1.

The area of right angled triangle is 500 cm^2. If the base of the triangle is 10 less than the altitude of the triangle, what are the dimensions of the triangle?(a) base = 32 cm, altitude = 42 cm(b) base = 42 cm, altitude = 62 cm(c) base = 76 cm, altitude = 42 cm(d) base = 32 cm, altitude = 55 cmThe question was posed to me in semester exam.This is a very interesting question from Solution of Quadratic Equation by Squaring Method in chapter Quadratic Equation of Mathematics – Class 10

Answer»

The correct answer is (a) base = 32 cm, ALTITUDE = 42 cm

Explanation: Let the altitude of the triangle be x cm.

The base will be x-10 cm.

The area of triangle is 500 cm^2

\(\frac {1}{2}\) × base × altitude = 672

\(\frac {1}{2}\) × (x-10) × x = 672

x^2-10X=1344

Adding \(\frac {b^2}{4}\) on both sides, where b=-10

x^2 – 10x + \(\frac {-10^2}{4}=\frac {-10^2}{4}\) + 1344

x^2 – 10x + \(\frac {100}{4}=\frac {100}{4}\) + 1344

x^2 – 10x + \(\frac {100}{4}=\frac {5476}{4}\)

\((x-\frac {10}{2})\)^2=\((\frac {74}{2})\)^2

x-\(\frac {10}{2}\)=±\(\frac {74}{2}\)

x = \(\frac {74}{2}+\frac {10}{2}=\frac {84}{2}\) = 42 and x = \(\frac {-74}{2}+\frac {10}{2}=\frac {-64}{2}\) = -32

Since, length cannot be NEGATIVE, hence x=42 cm

Base will be x-10=42-10=32 cm


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