## RD Sharma class 10 solutions Chapter 8 Quadratic Equations Ex 8.10

### RD Sharma Class 10 Solutions Quadratic Equations Exercise 8.10

Question 1.

The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm. Find the lengths of these sides.

Solution:

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

Question 2.

The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.

Solution:

Let shorter side of the rectangular field = x m

Then diagonal = (x + 60) m

and longer side = (x + 30) m

According to the condition,

(Diagonal)² = Sum of squares of the two sides

⇒ (x + 60)² = x² + (x + 30)²

⇒ x² + 120x + 3600 = x² + x² + 60x + 900

⇒ 2x² + 60x + 900 – x² – 120x – 3600 = 0

⇒ x² – 60x – 2700 = 0

⇒ x² – 90x + 30x – 2700 = 0

⇒ x (x – 90) + 30 (x – 90) = 0

⇒ (x – 90) (x + 30) = 0

Either x – 90 = 0, then x = 90

or x + 30 = 0, then x = – 30 which is not possible being negative

Longer side (length) = x + 30 = 90 + 30= 120

and breadth = x = 90 m

Question 3.

The hypotenuse of a right triangle is 3√10 cm. If the smaller leg is tripled and the longer leg doubled, new hypotenuse will be 9√5 cm. How long are the legs of the triangle ?

Solution:

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

Question 4.

A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that .the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it the possible to do so? If yes, at what distances from the two gates should the pole be erected ?

Solution:

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

### Exercise 8.10

#### RD Sharma Class 10 Solutions

- Chapter 8 Quadratic Equations Ex 8.1
- Chapter 8 Quadratic Equations Ex 8.2
- Chapter 8 Quadratic Equations Ex 8.3
- Chapter 8 Quadratic Equations Ex 8.4
- Chapter 8 Quadratic Equations Ex 8.5
- Chapter 8 Quadratic Equations Ex 8.6
- Chapter 8 Quadratic Equations Ex 8.7
- Chapter 8 Quadratic Equations Ex 8.8
- Chapter 8 Quadratic Equations Ex 8.9
- Chapter 8 Quadratic Equations Ex 8.10
- Chapter 8 Quadratic Equations Ex 8.11
- Chapter 8 Quadratic Equations Ex 8.12
- Chapter 8 Quadratic Equations Ex 8.13