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Class 10 Class 12

Find the area the quadrilateral LMNOP (as shown in the figure).


Diagonal LN = 8 cm
Perpendicular MX = 4.5 cm
Perpendicular OY = 3.5 cm
∵   Sum of the perpendiculars = MX + OY = 4.5 cm + 3.5 cm = 8 cm
∴   Area of the quadrilateral = 1 half cross times (A diagonal ) x (Sum of the lengths of the perpendiculars on it from opposite vertices)
                                            equals 1 half cross times space 8 space cross times space 8 space cm squared space equals space 32 space cm squared
                                             

949 Views

Find the length of the diagonal BD when, the area of the quadrilateral is 32 cm2.


Let the length of the diagonal BD = x cm
Area of the quadrilateral ABCD = 32 cm2
∵    Area of the quadrilateral  = 1 half x (A diagonal) x (Sum of the length of the perpendicular on the diagonal from the opposite vertices)
∴    Area of the quadrilateral ABCD = 1 half cross times BD cross times left parenthesis AP plus CQ right parenthesis
                                                        = 1 half cross times space straight x space cm cross times space left parenthesis 4.5 space cm space plus space 3.5 space cm right parenthesis

 
                                                        = 1 half cross times straight x cross times 8 space cm squared
Since area of the quadrilateral ABCD = 32 cm2
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                                                       = 8 cm
   Thus, the required length of the diagonal BD = 8 cm.
114 Views

The area of a rhombus and that of a square are equal. The side of the square is 6 cm. If one of the diagonal of the rhombus is 4 cm, then find the length of its other diagonal.

Here, side of the square = 6 cm
∴                   Area of the square =  Side x Side
                                                    = 6 cm x 6 cm
                                                    = 36 cm2
Since,         [Area of the rhombus]  = [Area of the square]
∴                  Area of the rhombus = 36 space cm squared

      One of the diagonal of the rhombus = 4 cm
Let the other diagonal of the rhombus  = d cm
∵                  Area of the rhombus  = 1 half cross times space Product space of space the space diagonals

∴                  Area of the rhombus  = space 1 half cross times 4 cross times straight d
rightwards double arrow                    1 half cross times 4 cross times straight d space equals space 36
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    Thus, the required length of the rhombus = 18 cm.
116 Views

The edge of a cube is 2 cm. Find the total surface area of the cuboid formed by three such cubes joined edge to edge.

The edge (side) of the given cube = 2 cm
Since, there such cubes are joined, then
Total length (l) = (2+2+2) cm
                         = 6 cm
Breadth (b) = 2 cm
Height (h) = 2 cm


∴   Total surface area of the resultant cuboid
                                                  = 2[lb + bh + hl]
                                                  = 2[6 x 2 + 2 x 2 + 2 x 6] cm2
                                                  = 2[28] cm2
                                                  = 56 cm2
Thus, the required total surface area of the cuboid = 56 cm2.

205 Views

A road roller makes 250 complete revolution to move once over to once level a road. Find the area of the road levelled if the diameter of the road roller is 84 cm and length is 1 m.

A road roller is a cylinder, such that
                Radius  = 84 over 2 space cm space equals space 42 space cm
            Length (h) = 1m = 100 cm
∵       Lateral surface area of a cylinder = space space 2 πrh
∴       Lateral surface area of the road roller
                                        space space space space equals space 2 cross times 22 over 7 cross times 42 cross times 100 space cm squared
                                             equals space 2 cross times 22 cross times 6 cross times 100 space cm squared
∴        Area of road levelled in 1 revolution = 26400 cm2
rightwards double arrow      Area of road levelled in 250 revolutions
                                                           equals space 250 space cross times space 26400 space cm squared

                                                           equals space fraction numerator 250 cross times 26400 over denominator 100 cross times 100 end fraction straight m squared

                                                            equals space fraction numerator 25 cross times 264 over denominator 10 end fraction space straight m squared space equals space 6600 over 10 space straight m squared

                                                             space equals space 660 space straight m squared.
 
137 Views

A rectangle and a square have equal perimeter. If the breadth of the rectangle is 10 cm and the side the square is 15 cm, then find the difference between their areas.

∵                     Side of the square = 15 cm
∴             Perimeter of the square = 4 x 15 cm = 60 cm
Since, [Perimeter of the rectangle] = [Perimeter of the square]
∴     [Perimeter of the rectangle] = 60 cm
            Let length of the rectangle = x cm
∴                  2[x + 10] = 60     [∵ Breadth of the rectangle = 10 cm]
rightwards double arrow                  2x + 20 = 60
rightwards double arrow                          2x = 60 - 20 = 40
rightwards double arrow                                   straight x equals 40 over 2 equals 20 space cm
rightwards double arrow          Length of the rectangle = 20 cm
Now, Area of the rectangle  = Length x Breadth
                                            = 20 x 10 cm2
                                            = 200 cm2
          Area of the square = Side x Side
                                        = 15 x 15 cm2
                                        = 225 cm2
∴ Difference in areas = 225 cm2 - 200 cm2 = 25 cm2

101 Views