Question

prove that
Ad=a√3/2 in Pythagorean theorem (of right triangle)

Answer 1

According to the question...


=>Let us consider a ∆ABC
=>on drawing the =>perpendicular AD from A to BC
=>where D is =90°
=>we get a ∆ADB and ∆ADC
=>on considering ;∆ADB be the right triangle
=>by applying the Pythagoras theorem ;we get
$=> {(AB)}^{2} = {(AD) }^{2} + {(BD)}^{2} \\ \\ =>{(AD)}^{2} ={(AB) }^{2} - {(BD)}^{2}\\ \\ =>{(AD)}^{2} ={(a)}^{2} - (\frac{ {a}^{2} }{2} ) \\ \\ =>{(AD)}^{2} = {(a)}^{2} - \frac{ {a}^{2} }{4}$

=> $\frac{ {4a}^{2} - {(a)}^{2} } {4}$

=> $4 {AD}^{2} = {3a}^{2}$

=> $AD = \sqrt{ \frac{ {3a}^{2} }{4} }$

=> $AD= a\frac{ \sqrt{3} }{2} \: units$

# be brainly

Answer 2

According to the definition, the Pythagoras Theorem formula is given as:

Hypotenuse2 = Perpendicular2 + Base2

c2 = a2 + b2

The side opposite to the right angle (90°) is the longest side (known as Hypotenuse) because the side opposite to the greatest angle is the longest.