Question

the hypotenuse of an isosceles right triangle is
$\sqrt{32}$
what is its area?

Answer 1

Let the base be a

Perpendicular = a

Hypotenuse = sqrt ( a^2 + a^2)

=> sqrt ( 32) = sqrt ( a^2 + a^2)

=> 32 = 2a^2

=> a^2 = 16

=> a = 4

Base = 4 cm

Perpendicular = 4 cm

Area = Base × Perpendicular / 2

= 4 × 4 / 2

= 8 cm^2

Answer 2

Any odd degree polynomial with real coefficients has at least one real root. In what follows, I ..... Oops - nice answer to the wrong question. 1 .... So necessary conditions are.