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# Find the unknown length x in the following figures (Fig 6.29):

Q. Find the unknown length x in the following figures (Fig $\small 6.29$):

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As we know in a Right-angled Triangle: By Pythagoras Theorem,

$(Hypotenus)^2=(Base)^2+(Perpendicular)^2$

So, using this theorem,

i)

$x^2=3^2+4^2$

$x^2=9+16$

$x^2=25$

$x=5$.

ii)

$x^2=6^2+8^2$

$x^2=36+64$

$x^2=100$

$x=10$

iii)

$x^2=15^2+8^2$

$x^2=225+64$

$x^2=289$

$x=17$

iv)

$x^2=24^2+7^2$

$x^2=576+49$

$x^2=625$

$x=25$

v) in this question as we can see from the figure, it is making the right angle with the half-length of x, so

$\left ( \frac{x}{2} \right )^2+12^2=37^2$

$\frac{x^2}{4} +12^2=37^2$

$\frac{x^2}{4} =37^2-12^2$

.$\frac{x^2}{4} =1369-144$

$\frac{x^2}{4} =1225$

$x^2=4\times 1225$

$x=2\times35$

$x=70.$

vi)

$x^2=12^2+5^2$

$x^2=144+25$

$x^2=169$

$x=13$

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