1. Home
  2. Mathematics grade 9
  3. Term 1
  4. Why all the fuss about pythagoras

1.2 Why all the fuss about pythagoras?  (Page 2/3)

<< Chapter < Page
  Mathematics grade 9     Page 2 / 3
Chapter >> Page >

Substitution:AH 2 + EH 2 = (6) 2 + (8) 2 = 36 + 64 = 100 cm 2 If E 2 = 100 cm 2 , then AE must be 10 cm.

2.1 Calculate the length of the third side of these triangles:

2.1.1 ΔDEF with D a right angle and e = 5 mm and f = 12 mm

2.1.2 ΔXYZ with Y a right angle and x = 3 cm and y = 5 cm.

3 Problem : What is the length of the shortest side (b) of the right–angled ΔABC if the other two sides are 6 cm and 9 cm? C is the right angle.

Solution : In a right–angled triangle the longest side is always the hypotenuse: the side opposite the right angle. Now we use the Theorem of Pythagoras in the other form.

  • If b is the shortest side, and C is a right angle, then c must be the longest side. So, you use:

b 2 = c 2 – a 2 (note carefully where b 2 is, and that we subtract)

b 2 = (9) 2 – (6) 2 = 81 – 36 = 45 cm 2 Calculator time!

b 2 = 45. Use the size 12{ sqrt {} } {} – button on the calculator to find the value of b.

  • Your calculator supplies the answer: b = 6,7082039 . . . et cetera. But is this a sensible answer? Discuss whether the approximated (rounded) answer of 6,7cm is good enough for our purposes.

3.1.1 Calculate the length of the hypotenuse of a triangle with both of the other sides equal to 9 cm. (Label the triangle yourself.)

3.1.2 ΔPQR is right–angled and isosceles. Calculate the length of PR, if the hypotenuse is 13,5 cm.

4 Problem : Is ΔGHK right–angled if GK = 24 cm, GH = 26 cm and HK = 10 cm?

Solution : In this problem we know all three the sides’ lengths. If we want to find out whether it is right–angled, we have to confirm whether (hypotenuse) 2 = (one side) 2 + (other side) 2 .

The hypotenuse is always the longest side. We have a very specific method whenever we have to confirm a result. We calculate the left–hand side and the right–hand side of the equation separately. Thus:

  • Left–hand side = (hypotenuse) 2 = 26 2 = 676 cm 2
  • Right–hand side = (one side) 2 + (other side) 2 = 24 2 + 10 2 = 576 + 100 = 676 cm 2
  • Because the left–hand side and right–hand side come out the same, we can conclude that the triangle is right–angled.
  • Is it possible to know which angle is the right angle? You give the answer!

4.1 Are the triangles with the given side lengths right–angled? Which angle is the right angle?

4.1.1 a = 30 mm, b = 40 mm and c = 50 mm.

4.1.2 p = 8 cm, q = 13 cm and r = 15 cm.

4.1.3 MN = 15,56 cm, and NP = MP = 11 cm.

end of CLASS WORK

HOMEWORK ASSIGNMENT

1 Find the third side of the following triangles:

1.1 ΔABC with C = 90° and b = 5 mm and c = 13 mm

1.2 ΔMNO with O the right angle and m = 6 cm and n = 8 cm.

2 Determine whether the following triangles are right–angled, and which angle is 90° .

2.1 a = 9 mm, b = 11 mm and c 13 cm

2.2 XZ = 85 mm, XY = 13 mm and YX = 86 mm.

end of HOMEWORK ASSIGNMENT

The connection between roots and exponents

CLASS WORK

1 Eight of the equations in this list must be filled in the second row of the table under the equation in the top row where each fits the best.

25 = 5 size 12{ sqrt {"25"} =5} {} ; b = b 2 size 12{b= sqrt {b rSup { size 8{2} } } } {} ; 9 = 3 size 12{ sqrt {9} =3} {} ; 64 6 = 2 size 12{ nroot { size 8{6} } {"64"} =2} {} ; a 3 3 = a size 12{ nroot { size 8{3} } {a rSup { size 8{3} } } =a} {} ; 8 3 = 2 size 12{ nroot { size 8{3} } {8} =2} {} ;

81 4 = 3 size 12{ nroot { size 8{4} } {"81"} =3} {} ; 64 = 8 size 12{ sqrt {"64"} =8} {} ; 49 = 7 size 12{ sqrt {"49"} =7} {}

Exponential form 2 3 = 8 size 12{2 rSup { size 8{3} } =8} {} 9 = 3 2 size 12{9=3 rSup { size 8{2} } } {} 25 = 5 2 size 12{"25"=5 rSup { size 8{2} } } {} 7 2 = 49 size 12{7 rSup { size 8{2} } ="49"} {} 3 4 = 81 size 12{3 rSup { size 8{4} } ="81"} {} b × b = b 2 size 12{b times b=b rSup { size 8{2} } } {} 64 = 2 6 size 12{"64"=2 rSup { size 8{6} } } {} a × a × a = a 3 size 12{a times a times a=a rSup { size 8{3} } } {}
Root form ........... .......... ............ ............. ............ ............... ............. ....................
<< Chapter < Page Page > Chapter >>

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Mathematics grade 9. OpenStax CNX. Sep 14, 2009 Download for free at http://cnx.org/content/col11056/1.1
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Mathematics grade 9' conversation and receive update notifications?

Ask