construct triangle DEF such that DE=5cm, DF=3, and m <EDF=90 DEGREE
solve question 9.
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adityakumar36:
thank you
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DE=5CM
DF=3CM
ED= ?
EDF=9O DEGREE
![{h}^{2} = {b}^{2} + {h}^{2} \\ {5}^{2} = {3}^{2} + {h}^{2} \\ 25 = 9 + {h}^{2} \\ 25 - 9 = {h}^{2} \\ 16 = {h}^{2} \\ \sqrt[2]{16} = h \\ 4 = h](https://tex.z-dn.net/?f=+%7Bh%7D%5E%7B2%7D++%3D++%7Bb%7D%5E%7B2%7D++%2B++%7Bh%7D%5E%7B2%7D++%5C%5C++%7B5%7D%5E%7B2%7D++%3D++%7B3%7D%5E%7B2%7D++%2B++%7Bh%7D%5E%7B2%7D++%5C%5C+25+%3D+9+%2B++%7Bh%7D%5E%7B2%7D++%5C%5C+25+-+9+%3D++%7Bh%7D%5E%7B2%7D++%5C%5C+16+%3D++%7Bh%7D%5E%7B2%7D++%5C%5C++%5Csqrt%5B2%5D%7B16%7D++%3D+h+%5C%5C+4+%3D+h+ "{h}^{2} = {b}^{2} + {h}^{2} \\ {5}^{2} = {3}^{2} + {h}^{2} \\ 25 = 9 + {h}^{2} \\ 25 - 9 = {h}^{2} \\ 16 = {h}^{2} \\ \sqrt[2]{16} = h \\ 4 = h")
DF=3CM
ED= ?
EDF=9O DEGREE
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