Answer:
the length of AB nearest to hundredth is 8.48 cm
Step-by-step explanation:
Pythagoras theorem states that the square of the hypotenuse side is equal to the sum of the other two sides.
Labelled the diagram as shown below.
In a right triangle BDC.
DC = 9 cm , BD = x units and BC = 12 units
then,
by Pythagoras theorem in triangle BDC
[tex]BD^2 +DC^2 = BC^2[/tex]
Substitute the given values we have;
[tex]x^2 + 9^2 = 12^2[/tex]
Simplify:
[tex]x^2+81=144[/tex]
Subtract 81 from both sides we get;
[tex]x^2 =63[/tex] cm
Now, in triangle BDA:
AD = 3 cm and BD = x cm
Using Pythagoras theorem, to solve for AB
[tex]BD^2 +AD^2 = AB^2[/tex]
Substitute the given values we have;
[tex]x^2+3^2=AB^2[/tex] .....[1]
Substitute the value of [tex]x^2 =63[/tex] in [1] we get;
[tex]63+9=AB^2[/tex]
or
[tex]AB^2= 72[/tex]
Simplify:
[tex]AB = \sqrt{72}=8.48528137[/tex] cm
Therefore, the length of AB nearest to hundredth is 8.48 cm
