PYTHAGORAS-1

Find the length  AC. [Diagram not to scale]

Suluhu

We will use the PYTHAGORAS theorem [ a² + b² = c² ]

a² + b² = c²

AB² + BC² = AC²

3² + 4² = AC²

9 + 16 = AC²

25 = AC² [Here, we find the square root of 25]

5 = AC

Hence AC = 5mm

CIRCLE PERIMETER-2

Find the perimeter of the circle below

Solution

Diameter = 2xr =2x70=140

So, D=140mm

Perimeter = ∏ x D

                = 22 x 140⁷⁰

                    7₁

                = 22 x 70

                = 1540

Hence perimeter is 1540 mm. 

GEOMETRY-1

Find the value of x in the following figure

Solution

2x + 2x + 8x + 8x = 360⁰

4x + 16x = 360⁰

20x = 360⁰

20x   = 360⁰

20         20

X = 18⁰

Hence x is 18⁰

ALGEBRA-3

AREA OF RECTANGLE-1

Find area of the following rectangle (diagram not to scale)

Solution

Area = Length x width

         = 38 x 20

         =760m²

Hence area is 760m²

LCM-2

Find LCM between 40 and 80

solution

Hence LCM is 80

AREA OF CIRCULAR PRISM-1

Find area of the following circular prism which is closed on both sides

Solution

Area = 2∏rh + 2∏ x r²

=(2 x 22 x 35⁵ x 100) + (2 x 22 x 35⁵ x 35)

           7₁                                7₁

=(2 x 22 x 5 x 100) + (2 x 22 x 5 x 35)

                                           

=(44 x 500) + (44 x 5 x 35)

=22000+ 7700

=29700

Hence area is 29700 cm²

PERIMETER OF SQUARE

Perimeter of the following rectangle is 44mm. Find the value of a.

solution

Perimeter = 4 x side

44 = 4(a - 5)

44 = 4a – 20

44 + 20 = 4a

64 = 4a

64 =  4a

 4      4

16 = a

Hence a is 16     

SQUARE ROOTS

Find the square root of 324 by using prime factors

Solution

Hence the square root is 18

SQUARE PERIMETER-1

Find perimeter of the square below

Solution

Perimeter = 4  x  side

                   = 4 x 200

                   = 800

Hence perimeter is 800 mm.