# Ex.4.2 Q5 Quadratic Equations Solutions - NCERT Maths Class 10

## Question

The altitude of right triangle is \(7\,\rm{ cm}\) less than its base. If the hypotenuse is \(13\,\rm{ cm},\) find the other two sides.

## Text Solution

**What is known?**

i) Altitude of right triangle is \(7\, \rm{cm}\) less than its base.

ii) Hypotenuse is \(13\, \rm{cm}\)

**What is Unknown?**

The measure of the two sides of a given right triangle.

**Reasoning:**

In a right triangle, altitude is one of the sides. Let the base be \(x\, \rm{cm.}\)

The altitude will be \((x - 7)\, \rm{cm.}\)

Next, we can apply the Pythagoras theorem to the given right triangle.

**Pythagoras theorem:**

\(\text{Hypotenuse}^2= \text{side 1}^2+ \text{side 2}^2\)

\[13{}^\text{2}\text{ }=\text{ }x{}^\text{2}\text{ }+\text{ }(x\text{ }-7)\text{ }{}^\text{2}\]

**Steps:**

\[\begin{align}13&= x + {(x - 7)^2}\\169 &= x + x - 14x + 49\\169 &= 2x - 14x + 49\\

2x - 14x + 49-169 &= 0\\2x - 14x - 120 &= 0\\&\frac{{2x - 14x - 120 = 0}}{2}\\x - 7x-60 &= 0\\

x - 12x + 5x-60 &= 0\\x\left( {x - 12} \right) + 5\left( {x - 12} \right) &= 0\\(x + 5)(x - 12) &= 0\\

x-12 = 0 &\qquad x + 5 = 0\\x = 12 &\qquad \quad\;\; x = - 5\end{align}\]

We know that the value of the base cannot be negative.

\(\therefore\) Base \(= x = 12 \,\rm{cm}\)

Altitude \(= 12 -7 = 5 \,\rm{cm}\)

Lengths of two sides are \(12\, \rm{cm}\) and \(5\,\rm{ cm.}\)