Question #373aa

1 Answer
Mar 18, 2017

#7/2(2-sqrt2)~~2.051 cm (3dp).#

Explanation:

Let each of the congruent side be #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.!