This page 10th cbse maths solution for exercise 4.3 part 3 is going to provide you solution for every problems that you find in the exercise no 4.3 part 3

(6) 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 'x' be the length of shorter side of the rectangle

length of diagonal = x + 60

length of longer side = x + 30

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

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

**x****² + 3600 ****+ 120 x = 2 x****² +**** 60 x + 900**

**2 x****² - ****x****² + 60 x - 120 x + 900 - 3600 = 0**

**x****² - 60 x - 2700 = 0**

**x****² - 90 x + 30 x - 2700 = 0**

**x (x - 90) + 30 (x - 90) = 0**

** (x + 30) (x - 90) = 0 **

** x + 30 = 0 x - 90 = 0**

** x = -30 x = 90**

**breadth of rectangle = 90 m**

**length of rectangle = 90 + 30 = 120 m**

(7) The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Solution:

Let 'x' be the larger number

let 'y' be the smaller number

y**² = 8 x **

**The difference of squares of two numbers = 180**

** x****² - y****² = 180**

** ** **x****² - 8 x**** = 180**

** ** **x****² - 8 x**** - 180 = 0**

** ** **x****² - 18 x + 10 x**** - 180 = 0**

**x (x - 18) + 10 (x - 18) = 0**

** (x - 18) (x + 10) = 0 **

** x - 18 = 0 x + 10 = 0**

** x = 18 x = -10**

**y****² = 8 (18)**

** y = **√
**8 (18)**

** y = **√**2 x 2 x 2 x 3 x 3 x 2**

** y = 2 x 2 x 3**

** y = 12**

**Therefore the larger number = 18**

**and smaller number = 12**

(8) A train travels 360 km at a uniform speed. If the speed had been 5 km/h more,it would have taken 1 hour less for the same journey. Find the speed of the train.

Solution:

Let 'x' be the speed of the train

If the speed had been 5 km/h more,it would have taken 1 hour less for the same journey

Time = distance/speed

Distance to be covered = 360 km

T₁ = 360/x

T₂ = 360/(x + 5)

T₁ - T₂ = 1 hour

[360/x] - [360/(x + 5)] = 1

360 [ (1/x) - (1/(x + 5))] = 1

360 [ (x + 5 - x)/x(x + 5) ] = 1

360[5/(x**² + 5 x)]** = 1

1800/(x**² + 5 x) = 1**

**1800 = **(x**² + 5 x)**

x**² + 5 x - 1800 = 0**

x**² - 40 x + 45 x - 1800 = 0**

**x (x - 40) + 45 (x - 40) = 0**

**(x - 40) (x + 45) = 0**

** x - 40 = 0 x + 45 = 0 **

** x = 40 and x = -45 **

**Therefore speed of the train = 40 km/hr**

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