An isosceles triangle has height $15 \text{ cm}$ with respect to unequal side. It's height is diminished to $5 \text{ cm}$ keeping equal side constant so that the base increases. How much the base is increased??
How base of isosceles triangle change with change in height
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$\begingroup$
geometry
euclidean-geometry
1 Answers
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You cannot tell. Let $x$ be the length of the equal sides and let $2b$ be the base when the height is 15 and $2B$ be the base when the height is 5.
Then $$ x^2 = 15^2 + b^2 = 5^2 + B^2. $$ Hence $$ B-b = \sqrt{x^2-5^2}-\sqrt{x^2-15^2} $$ which is not constant in $x$.
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0But the increased length should be unique.So there should be a way – 2017-01-09
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0@Anonymous $B-b$ depends by the length of the equal sides, so it is not unique in this sense. – 2017-01-09