Answer to Question #228621 in Algebra for Keleko Pierre Mich

Question

, Let ABC be a right-angle triangle in B and of area S. Let a be the internal angle with vertex in A and we consider the point D on the side BC such that the segment AD can form an angle of width a/2 with AB then the area of triangle ABC is ?

A.     Greater than S/2

B.     Equal to S/2

C.     Less than S/2

None of the above

Answer 1

Area of triangle ABC =s

$S=\frac{1}{2}×AB×BC....(1)$

Area of triangle ABD=s

$S_1=\frac{1}{2}×AB×BD....(2)$

From DABC

$tan \space a=\frac{BC}{AB}\\BC=AB=AB \space\space tan \space a.....(3)$

From DABD

$tan\frac{a}{2}=\frac{BD}{AB}\\BD=AB=AB \space tan \frac {a}{2}....(4)$

We put value of BC and BD in equition 1 and 2

$S=\frac{1}{2}×AB×AB\space tan \space a$

$S=\frac{1}{2}(AB)^2 tan\space a......(5)\\S_1=\frac{1}{2}(AB)^2 tan\frac{a}{2}......(6)$

Dividing equition 5 by 6 we get

$\frac{s}{s_1}=\frac{\frac{1}{2}(AB)^2 tan\space a}{\frac{1}{2}(AB)^2 tan\frac{ a}{2}}\\s_1=\frac{tan\frac{a}{2}}{tan \space a}s\\s_1=\frac{tan\frac{a}{2}}{\frac{2tan\frac{a}{2}}{1-tan^2\frac{a}{2}}}s\\s_1=(1-tan^2\frac{a}{2})\frac{s}{2}....(7)$

Angle

$a(0°,90°),\frac{a}{2}(0°,45°)$

And

$tan \frac{a}{2}(0,1)\\1-tan^2\frac{a}{2}(0,1)$

Since $1-tan^2\frac{a}{2}$ belongs from 0 to 1

Therefore from equition 7 we can conclude that s₁ is less than $\frac{s}{2}$