Question

2. Find AC using the Pythagorean theorem

Answer 1

`{\huge{\pink{\underline{\underline{Solution:}}}}}`

In triangle,ADB

p(perpendicular)=BD=12 cm

h(hypotenuse)=AB=15cm

b(base)=AD=??

Let the AD be x

Using Pythagorean theorem,

`{p}^(2) + {b}^(2) = {h}^(2) \\</p><p>=>{12}^(2) + {x}^(2) = {15}^(2) \\</p><p>=>144+{x}^(2)=225\\</p><p>=>{x}^(2)=225-144\\</p><p>=>{x}^(2)=81\\</p><p>=>x= √(81) \\</p><p>=>x= √(9×9)`

Hence,x=AD=9

So,AC=AD+DC=9+16=25.

Answer 2

ad=x
12^2+x^2=15^2
144+x^2=225
X^2=225-144
X^2=81
X=9
ad=9
ac=ad+dc
ac=9+16=25