#### Exercise 8.10

**Q.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.**

**Sol:** let the length of one side of the right triangle be x cm then,

the other side be = (x + 5) cm

and given that hypotenuse = 25 cm

By using Pythagoras Theorem,

x^{2} + (x + 5)^{2} = 25^{2}

x^{2} + x^{2 }+ 10x + 25 = 625

2x^{2 }+ 10x + 25 – 625 = 0

2x^{2} + 10x – 600 = 0

x^{2} + 5x – 300 = 0

x^{2} – 15x + 20x – 300 = 0

x(x – 15) + 20(x -15) = 0

(x – 15)(x + 20) = 0

x = 15 or x = – 20

Since, the side of triangle can never be negative

Therefore, when, x = 15

And, x + 5 = 15 + 5 = 20

Therefore, length of side of right triangle is = 15 cm and other side is = 20 cm

**Q.2: The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.**

**Sol:**

let the length of smaller side of rectangle be x metres then, the larger side be (x + 30) metres and

diagonal be = (x + 60) metres

By using Pythagoras theorem,

x^{2} + (x + 30)^{2} = (x + 60)^{2}

x^{2} + x^{2} + 60x + 900 = x^{2} + 120x + 3600

2x^{2} + 60x + 900 – x^{2} – 120x – 3600 = 0

x^{2} – 60x – 2700 = 0

x^{2} – 90x + 30x – 2700 = 0

x(x – 90) + 30(x – 90) = 0

(x – 90)(x + 30) = 0

x = 90 or x = -30

Since, the side of rectangle can never be negative

Therefore, x = 90

x + 30 = 90 + 30 = 120

Therefore, the length of smaller side of rectangle is = 90 metres and larger side is = 120 metres.

** **

**Q.3: The hypotenuse of a right triangle is \(3\sqrt{10}\) cm. If the smaller leg is tripled and the longer leg doubled, new hypotenuse will be \(9\sqrt{5}\) cm. How long are the legs of the triangle?**

**Sol:**

let the length of smaller side of right triangle be = x cm then large side be = y cm

By using Pythagoras theorem,

\(x^{2} + y^{2} = (3\sqrt{10})^{2}\)

x^{2 }+ y^{2} = 90 ….eqn.(1)

If the smaller side is triple and the larger side is doubled, the new hypotenuse is \(9\sqrt{5}\)

Therefore,

(3x)^{2} + (2y)^{2} = \((9\sqrt{5})^{2}\)

9x^{2} + 4y^{2} = 405 ….eqn.(2)

From equation (1) we get,

y^{2} = 90 – x^{2}

Now putting the value of y^{2} in eqn. (2)

9x^{2} + 4(90 – x^{2}) = 405

9x^{2} + 360 – 4X^{2} – 405 = 0

5x^{2} – 45 = 0

5(x^{2} – 9) = 0

x^{2} – 9 = 0

x^{2} = 9

\(x = \sqrt{9}\)

\(x = \pm 3\)

Since, the side of triangle can never be negative

Therefore, when x = 3

Then, y^{2} = 90 – x^{2 }= 90 – (3)^{2 }= 90 – 9 = 81

\(y = \sqrt{81}\)

\(y = \pm 9\)

Hence, the length of smaller side of right triangle is = 3cm and larger side is = 9cm

** **

**Q.4) A pole has to be erected at a point on the boundary of a circular park of diameter meters in such a way that the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 meters. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?**

**Sol: **

let P be the required location on the boundary of circular park such that its distance from the gate B is x metres that is BP = x metres

Then, AP = x + 7

In right triangle ABP, by using Pythagoras theorem,

AP^{2} + BP^{2} = AB^{2}

(x + 7)^{2} + x^{2 } = 13^{2}

x^{2 }+ 14x + 49 + x^{2} = 169

2x^{2 } + 14x + 49 – 169 = 0

2x^{2} + 14x – 120 = 0

2(x^{2 }+ 7x – 60) = 0

x^{2} + 12x – 5x – 60 = 0

x(x + 12) – 5(x + 12) = 0

(x + 12)(x – 5) = 0

x = – 12 or x = 5

Since, the side of triangle can never be negative

Therefore, P is at a distance of 5 metres from the gate B