Properties of Triangles 2 Question 12

12. In a triangle of base a, the ratio of the other two sides is r(<1). Show that the altitude of the triangle is less than or equal to ar1r2.

(1991, 4M)

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Solution:

  1. Let ABC be a triangle with base BC=a and altitude AD=p, then

Area of ABC=12bcsinA

Also, area of ABC=12ap

12ap=12bcsinA

p=bcsinAa

p=abcsinAa2

p=abcsinA(sin2Bsin2C)a2(sin2Bsin2C)

=abcsinAsin(B+C)sin(BC)(b2sin2Ac2sin2A) using sine rule, asinA=bsinB=csinC=abcsin2Asin(BC)(b2c2)sin2A=abcsin(BC)b2c2=ab2rsin(BC)b2b2r2=arsin(BC)1r2par1r2[sin(BC)1]



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