WorkSHEET 5.1 Pythagoras’ theorem Name:

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WorkSHEET 5.1 Pythagoras’ theorem Name: ___________________________ 1 An aeroplane takes off and reaches a height of

1500 m after travelling an overland distance of 12 km. What is the distance covered by the plane, in km?

Let the distance covered be x km.

c2 = a2 + b2 x2 = 122 + (1.5)2 x2 = 144 + 2.25 x2 = 146.25 x = x » 12.09 km The plane covers about 12.09 km.

2 If sin 𝑥 = !" , then what is the value of cos 𝑥 ? 𝑠𝑖𝑛𝑒 =

𝑂𝐻

therefore have a triangle with an opposite side length of O=3 and a Hypotenuse H=5 By Pythagoras we have an adjacent length of A=4

Therefore, given sin 𝑥 = !" we have:

𝑂 = 3

𝐴 = 4

𝐻 = 5

Now, 𝑐𝑜𝑠𝑖𝑛𝑒 = #$

Therefore we can say:

cos 𝑥 =45

3 If tan 𝑥 = "%&

, then calculate the value of sin 𝑥 ? Similar to above, by 𝑡𝑎𝑛 = '# , we have:

𝑂 = 5

𝐴 = 12 by Pythagoras

𝐻 = 13

and 𝑠𝑖𝑛𝑒 = '$

so,

sin 𝑥 =513

25.146

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4 (a) If the area of a square is 25 cm2, find the length of the diagonal. Leave your answer in exact form.

(b) If the area of a square is 100 m2, find the

length of the diagonal. Leave your answer in exact form.

Let the length of the diagonal be x cm. (a)

c2 = a2 + b2

x2 = 52 + 52

x2 = 25 + 25 x2 = 50 x = cm The diagonal is cm long.

(b)

c2 = a2 + b2

x2 = 102 + 102

x2 = 100 + 100 x2 = 200 x = m

The diagonal is m long.

5 Josef and Maria walk to school every morning as shown in the figure. Josef decides to take the shortcut through the park. How much distance does he save (in km)?

Let the diagonal length be x km. c2 = a2 + b2 x2 = (2.4)2 + (1.33)2 x2 = 5.76 + 1.7689 x2 = 7.5289 x = x » 2.74 km Maria’s total distance travelled without using the shortcut is 2.4 + 1.33 = 3.73 km. Distance saved by Josef using the shortcut is 3.73 - 2.74 = 0.99 km.

5050

200200

5289.7

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6

Find the length of GD and hence find the length of ED correct to 2 decimal places.

GD2 = GC2 + CD2 ED2 = EG2 + GD2 = 122 + 82 = 252 + (14.42)2 = 144 + 64 = 625 + 208 = 208 = 883 GD = ED = » 14.42 cm » 28.86 cm The length of GD is 14.42 cm. The length of ED is 28.86 cm.

7

By first finding the length of ED, find the length of AE, and hence find the length of AC.

ED2 = CD2 + CE2 = (6.4)2 + (4.5)2 = 40.96 + 20.25 = 61.21 ED = » 7.82 m The length of ED is 7.82 m. AE2 = AD2 – ED2 = (9.2)2 – ( )2 = 84.64 – 61.21 = 23.43 AE = » 4.84 m The length of AE is 4.84 m. AC2 = CE2 + AE2 = (4.5)2 + ( )2 = 20.25 + 23.43 = 43.68 AC = » 6.61 m The length of AC is 6.61 m.

208 883

21.61

21.61

43.23

43.23

68.43

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8 The sloping side of a cone is 15 cm and the height is 9 cm. Find the radius of the base.

a2 = c2 – b2 r2 = 152 – 92 r2 = 225 – 81 r2 = 144 r = r = 12 cm The radius of the base of the cone is 12 cm.

9 Beth places her straw in a can of soft drink. The height of the can is 19 cm and the diameter is 8 cm. If the length of her straw is 26 cm, how much is it above the top of the can?

c2 = a2 + b2 x2 = 192 + 82 x2 = 361 + 64 x2 = 425 x = x » 20.6 cm Diagonal length = 20.6 cm 26 – 20.6 = 5.4 cm Length of straw out of can is 5.4 cm.

144

425

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10 A square-based pyramid has a base length of 18 cm and a sloping face 60 cm high. Find the height of the pyramid.

Let the height of the pyramid be x cm.

a2 = c2 – b2 x2 = 602 – 92 x2 = 3600 – 81 x2 = 3519 x = x » 59.32 cm The pyramid is 59.32 cm high.

3519

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11

(a) Find the length of the sloping edge of the

pyramid. (b) Find the length of the sloping face of the

pyramid.

(a) Need to find z.

Length of sloping edge of pyramid is 7.04 cm.

(b)

Length of sloping face of pyramid is 6.1 cm.

cm .0475.49

5.49255.24595.4

Finding 2Step

cm 4.952cm90.9

98

494977

Finding 1 Step

2

2

222

222

2

222

222

»=

=

+=

+=

+=

=÷»=

+=

+=

+=

zz

zzz

bacz

yyy

yy

bacy

cm 1.6 cm1.625.37 25.37

25.37 25.3725.1225 25.125.49

5.35 5.304.7or

Finding

22

22

222222

222222

»»==

==

+=-=

+=-=

+=-=

xxxx

xxxxxx

bacbcax

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12 A piece of cheese is the shape of a right-angled triangular prism with the dimensions shown.

(a) Find the length of the sloping sides

(in cm), correct to 2 decimal places. (b) Find the area of the 2 sloping rectangles

on the top surface of the cheese, to the nearest cm2.

Let the sloping sides be x cm long.

(a) c2 = a2 + b2

x2 = 122 + (7.5)2

x2 = 144 + 56.25 x2 = 200.25 x = x » 14.15 cm The sloping sides are 14.15 cm long.

(b) Area = 14.15 ´ 20 ´ 2 = 566 cm2 The area of the two sloping rectangles is 566 cm2.

25.200