In Delta ABC,m angle B = 90^{circ}. If AB = 1 | vedclass

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(A) In a right-angled triangle $\Delta ABC$,where $\angle B = 90^{\circ}$,the sides $AB$ and $BC$ are the legs,and $AC$ is the hypotenuse.According to

Vedclass · Accepted Answer

$132$

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(A) In a right-angled triangle $\Delta ABC$,where $\angle B = 90^{\circ}$,the sides $AB$ and $BC$ are the legs,and $AC$ is the hypotenuse.According to the Pythagorean theorem: $AC^2 = AB^2 + BC^2$.Given $AB = 11$ and $BC = 60$,we have $AC^2 = 11^2 + 60^2$.$AC^2 = 121 + 3600 = 3721$.Taking the square root,$AC = \sqrt{3721} = 61$.The perimeter of $\Delta ABC$ is the sum of all its sides: $P = AB + BC + AC$.$P = 11 + 60 + 61 = 132$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$. If $AB = 11$ and $BC = 60$,find the perimeter of $\Delta ABC$.

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