Job Roles

Let the breadth of the given rectangle be x then length is 2x.
thus area of the given rect is 2x2

after dec 5cm from length and inc 5cm breadth , new lenght becomes 2x-5 and breadth is x+5.thus new area =(2x-5)(x+5)=2x2+5x-25

since new area is 75 units greater than original area thus
2x2+75 = 2x+5x-25
5x=75+25
5x=100
therefore x=20
hence length of the rectangle is 40 cm.

Length of the wire fencing = perimeter = 2(90 + 50) = 280 metres

Two poles will be kept 5 metres apart. Also remember that the poles will be placed along the perimeter of the rectangular plot, not in a single straight line which is very important.

Hence number of poles required = 280 / 5 = 56

Area = 5.5 × 3.75 sq. metre.

Cost for 1 sq. metre. = Rs. 800

Hence total cost = 5.5 × 3.75 × 800 = 5.5 × 3000 = Rs. 16500

Area of the sheet = 20×30cm2  

Area used for typing = 20-4×30-6cm2   

required % =384600×100=64%

Given lb =96

Area of pathway = [(l-4)(b-4)-lb] = 16-4(l+b)

Which cannot be determined. so, data is inadequate.

Let the areas of the two parts be x and (700-x) hectares

therefore, x-700-x=15x+700-x2

 ⇒2x-700=70

 ⇒x=385

So, the two parts are 385 and 315.

 Hence, Area of the smaller = 315 hectares

Given that length and area, so we can find the breadth.

 Length  x  Breadth = Area 

 => 20  x  Breadth = 680 

=> Breadth = 34 feet 

Area to be fenced = 2B + L = 2 (34) + 20 = 88 feet

Let the triangle and parallelogram have common base b, 

let the Altitude of triangle is h1 and of parallelogram is h2(which is equal to 100 m), then  

Area of triangle    = 12×b×h1 

Area of rectangle  = b*h2 

As per Given,

12*b*h1=b*h2

12*b*h1=b*100

 h1=200