Pythaguras | Revision Server | Page 1

valid summit Feb 21, 2023, 3:32 PM

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How would you figure out the permitted and area of this triangle to 2dp

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The width says 4cm and height is 9cm

merry prawn Feb 21, 2023, 3:34 PM

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Area ½ × 4 × 9

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For Perimeter you need hypotenuse

valid summit Feb 21, 2023, 3:35 PM

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So 4 * 9 = 36/2 = 18cm^2 as area?

merry prawn Feb 21, 2023, 3:35 PM

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Yep

valid summit Feb 21, 2023, 3:35 PM

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merry prawn For Perimeter you need hypotenuse

Aren't both sides equal cause of the line symbol

merry prawn Feb 21, 2023, 3:36 PM

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Square root 9²+ 4²

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So u get hypotenuse

valid summit Feb 21, 2023, 3:36 PM

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Would you squareroot the answer?

merry prawn Feb 21, 2023, 3:37 PM

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Yeah

valid summit Feb 21, 2023, 3:37 PM

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And that'd be the permiter?

merry prawn Feb 21, 2023, 3:37 PM

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No

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So whats your hypotenuse?

valid summit Feb 21, 2023, 3:38 PM

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merry prawn So whats your hypotenuse?

48.5cm

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After squareroot

merry prawn Feb 21, 2023, 3:39 PM

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Just wait a min

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So perimeter is 23.70 cm (2d.p)

valid summit Feb 21, 2023, 3:43 PM

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Ah so you do square root of 97 for both side added together along with 4

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Makes sense thanks

merry prawn Feb 21, 2023, 3:44 PM

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Np👍

bright prawn Feb 21, 2023, 6:55 PM

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That’s wrong

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It’s not 4^2 it’s 2^2

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@valid summit

valid summit Feb 21, 2023, 6:56 PM

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bright prawn It’s not 4^2 it’s 2^2

To get hypotenuse?

bright prawn Feb 21, 2023, 6:58 PM

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Ye

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The length I’ve circled is half the base not the full base

merry prawn Feb 21, 2023, 8:18 PM

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Bruh my bad

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