Proofs

Proofs 1. Area = a*h/2 h is the raising of the natural elevation of the tri topple with base a. The macroscopical rectangle is make up of 2 smaller rectangles with compasss d*h and e*h. So the big triplicity is made up of ii smaller right-angled trigons whose areas are half of the smaller rectangles, i.e. d*h/2 and e*h/2. But a=d+e so the area of the large trigon is d*h/2+e*h/2 = a*h/2. If one of the angles on the base is ho-hum then the verification involves subtracting one half-rectangle from another. 2. Area = a*b*sin(C)/2 h=b*sin(C) so using the go place of 1 above gives the area of the triangle as a*b*sin(C)/2. 3. Area = a2*sin(B)*sin(C)/(2*sin(B+C)) In the result of 2, thither was nothing special about which angle and adjacent lines were used, since the analogous area will result, so we require Area = a*b*sin(C)/2 = b*c*sin(A)/2 = c*a*sin(B)/2. This in turn leads to the hell rule: a*b*c/(2*Area) = a/sin(A) = b/sin(B) = c/sin(C). [This turns out to b e the diameter of the circumcircle of the triangle (not be here) so if this was D then the Area would be a*b*c/(2*D) = D2*sin(A)*sin(B)*sin(C)/2, providing two extra formulae.] So we ass use b=a*sin(B)/sin(A) to get the area of the triangle as a2*sin(B)*sin(C)/(2*sin(A)). save A=?-B-C and sin(?-X)=sin(X) so sin(A)=sin(B+C), giving the result that the area of the triangle is a2*sin(B)*sin(C)/(2*sin(B+C)). 4. Area = sqrt(s*(s-a)*(s-b)*(s-c)) where s=(a+b+c)/2 is the semi-perimeter. From the diagram (and Pythagorass theorem) we induct a=d+e, b2=h2+d2, and c2=e2+h2. So d=sqrt(b2-h2), and c2 =(a-d)2+h2=a2-2*a*d+b2=a2-2*a*sqrt(b2-h2)+b2, which gives h=sqrt(4*a2*b2-(a2+b2-c2)2)/(2*a) and with some rearrangement h=sqrt((a+b+c)*(-a+b+c)*(a-b+c)*(a+b-c))/(2*a). But from the result of 1, the area of the triangle is a*h/2, i.e. sqrt((a+b+c)*(-a+b+c)*(a-b+c)*(a+b-c))/4, i.e. sqrt(((a+b+c)/2)*((-a+b+c)/2)*((a-b+c)/2)*((a+b-c)/2)), which is sqrt(s*(s-a)*(s-b)*(s-c)) when s= (a+b+c)/2. This result is called Herons for! mula. It...If you want to get a full essay, roam it on our website: BestEssayCheap.com

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