Question

Find the area of an isosceles triangle each of two equal sides being ‘α’ and the angle

included between them 45°
pls help me urgently ​

Answer 1

Answer:

$area \: of \: triangle \:$

By SAS (side - angle - side )

$\frac{1}{2} \times a \times b \times \sin({ \theta}) \\ where \: { \theta} \: is \: angle \: between \: \\ side \: a \: and \: b$

$where \: a \: = b \: = \alpha \\ { \theta} = 45 {}^{o} \\ area \: = \frac{1}{2} \times \alpha \times \alpha \times \sin( {45}^{o} ) \\ < < \sin(45 {}^{o} ) = \frac{1}{ \sqrt{2} } = \frac{ \sqrt{2} }{2} > > \\ = \frac{1}{2} { \alpha }^{2} \times \frac{1}{ \sqrt{2} } \\ = \frac{ \alpha {}^{2} }{2 \sqrt{2} }$