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ABC is an isosceles triangle inscribed in a circle of radius r. If AB=AC and h is the altitude form A to BC. If P is perimeter and A is the area of the triangle then find the valueof lim_(hto0)(A)/(P^(3)). |
Answer» Solution :In `DeltaABC, AB=AC, AD_|_BC(D" is midpoint of "BC)` Let r= radius of CIRCUMCIRCLE `:.""OA=OB=OC=r` Now `BD=sqrt(BO^(2)-OD^(2))=sqrt(r^(2)-(h-r)^(2))=sqrt(2rh-h^(2))` `:.""BC=2sqrt(2rh-h^(2))` Also` AB^(2)=BD^(2)+AD^(2)=2hr-h^(2)+h^(2)=2hr`. `:.""AB=AC=sqrt(2hr)` `:."""PERIMETER",P=2sqrt(2rh-h^(2))+2sqrt(2hr)` `:."""Area of "DeltaABC=(1)/(2)xxBCxxAD=hsqrt(2rh-h^(2))` So,`""underset(hto0)lim(A)/(P^(3))=(hsqrt(2rh-h^(2)))/(8(sqrt(2rh-h^(2))+sqrt(2hr))^(3))` `=underset(hto0)lim(h^(3//2)sqrt(2r-h))/(8h^(3//2(sqrt(2r-h)+sqrt(2r))^(3)))` `=underset(hto0)lim(sqrt(2r-h))/(8[sqrt(2r-h)+sqrt(2r)]^(3))` `=(sqrt(2r))/(8(sqrt(2r)+sqrt(2r))^(3))=(1)/(128r)` |
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