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?? ????? ??????? ?? ?????? 12 cm ?? ?? ???? ???? 5 cm ????? ?? ? ??????? ?? ????? ?????? || ??? ????????? ????? ???? ? ??????? ?????? ?? ???? ???? ?? ????? ?? ???? ??

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Given;that;hypotaneous=5;cm;and;perimeter=12;cm\*Let;base;and;height;are;a;and;b;respectively\* perimeter=a+b+c=12\*Rightarrow a+b=12-5=7\*Rightarrow a+b=7...........eq(1)\* Using;pythagorus;theorem ;c^2=a^2+b^2\*25=a^2+b^2\*Rightarrow a^2+b^2+2ab=25-2ab\* Rightarrow (a+b)^2=25-2ab\*from;eq(1);we;have;a+b=7\* Rightarrow (7)^2=25-2ab\*Rightarrow ab=12\*Rightarrow b=frac12a\* Now,;put;value;b;in;eq(1)\*Rightarrow a+frac12a=7\* Rightarrow a^2-7a+12=0\*Rightarrow (a-4)(a-3)=0\*Rightarrow a=4;or;a=3\*So,;other;two;sides;are;a=4;cm;and;b=3;cm;or;vice;verca\*Area;of;triangle=frac12	imes a	imes b=frac12	imes 4	imes 3=6;cm^2\* Using;herons;formula\*Area;of;triangle=sqrts(s-a)(s-b)(s-c)\* s=fraca+b+c2=frac3+4+52=6\* Area;of;triangle=sqrt6(6-3)(6-4)(6-5)=sqrt36=6;cm^2\* Hence,;siders;are;3;cm;and;4;cm;;and;area;of;triangle;is;6;cm^2

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Deependra Verma

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