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    I Cannot seem to answer question 9 from clip 191-MATHSWATCH. It is worth 4 marks and is really tough - you will have to apply your knowledge to get it right.
    A right angled triangle has sides of (x+5), (x-3) and 9cm
    work out the value of x to 3sf?
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    Do you know which one is the hypotenuse?
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    Hint: Start by drawing a diagram labeling each side. Perhaps Pythagoras may come in to use *cough cough*
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    Does it say which is the hypotenuse?
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    (Original post by Sammyboy1234)
    I Cannot seem to answer question 9 from clip 191-MATHSWATCH. It is worth 4 marks and is really tough - you will have to apply your knowledge to get it right.
    A right angled triangle has sides of (x+5), (x-3) and 9cm
    work out the value of x to 3sf?
    Use Pythagoras and you will formulate a quadratic equation. I'd do it for you now but you haven't told us which is the hypotenuse. Essentially you will have to Square each one, and either add them or subtract them from each other depending on which side is the hypotenuse (the equation will be equal to this as it's c) then you expand everything out, and make it equal to 0. From there you just put the values of the coefficients into the quadratic formula.
    🤙🏼
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    the hypotenuse is is 9cm and the adjacent (x-3)
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    (Original post by BananaBusiness)
    Use Pythagoras and you will formulate a quadratic equation. I'd do it for you now but you haven't told us which is the hypotenuse. Essentially you will have to Square each one, and either add them or subtract them from each other depending on which side is the hypotenuse (the equation will be equal to this as it's c) then you expand everything out, and make it equal to 0. From there you just put the values of the coefficients into the quadratic formula.
    🤙🏼
    hypotenuse is 9 and the adjacent in (X-3)
 
 
 
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