8.11
- RD Chapter 8- Quadratic Equations Ex-8.12
- RD Chapter 8- Quadratic Equations Ex-8.13

RD Chapter 8- Quadratic Equations Ex-8.1 |
RD Chapter 8- Quadratic Equations Ex-8.2 |
RD Chapter 8- Quadratic Equations Ex-8.3 |
RD Chapter 8- Quadratic Equations Ex-8.4 |
RD Chapter 8- Quadratic Equations Ex-8.5 |
RD Chapter 8- Quadratic Equations Ex-8.6 |
RD Chapter 8- Quadratic Equations Ex-8.7 |
RD Chapter 8- Quadratic Equations Ex-8.8 |
RD Chapter 8- Quadratic Equations Ex-8.9 |
RD Chapter 8- Quadratic Equations Ex-8.11 |
RD Chapter 8- Quadratic Equations Ex-8.12 |
RD Chapter 8- Quadratic Equations Ex-8.13 |

**Answer
1** :

Length of the hypotenuse of a let right ∆ABC = 25 cm

Let length of one of the other two sides = x cm

Then other side = x + 5 cm

According to condition,

(x)² + (x + 5)² = (25)² (Using Pythagoras Theorem)

=> x² + x² + 10x + 25 = 625

=> 2x² + 10x + 25 – 625 = 0

=> 2x² + 10x – 600 = 0

=> x² + 5x – 300 = 0 (Dividing by 2)

=> x² + 20x – 15x – 300 = 0

=> x (x + 20) – 15 (x + 20) = 0

=> (x + 20) (x – 15) = 0

Either x + 20 = 0, then x = -20, which is not possible being negative

or x – 15 = 0, then x = 15

One side = 15 cm

and second side = 15 + 5 = 20 cm

**Answer
2** :

Let the smaller leg of right triangle = x cm

and larger leg = y cm

Then x² + y² = (3√10)² (Using Pythagoras Theorem)

x² + y² = 90 ….(i)

According to the second condition,

(3x)² + (2y)² = (9√5)²

=> 9x² + 4y² = 405 ….(ii)

Multiplying (i) by 9 and (ii) by 1

y = 9

Substituting the value of y in (i)

x² + (9)² = 90

=> x² + 81 = 90

=> x² = 90 – 81 = 9 = (3)²

x = 3

Length of smaller leg = 3 cm

and length of longer leg = 9 cm

**Answer
3** :

In a circle, AB is the diameters and AB = 13 m

Let P be the pole on the circle Let PB = x m,

then PA = (x + 7) m

Now in right ∆APB (P is in a semi circle)

AB² = AB² + AP² (Pythagoras Theorem)

(13)² = x² + (x + 7)²

=> x² + x² + 14x + 49 = 169

=> 2x² + 14x + 49 – 169 = 0

=> 2x²+ 14x – 120 = 0

=> x2 + 7x – 60 = 0 (Dividing by 2)

=> x² + 12x – 5x – 60 = 0

=> x (x + 12) – 5 (x + 12) = 0

=> (x + 12) (x – 5) = 0

Either x + 12 = 0, then x = -12 which is not possible being negative

or x – 5 = 0, then x = 5

P is at a distance of 5 m from B and 5 + 7 = 12 m from A

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