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P6 MATHS © Singapore Asia Publishers Pte Ltd Website: www.sapgrp.com | Facebook: Singapore-Asia-Publishers SAPMATP6_W50 Page 1/5 Taken from Learning Maths Book 6B Revision: Circles Do these word problems. Show your working clearly in the space provided. 1. A semicircle and a quadrant are enclosed in a rectangle as shown below. Find the area of the shaded part. (Take π = 3.14) 10 cm 10 cm O 2. The figure below shows two identical circles and a rectangle. The radius of each circle is 28 cm. Find the area of the shaded part. (Take π = 22 ___ 7 ) O O 3. The figure below shows a semicircle and three identical quadrants removed from a square. The length of the square is four times the diameter of the semicircle. (Take π = 3.14) (a) Find the area of the shaded part. Round off your answer to the nearest whole number. (b) Find the perimeter of the shaded part. Round off your answer to the nearest whole number. 5 cm 5 cm 5 cm 5 cm
Transcript

P6 MATHS

© Singapore Asia Publishers Pte LtdWebsite: www.sapgrp.com | Facebook: Singapore-Asia-Publishers

SAPMATP6_W50Page 1/5

Taken from Learning Maths Book 6B

Revision: Circles

Do these word problems. Show your working clearly in the space provided.

1. Asemicircleandaquadrantareenclosedinarectangleasshownbelow.Findtheareaoftheshadedpart.(Takeπ=3.14)

10cm

10cm

O

2. Thefigurebelowshowstwoidenticalcirclesandarectangle.Theradiusofeachcircleis28

cm. Findtheareaoftheshadedpart.(Takeπ= 22___ 7 )

O O

3. Thefigurebelowshowsasemicircleandthreeidenticalquadrantsremovedfromasquare.Thelengthofthesquareisfourtimesthediameterofthesemicircle.(Takeπ=3.14)

(a) Findtheareaoftheshadedpart.Roundoffyouranswertothenearestwholenumber.

(b) Find theperimeterof the shadedpart. Roundoff youranswer to thenearestwholenumber.

5cm 5cm

5cm

5cm

P6 MATHS

© Singapore Asia Publishers Pte LtdWebsite: www.sapgrp.com | Facebook: Singapore-Asia-Publishers

1. Areaoftherectangle=30×20=600cm2

Areaofthesemicircle= 1__ 2 ×�r2= 1__ 2 ×3.14×10×10=157cm2

Areaofthequadrant= 1__ 4 ×�r2= 1__ 4 ×3.14×10×10=78.5cm2

600–157–78.5=364.5cm2

Theareaoftheshadedpartis364.5 cm2.

2. Areaoftherectangle=(2×28)×28=1568cm2

Areaofeachquadrant= 1__ 4 ×�r2= 1__ 4 × 22___ 7 ×28×28=616cm2

1568–616–616=336cm2

Theareaoftheshadedpartis336 cm2.

3. (a) Lengthofthesquare=4×5=20cm

Areaofthesquare=20×20=400cm2

Areaofthesemicircle= 1__ 2 ×�r2= 1__ 2 ×3.14× 5__ 2 × 5__ 2 =9.8125cm2

Areaofeachquadrant= 1__ 4 ×�r2= 1__ 4 ×3.14×5×5=19.625cm2

400 – 9.8125 – (3 × 19.625) = 331.3125 ≈ 331 cm2

Theareaoftheshadedpartis331 cm2.

(b) Circumferenceofeachquadrant= 2�r___ 4 = 2×3.14×5___________ 4 =7.85cm

Circumferenceofthesemicircle= �d___ 2 = 3.14×5________ 2 =7.85cm

7.85 + 10 + 7.85 + 10 + 7.85 + 10 + 7.85 + 15 = 76.4 ≈ 76 cm

Theperimeteroftheshadedpartis76 cm.

Solutions to Revision: Circles

SAPMATP6_W50Page 2/5

Taken from Learning Maths Book 6B

P6 MATHS

© Singapore Asia Publishers Pte LtdWebsite: www.sapgrp.com | Facebook: Singapore-Asia-Publishers

SAPMATP6_W50Page 3/5

Taken from Learning Maths Book 6B

Revision: Pie Charts

Write your answers on the lines provided.

Thepiechartbelowshowsthedifferentactivitiesthataclassofstudentsenjoysdoingduringfreetime.Studyitcarefullyandanswerquestions1to3.

Jogging

Reading

Collectingstamps

Watchingmovies

25%

15%

Cycling

30%

1. If12studentslikecycling,howmanystudentsarethereintheclass?

2. Howmanystudentslikewatchingmovies?

3. Ifthenumberofstudentswholikereadingistwicethenumberofstudentswholikejogging,howmanystudentsliketoread?

1.40 students30%→1210%→12÷3=4

100%→10×4=40

2.10 students 25 ___ 100×40=10

3.8 students100%–25%–15%–30%=30%20%ofthestudentslikereading.

20 ___ 100×40=8

Solutions:

P6 MATHS

© Singapore Asia Publishers Pte LtdWebsite: www.sapgrp.com | Facebook: Singapore-Asia-Publishers

SAPMATP6_W50Page 4/5

Taken from Learning Maths Book 6B

Revision: Pie Charts

Do these word problems. Show your working clearly in the space provided.

Thepiechartbelowshowsthenumberoffruitsoldbyashopkeeper.Studyitcarefullyandanswerquestions1and2below.

Papayas

Watermelons

20%

Bananas

Starfruit

25%

1. Iftheshopkeepersold300fruitaltogether,howmanywatermelonsdidhesell?

2. Howmuchmoneywouldhemakefromthesaleofstarfruitandpapayas ifhesoldeachstarfruitat$1.30andeachpapayaat$1.60?

1.100%–20%–25%–25%=30%

30 ___ 100 ×300=90

Hesold90watermelons.

2. 25 ___ 100 ×300=75

Hesold75starfruit.

20 ___ 100×300=60

Hesold60papayas.(75×$1.30)+(60×$1.60)=$193.50Hewouldmake$193.50.

Solutions:

P6 MATHS

© Singapore Asia Publishers Pte LtdWebsite: www.sapgrp.com | Facebook: Singapore-Asia-Publishers

SAPMATP6_W50Page 5/5

Taken from Mental Mathematics Book 6

Mental Maths: GENERAL REVISION 18

Do these sums mentally.

1. 64÷1 1__3 =

2. 882=

3. √_______

17161 =

4. 3√_______

42875 =

5. 347×99=

6. 49×111=

7. 33 1__3 %of174=

8. 182×429=

9. 212–202=

10. 17+18+19+20+21+...+54=

Solutions:1.482.77443.1314.355.343536.54397.588.780789.4110.1349


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