I have following sides(PQRST) of a Polygon where PQ=13, QR=22, RS=8, ST=?, PT= 10 ... i need to find out ST? i don't have any angle i just have the shape? And for calculating perimeter i need to find out the ST length of polygon!
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2Presumably, all the angles that look like right angles are supposed to be taken to be right angles, in which case a little insight and a use of Pythagoras should answer your questions. – Gerry Myerson Mar 12 '14 at 06:22
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I found its solution done by somebody using Pythagoras Theorem. and he got the missing side with length 13 and calculated perimeter 66 if am not wrong. But, i don't know how he has done. Because i don't have much knowledge about polygons and new here. I need some sort of a helping hand to solve it – Nomiluks Mar 12 '14 at 06:44
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We can make use of the symmetry here and use Pythogoras theorm to Solve:
Observe that:
QR = PO = 22 {Opposite sides of a rectangle}
PT + TO = 22
TO = 22- PT = 22 - 10 = 12
Similarly,
PQ = OR = 13 {Opposite sides of a rectangle}
OS + SR = 13
OS = 13 - SR = 13 - 8 = 5
Now In right Traingle TOS, rt angled at O,
$TO^2 + OS^2 = ST^2 {Using Pythogoras Theorm}$
$12^2 + 5^2 = TS^2$
$TS^2 = 144 + 25 = 169$
Hence, TS = 13
Now you can find the perimeter.
Kailas
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Why don't you draw perpendicular from $T$ to $QR$ and from $S$ to $PQ$ ? Pythagoras Theorem.
Complete rectangles by drawing sides I told you about. Use the fact that opposite sides of rectangles are equal You should be able to get a right angled triangle with hypotenuse as $ST$ and other 2 sides as $13-8$ and $22-10$.
evil999man
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I found its solution done by somebody using Pythagoras Theorem. and he got the missing side with length 13 and calculated perimeter 66 if am not wrong. But, i don't know how he has done. Because i don't have much knowledge about polygons and new here – Nomiluks Mar 12 '14 at 06:39
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@Nomi Can't u just think of this polygon as some rectangles and a triangle put together and deleted all lines except the perimeter? You just gotta reconstruct those lines which question maker deleted. – evil999man Mar 12 '14 at 06:49
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