Question #373aa

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Answer 1 · 2017-03-18T10:39:12

$\frac{7}{2} (2 - \sqrt{2}) \approx 2.051 c m (3 d p) .$ $7/2(2-sqrt2)~~2.051 cm (3dp).$

Let each of the congruent side be $a = b = 7 c m .$ $a=b=7cm.$

$:." The Hypo. c="sqrt(7^2+7^2)=7sqrt2.$

$:." the Area of the "Delta="1/2*7*7=49/2 (cm)^2.$

If, $2s$ is the perimeter of $Delta," then, "2s=7+7+7sqrt2$

$rArr s=7/2(2+sqrt2)$

Hence, if $r$ is the Inradius, then, from Trigo ., we know that,

$Area=rs rArr 49/2=r{7/2(2+sqrt2)}$

$:. r=7/{(2+sqrt2)}={7(2-sqrt2)}/{(2+sqrt2)(2-sqrt2)}=7/2(2-sqrt2).$

In fact, there is a Direct Formula for $r$ of a Right-triangle

having sides $a, b$ and Hypo. $c : r=(a+b-c)/2.$

In our Problem, $r=(7+7-7sqrt2)/2=7/2(2-sqrt2).$

Taking, $sqrt2~~1.414 (3dp)$ , $r~~2.051 cm (3dp).$

Enjoy Maths.!