Given :
• Hypotenuse (H) = 15 in
• Base (B) = 12 in
To calculate :
• Perpendicular (P) [x]
Calculation :
By using pythagoras property,
[tex] \longmapsto \sf { H^2 =B^2 + P^2 } [/tex]
[tex] \longmapsto \sf { (15)^2 =(12)^2 +(x)^2 } [/tex]
[tex] \longmapsto \sf { x^2 = (15)^2 - (12)^2 } [/tex]
[tex] \longmapsto \sf { x^2 = 225 - 144 } [/tex]
[tex] \longmapsto \sf { x^2 = 81 } [/tex]
[tex] \longmapsto \sf { x = \sqrt{81} } [/tex]
[tex] \longmapsto \sf \red { x = 9 \: in} [/tex]
So, the value of x (perpendicular) is 9 in.
