The base and altitude of right angled triangle are $119$ cm and $120$ cm respectively. If the hypotaneous of triangle be $x$ where $x$ is an integer. find out the value of $\sqrt{x}$ .

169

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\sqrt{13}

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13

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\sqrt{239}

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Solution

The correct option is C $13$
We know,
Hypotenuse $= \sqrt{b a s e^{2} + a l t i t u d e^{2}} x = \sqrt{119^{2} + 120^{2}} x = \sqrt{14400 + 14161} x = \sqrt{28561} x = 169$

And $\sqrt{x} = \sqrt{169} = 13$
So, option $(c)$ correct.

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The base and altitude of right angled triangle are 119 cm and 120 cm respectively. If the hypotaneous of triangle be x where x is an integer. find out the value of √x.

The correct option is C 13We know, Hypotenuse =√base2+altitude2x=√1192+1202x=√14400+14161x=√28561x=169 And √x=√169=13 So, option(c) correct.

The base and altitude of right angled triangle are $119$ cm and $120$ cm respectively. If the hypotaneous of triangle be $x$ where $x$ is an integer. find out the value of $\sqrt{x}$ .

The correct option is C $13$
We know,
Hypotenuse $= \sqrt{b a s e^{2} + a l t i t u d e^{2}} x = \sqrt{119^{2} + 120^{2}} x = \sqrt{14400 + 14161} x = \sqrt{28561} x = 169$

And $\sqrt{x} = \sqrt{169} = 13$
So, option $(c)$ correct.