Mensuration - Area

Sponsor Area

Question
SSCCGLENQA12042730

Perimeter of a rhombus is 2p unit and sum of length of diagonals is m unit, then area of the rhombus is
  • 1 fourth straight m squared straight p space sq space unit
  • 1 fourth mp squared space sq space unit
  • 1 fourth left parenthesis straight m squared minus straight p squared right parenthesis space sq space unit
  • 1 fourth left parenthesis straight p squared minus straight m squared right parenthesis space sq space unit

Solution

C.

1 fourth left parenthesis straight m squared minus straight p squared right parenthesis space sq space unit

Let d1, d2 be the diagonals of the rhombus
Perimeter of rhombus (P) = 2 square root of straight d subscript 1 squared plus straight d subscript 2 squared end root
                         Perimeter = 2p(given)
∴       2 straight p space equals space 2 square root of straight d subscript 1 squared space plus space straight d subscript 2 squared end root
Squaring both the sides, we get
         straight p squared space equals space straight d subscript 1 squared space plus space straight d subscript 2 squared       ...(i)
Now, sum of diagonals (d1 + d2) =  m (given)
  Again, Squaring both the sides, we get
               left parenthesis straight d subscript 1 space plus space straight d subscript 2 right parenthesis squared space equals space straight m squared
         rightwards double arrow space space straight d subscript 1 squared space plus space straight d subscript 2 squared space plus space 2 space cross times space straight d subscript 1 space cross times space straight d subscript 2 space equals space straight m squared
         rightwards double arrow space space space straight p squared space plus space 2. straight d subscript 1. straight d subscript 2 space equals space straight m squared space space space space left square bracket using space left parenthesis straight i right parenthesis right square bracket
rightwards double arrow space space space space space space space space space space 2. space straight d subscript 1. space straight d subscript 2 space equals space straight m squared space minus space straight p squared
rightwards double arrow space space space space space space space space space space space straight d subscript 1. space straight d subscript 2 space equals space fraction numerator straight m squared minus straight p squared over denominator 2 end fraction space.... left parenthesis ii right parenthesis
Area of rhombus = 1 half cross times space straight d subscript 1 space cross times space straight d subscript 2
                        equals space 1 half space cross times space fraction numerator straight m squared minus straight p squared over denominator 2 end fraction space space space left square bracket using space left parenthesis ii right parenthesis right square bracket
equals space 1 fourth left parenthesis straight m squared minus straight p squared right parenthesis
                          
 
         

Sponsor Area

Question
SSCCGLENQA12043005

The area of the square inscribed in a circle of radius 8 cm is
  • 256 sq. cm.

  • 250 sq. cm.

  • 128 sq. cm.

  • 125 sq. cm.

Solution

C.

128 sq. cm.

When a square is inscribed in a circle
rightwards double arrow space space space straight a space equals space square root of 2 straight r
rightwards double arrow space space space straight a space equals space square root of 2 cross times 8 space equals space 8 square root of 2
Area of the square = left parenthesis side right parenthesis squared = 8 square root of 2 space cross times space 8 square root of 2 space equals space 128 space cm squared

Question
SSCCGLENQA12042988

A circle and a rectangle have the same perimeter. The sides of the rectangle are 18 cm and 26 cm. The area of the circle is open square brackets Take space straight pi space equals space 22 over 7 close square brackets

  • 125 cm2

  • 230 cm2

  • 550 cm2

  • 616 cm2

Solution

D.

616 cm2

Circumference = 2 πr
Perimeter of rectangle = 2(l + b)
According to the question,
rightwards double arrow space space space space 2 πr space equals space space 2 space left parenthesis 18 space plus space 26 right parenthesis
rightwards double arrow space space 2 space cross times space 22 over 7 space cross times space straight r space equals space 44 space cross times space 2
rightwards double arrow space space space space space straight r space equals space 14 space cm
therefore space space space space space space Area space of space circle space space equals space πr squared
space space space space space space space equals space 22 over 7 space cross times space 14 space cross times space 14 space equals space 616 space cm squared

Question
SSCCGLENQA12042786

A lawn is in the form of a rectangle having its breadth and length in the ratio 3 : 4. The area of the lawn is 1 over 12 hectare. The breadth of the lawn is

  • 25 metres

  • 50 metres

  • 75 metres

  • 100 metres

Solution

A.

25 metres

Let the length be 3x metre
          breadth be 4x metre
Area of the lawn  = 1 over 12 hectare  = 1 over 12 cross times 10000 space straight m squared equals space space 2500 over 3 straight m squared
therefore space space space space 3 straight x space cross times space 4 straight x space equals space 2500 over 3
rightwards double arrow space space space straight x squared space equals space fraction numerator 2500 over denominator 3 cross times 3 cross times 4 end fraction space space rightwards double arrow space space straight x space equals space 50 over 6
rightwards double arrow space Breadth space equals space 3 straight x space equals space 3 space cross times space 50 over 6 space equals space 25 space metre

Question
SSCCGLENQA12042492

A piece of wire 132 cm long is bent successively in the shape of an equilateral triangle, a square and a circle. Then area will be longest in shape of

  • Circle

  • Equilateral triangle

  • Square

  • Equal in all the shapes

Solution

A.

Circle

Side of square = 132/4 = 33 cm.
Area of square = 33 x 33 = 1089 cm2
Side of equilateral triangle  = 132/3 = 44 cm.
Area of equilateral triangle
                        equals space open parentheses fraction numerator square root of 3 over denominator 4 end fraction space cross times space 44 space cross times space 44 close parentheses space cm squared
equals space 484 space square root of 3 space cm squared
space equals space 484 space cross times space 1.7 space equals space 822.8 space cm squared
Circumference of circle  = 132
rightwards double arrow space space space 2 space cross times space 22 over 7 space cross times space straight r space equals space 132
rightwards double arrow space space space space space space straight r space equals space fraction numerator 132 space cross times 7 over denominator 2 space cross times space 22 end fraction space equals space 21 space cm
∴     Area of circle
               equals space 22 over 7 space cross times space 21 space cross times space 21 space equals space 1386 space cm squared