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2KGT - H7 - Doorsnede berekenen

Mathematics

7th - 8th Grade

Used 8+ times

2KGT - H7 - Doorsnede berekenen
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6 questions

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1.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

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Wat is de juiste berekening (met Pythagoras) van de diagonaal AC?

32 + 32 = 9 + 9 = 18

√18 ≈ 4,24 cm

3 + 3 = 6

√6 ≈ 2,45 cm

3 x 3 = 9

9 : 2 = 4,50 cm

3 + 3 = 6

62 + 62 = 36 + 36 = 72

√72 ≈ 8,49 cm

2.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

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Wat is de juiste berekening (met Pythagoras) van de diagonaal BG?

62 + 62 = 36 + 36 = 72

√72 ≈ 8,49 cm

6 + 6 = 12

√12 ≈ 3,46 cm

6 x 6 = 36

36 : 2 = 18 cm

6 + 6 = 12

122 + 122 = 144 + 144 = 288

√288 ≈ 16,97 cm

3.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

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Bereken met Pythagoras de lengte van diagonaal QT.

√25 = 5,0 cm

√7 ≈ 2,65 cm

√12 ≈ 3,46 cm

√29 = 5,39 cm

4.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

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Bereken met Pythagoras de lengte van diagonaal BG.

√50 ≈ 7,07 cm

√25 = 5 cm

√10 ≈ 3,16 cm

√100 = 10,0 cm

5.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Media Image

Bereken met Pythagoras de lengte van diagonaal QV.

√128 ≈ 11,31 cm

√64 = 8,0 cm

√16 = 4,0 cm

√512 ≈ 22,63 cm

6.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Media Image

Bereken met Pythagoras de lengte van diagonaal BE.

√80 ≈ 8,94 cm

√32 ≈ 5,66 cm

√160 ≈ 12,65 cm

√89 ≈ 9,43 cm

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