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1 Answer
Jim G.
Jan 31, 2018

#p=2sqrt10#

Explanation:

#angleBCA=90^@tocolor(blue)"angle in a semicircle"#

#"calculate the lengths of the sides AB, BC and AC"#

#"since the x-coordinates of A and B are zero then"#

#AB=10-(-4)=14#

#"calculate BC and AC using the "color(blue)"distance formula"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(d=sqrt((x_2-x_1)^2+(y_2-y_1)^2))color(white)(2/2)|)))#

#BC" let "(x_1,y_1)=(0,10)" and "(x_2,y_2)=(p,0)#

#rArrBC=sqrt((p-0)^2+(0-10)^2)=sqrt(p^2+100)#

#AC" let "(x_1,y_1)=(0,-4)" and "(x_2,y_2)=(p,0)#

#rArrAC=sqrt((p-0)^2+(0+4)^2)=sqrt(p^2+16)#

#"using "color(blue)"Pythagoras' theorem ""in "triangleABC#

#BC^2+AC^2=AB^2#

#rArr(sqrt(p^2+100))^2+(sqrt(p^2+16))^2=14^2#

#rArrp^2+100+p^2+16=196#

#rArr2p^2+116=196#

#"subtract 116 from both sides"#

#rArr2p^2=196-116=80#

#"divide both sides by 2"#

#rArrp^2=40#

#color(blue)"take square root of both sides"#

#rArrp=+-sqrt40larrcolor(blue)"note plus or minus"#

#"since "p>0#

#"then "p=sqrt40=sqrt(4xx10)=2sqrt10" in simplest form"#

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