Maths( Grade 10)

5. A circle of radius 5 cm is inscribed inside a square as shown. Find the area shaded.

(the radius is not exactly 5 cm in the picture)

Shaded area   =   area of square −  area of circle  =      [10 cm]^2   − pi [5cm]^2  =

100 cm^2   −  25 pi  cm^2   =    [ 100  −  25 pi]  cm^2  ≈  21.46 cm^2

1. A kite of area 40 m^2 has one diagonal 2 m longer than the other. Find the lengths of the diagonals.

Area_kite  = D1 * D2  / 2    let  D2  = D1 + 2

40  =  D1 * [ D1 + 2]  / 2         multiply both sides by 2

80 =  D1 *  [D1 + 2] 

80 =  [D1]^2  + 2D1         rearrange

[D1]^2  + 2D1  −  80   = 0       factor

( D1   +  10)  (D1 −  8 )  =  0        set the factors to 0  →  D1 = −10m  [reject ]     or   D1  = 8 m

And D2 =  D1  + 2   =  8 + 2    = 10 m

2. Find the length of a side of an equilateral triangle of area 10.2 m^2.

Area  =  (1/2)* side^2 * [√3 / 2 ]

10.2 = [√3/4 ] * side^2     multiply both sides by   4/ √3

[10.2 * 4 ]  /  √3    =   side^2

40.8 / √3  = side^2   take the square root  of both sides

√ [  40.8 / √3 ]  = side  ≈ 4.853 m

It comes from the following "formula"  for the area of an equilateral triangle

Area =  (1/2) *  side^2  *  sin(60°)       and  sin(60°)  = [√3 / 2 ]

So we have

Area =  (1/2) *  side^2  * [√3 / 2 ]

4. Find the area of a regular pentagon of side 8 cm.

Area  ≈  1.720 * side^2   =  1.720 * [8 cm]^2  =  [1.720 * 64 ] cm^2   ≈  110.08 cm^2

3. A rhombus has an area of 40 cm^2 and adjacent angles of 50 degrees and 130 degrees. find the length of the side of the rhombus.

We can use this :

(1/2) Area  =  (1/2) side^2  sin(50°)    multiply by 2 on both sides

Area  =  side^2  sin(50°)

40  = side^2 sin(50°)       divide both sides by sin(50)

40 / sin (50°)   = side^2       take the sq root of both sides

√ [40 / sin (50°) ]   = side ≈    7.226  cm

Rauhan.....See the approximate pentagon area formula I used here :

https://en.wikipedia.org/wiki/Pentagon