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The area of a right triangle is 600 \mathrm{~cm}^{2}. If the base of the triangle exceeds the altitude by 10 \mathrm{~cm}, find the dimensions of the triangle.
CBSE | Class 10 | Exercise 10e | Maths | Quadratic Equations | RS Aggarwal
The area of a right triangle is 600 \mathrm{~cm}^{2}. If the base of the triangle exceeds the altitude by 10 \mathrm{~cm}, find the dimensions of the triangle.

Let the altitude of the triangle be x \mathrm{~cm}

Therefore, the base of the triangle will be (x+10) \mathrm{cm}

\begin{array}{l} \text { Area of triangle }=\frac{1}{2} x(x+10)=600 \\ \Rightarrow(x+10)=1200 \\ \Rightarrow x^{2}+10 x-1200=0 \\ \Rightarrow x^{2}+(40-30) x-1200=0 \\ \Rightarrow x^{2}+40 x-30 x-1200=0 \\ \Rightarrow x(x+40)-30(x+40)=0 \\ \Rightarrow(x+40)(x-30)=0 \\ \Rightarrow x=-40 \text { or } x=30 \end{array}

\Rightarrow x=30 \quad[\because \text { Altitude cannot be negative]}

Thus, the altitude and base of the triangle are 30 \mathrm{~cm} and (30+10=40) \mathrm{cm}, respectively.

\begin{array}{l} (\text { Hypotenuse })^{2}=(\text { Altitude })^{3}+(\text { Base })^{2} \\ \Rightarrow(\text { Hypotenuse })^{2}=(30)^{2}(40)^{2} \\ \Rightarrow(\text { Hypotenuse })^{2}=900+1600=2500 \\ \Rightarrow(\text { Hypotenuse })^{2}=(50)^{2} \\ \Rightarrow(\text { Hypotenuse })=50 \end{array}

Thus, the dimensions of the triangle are:

\begin{array}{l} \text { Hypotenuse }=50 \mathrm{~cm} \\ \text { Altitude }=30 \mathrm{~cm} \\ \text { Base }=40 \mathrm{~cm} \end{array}

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