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

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## 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.

Video Solution
Ex 4.2 | Question 5

## 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 \end{align}

\begin{align} 2x \!\!- \!\!14x \!\!+ \!49 \!- \!169 & \!=\! 0\\ 2x - 14x \\ - 120 &= 0\\ \frac{{ 2x - 14x - 120 }}{2}&= 0\\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\\ \end{align} \begin{align} 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 $$5\,\rm{ cm.}$$

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