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# Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.10

• Last Updated : 13 Jan, 2021

### 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:

Let the length of other two sides of the triangle be x and y.

Therefore, according to the question,

x – y = 5

or,

x = y + 5

Now the two sides are of length y and y + 5.

Applying Pythagoras theorem on the right-angled triangle, we get,

252 = y2 + (y + 5)2

625 = y2 + y2 + 25 + 10y

2y2 + 25 + 10y = 625

2y2 + 10y = 625 – 25

2y2 + 10y – 600 = 0

y2 + 5y – 300 = 0

y2 + 20y – 15y – 300 = 0 [By middle term splitting]

y(y + 20) – 15(y + 20) = 0

(y – 15)(y + 20) = 0

y -15 = 0 or y + 20 = 0

y = 15 or y = -20

As the length of a side is positive, we will take y = 15.

Therefore,

x = y + 5

x = 15 + 5

x = 20

The two sides of the triangle are 15 cm and 20 cm.

### Question 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.

Solution:

Let the shorter side of the rectangle be x.

Now, according to the question, length of rectangle is x + 60 and length of the longer side of rectangle is x + 30.

Now, we will apply Pythagoras theorem on the right-angled triangle formed by the diagonal and the two sides,

(x + 60)2 = (x + 30)2 + x2

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

3600 – 900  = x2 + 60x – 120x

2700 = x2 – 60x

x2 – 60x – 2700 = 0

x2 – 90x + 30x – 2700 = 0

x(x – 90) + 30(x – 90) = 0

(x + 30)(x – 90) = 0

x + 30 = 0 or x – 90 = 0

x = -30 or x = 90

We will take positive side, i.e. x = 90

Now,

length of shorter side is 90 m

length of diagonal is 90 + 60 = 150 m

length of longer side is 90 + 30 = 120 m

### Question 3. The hypotenuse of a right triangle is 3âˆš10 cm. If the smaller leg is tripled and the longer leg doubled, new hypotenuse will be 9âˆš5 cm. How long are the legs of the triangle?

Solution:

Let the smaller and longer side of the triangle be x and y.

Now applying Pythagoras theorem, we get,

(3âˆš10)2 = x2 + y2

x2 + y2 = 90     ……… (i)

Now the smaller and longer sides are tripled and doubled respectively,

Therefore, the new sides of the triangle will be 3x and 2y.

Applying Pythagoras theorem,

(9âˆš5)2 = (3x)2 + (2y)2

9x2 + 4y2 = 405     ……… (ii)

Now multiplying (i) by 4 and then subtracting (i) from (ii), we get

9x2 – 5x2 + 4y2 – 4y2= 405 – 360

5x2 = 45

x2 = 9

x = 3

Substituting value of x in (i),

9 + y2 = 90

y2 = 81

y = 9

Length of smaller side is 3 cm.

Length of longer side is 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 possible to do so? If yes, at what distances from the two gates should the pole be erected?

Solution:

In the circle, let P be the pole on the circumference of the circle and points A and B be the two diametrically opposite fixed gates.

Let length of PB be x.

Given,

PA – PB = 7

PA = x + PB

PA = x + 7

The triangle PQR is a right-angled triangle at P, as AB is the diameter of the circle.

Now applying Pythagoras theorem,

AB2 = PA2 + PB2

132 = (x + 7)2 + x2

169 = x2 + 49 + 14x + x2

169 = 2×2 + 14x + 49

2x2 + 14x + 49 – 169 = 0

2x2 + 14x – 120 = 0

x2 + 7x – 60 = 0

x2 + 12x – 5x – 60 = 0

x(x + 12) – 5(x + 12) = 0

(x – 5)(x + 12) = 0

x = 5 or x = -12

Only x = 5 is possible.

PB = 5 m

PA = 5 + 7 = 12 m

P should be erected at a distance of 5 m from PB and 12 m from PA.

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