Question #24916

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Answer 1 · 2018-02-03T06:51:39

The length of AB $=sqrt(AC^2+BC^2)=sqrt(4^2_3^2)=5cm$

If $AD=BE=CF=h$ cm be the height of the prism then surface area

$=h(AB+BC+CA)+2DeltaABC$
$=h(5+3+4)+2*1/2*4*3$ $cm^2$

$=12h+12$ $cm^2$

By the problem

$12h+12=108$

$=>h=8cm$

Now $AF=sqrt(AC^2+CF^2)=sqrt(4^2+8^2)=4sqrt5cm=$

And $BF=sqrt(BC^2+CF^2)=sqrt(3^2+8^2)=sqrt73cm=$

Now three sides of $Delta ABF$ are known . So area of the surface $ABF$ can be found out using Heron's formula given below

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