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STEP II 1995 question 7 solutionTSR Wiki > Study Help > Exams and Qualifications > A Levels > STEP > STEP II 1995 question 7 solution Area of Area of Area of (Due to Hence: Area of
Hence:
Hence:
Hence:
Hence:
Now consider the following:
Hence:
Hence:
Consider that the sphere will rest in the incircle of the triangle, and hence will satisfy:
(Where "d" is the distance to be calculated. [A diagram may help visualisation of this property].). Consider also:
Hence:
Hence:
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![\triangle ABC = \frac{1}{2} r \left[ a + b + c \right] = sr](http://www.thestudentroom.co.uk/latexrender/pictures/b4/b4830cbf1e08c9f5e8dc4f47a4ffe1f5.png "\triangle ABC = \frac{1}{2} r \left[ a + b + c \right] = sr")
![\triangle ^{2} = \frac{1}{16} \left( 4b^{2} c^{2} - 4b^{2} c^{2} \cos ^{2} { \alpha } \right) = \frac{1}{16} \left( 4b^{2} c^{2} - \left[ 2bc \cos{ \alpha } \right] ^{2} \right)](http://www.thestudentroom.co.uk/latexrender/pictures/44/445cf9e306760753f0fe86bd73d6e621.png "\triangle ^{2} = \frac{1}{16} \left( 4b^{2} c^{2} - 4b^{2} c^{2} \cos ^{2} { \alpha } \right) = \frac{1}{16} \left( 4b^{2} c^{2} - \left[ 2bc \cos{ \alpha } \right] ^{2} \right)")
![\triangle ^{2} = \frac{1}{16} \left( 4b^{2} c^{2} - \left[ b^{2} + c^{2} - a^{2} \right] ^{2} \right) = \frac{1}{16} \left( \left[ 2bc - b^{2} - c^{2} + a^{2} \right] \left[ 2bc + b^{2} + c^{2} - a^{2} \right] \right)](http://www.thestudentroom.co.uk/latexrender/pictures/db/dbd5df6958d48208a2eb88189f59033b.png "\triangle ^{2} = \frac{1}{16} \left( 4b^{2} c^{2} - \left[ b^{2} + c^{2} - a^{2} \right] ^{2} \right) = \frac{1}{16} \left( \left[ 2bc - b^{2} - c^{2} + a^{2} \right] \left[ 2bc + b^{2} + c^{2} - a^{2} \right] \right)")
![\triangle ^{2} = \frac{1}{16} \left( \left[ a^{2} - \left( b - c \right) ^{2} \right] \left[ \left( b + c \right) ^{2} - a^{2} \right] \right)](http://www.thestudentroom.co.uk/latexrender/pictures/51/51f9134c111fef3bc4fda447b5c107d5.png "\triangle ^{2} = \frac{1}{16} \left( \left[ a^{2} - \left( b - c \right) ^{2} \right] \left[ \left( b + c \right) ^{2} - a^{2} \right] \right)")
![\triangle ^{2} = \frac{1}{16} \left( \left[ \left( a - (b - c) \right) \left( a + (b - c) \right) \right] \left[ \left( (b + c) - a \right) \left( (b + c) + a \right) \right] \right)](http://www.thestudentroom.co.uk/latexrender/pictures/2f/2fd0093750e71aaa2bb1bef2109edeeb.png "\triangle ^{2} = \frac{1}{16} \left( \left[ \left( a - (b - c) \right) \left( a + (b - c) \right) \right] \left[ \left( (b + c) - a \right) \left( (b + c) + a \right) \right] \right)")
![\triangle ^{2} = \frac{1}{16} \left( \left[ \left( 2(s - b) \right) \left( 2(s - c) \right) \right] \left[ \left( 2(s - a) \right) \left( 2s \right) \right] \right) = s(s - a)(s - b)(s - c)](http://www.thestudentroom.co.uk/latexrender/pictures/5c/5cca3764e9155652935c5eb2b218c98f.png "\triangle ^{2} = \frac{1}{16} \left( \left[ \left( 2(s - b) \right) \left( 2(s - c) \right) \right] \left[ \left( 2(s - a) \right) \left( 2s \right) \right] \right) = s(s - a)(s - b)(s - c)")
