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Question-189701




Question Number 189701 by DAVONG last updated on 20/Mar/23
Commented by Rasheed.Sindhi last updated on 23/Mar/23
16) 59 rectangles(including squares)    having sum multiple of 5.  I will write detailed solution if you  give some response.
$$\left.\mathrm{16}\right)\:\mathrm{59}\:{rectangles}\left({including}\:{squares}\right)\: \\ $$$$\:{having}\:{sum}\:{multiple}\:{of}\:\mathrm{5}. \\ $$$${I}\:{will}\:{write}\:{detailed}\:{solution}\:{if}\:{you} \\ $$$${give}\:{some}\:{response}. \\ $$
Answered by Rasheed.Sindhi last updated on 21/Mar/23
17)   Contradiction!  As the triangle is right-angled  the largest angle is 90°  • cos 90=((a+(√b))/c)=0       a+(√b) =0       a=−(√b)       ∵ b∈P      ∴ a∉Z
$$\left.\mathrm{17}\right)\:\:\:\mathrm{Contradiction}! \\ $$$${As}\:{the}\:{triangle}\:{is}\:{right}-{angled} \\ $$$${the}\:{largest}\:{angle}\:{is}\:\mathrm{90}° \\ $$$$\bullet\:\mathrm{cos}\:\mathrm{90}=\frac{{a}+\sqrt{{b}}}{{c}}=\mathrm{0} \\ $$$$\:\:\:\:\:{a}+\sqrt{{b}}\:=\mathrm{0} \\ $$$$\:\:\:\:\:{a}=−\sqrt{{b}}\: \\ $$$$\:\:\:\:\because\:{b}\in\mathbb{P} \\ $$$$\:\:\:\:\therefore\:{a}\notin\mathbb{Z} \\ $$

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