Question

(x-4)^2+ 5^2 = 13^2​

Answer 1

$\Large{\underline{\underline{\bf{Solution :}}}}$

As, we have to find the value of x in given below equation

$\sf{(x - 4)^2 + (5)^2 = (13)^2}$

So,

$\sf{→(x - 4)^2 + (5)^2 = (13)^2} \\ \\ \sf{→(x - 4)^2 + 25 = 169} \\ \\ \sf{→(x - 4)^2 = 169 - 25} \\ \\ \sf{→(x - 4)^2 = 144} \\ \\ \sf{→x - 4 = \sqrt{144}} \\ \\ \sf{→x - 4 = (\sqrt{12})^2} \\ \\ \sf{→x - 4 = 12} \\ \\ \sf{→x = 12 + 4} \\ \\ \sf{→x = 16} \\ \\ \Large{\implies{\boxed{\boxed{\sf{x = 16}}}}}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\put(1,1.2){\line(1,0){6.5}}\end{picture}$

$\setlength{\unitlength}{0.19 cm}}\begin{picture}(12,4)\thicklines\put(12,40){ \#\:STAY\:HOME }\put(11,35){ \#\:STAY\: SAFETY }\put(20,20){\circle{10}}\put(19.5,21){\line(-1,0){1.5}}\put(20.5,21){\line(1,0){1.5}}\put(18.8,20.3){\circle*{0.7}}\put(21.1,20.3){\circle*{0.7}}\put(20,20.4){\line(0,-1){1.4}}\put(20.7,18.4){\line(-1,0){1.57}}\put(18.6,16.56){\line(0,-1){2}}\put(21.4,16.56){\line(0,-1){2}}\put(19,14.55){\line(-1,0){5}}\put(21,14.55){\line(1,0){5}}\put(14,14.55){\line(0,-1){16}}\put(26,14.55){\line(0,-1){16}}\put(14,14.55){\line(-4,3){5}}\put(26,14.55){\line(4,3){5}}\put(14,11){\line(-4,3){7.5}}\put(26,11){\line(4,3){7.5}}\put(9,18.3){\line(0,1){8.3}}\put(31,18.3){\line(0,1){8.3}}\put(33.5,16.6){\line(0,1){10}}\put(6.5,16.6){\line(0,1){10}}\put(3.5,26.5){\line(1,0){33}}\put(3.5,26.5){\line(0,1){20}}\put(3.5,46.5){\line(1,0){33}}\put(36.5,26.5){\line(0,1){20}}\put(14,-0.6){\line(1,0){12}}\put(14,-1.5){\line(1,0){12}}\put(14.4,-1.6){\line(0,-1){18.5}}\put(25.6,-1.6){\line(0,-1){18.5}}\put(20,-1.6){\line(0,-1){3}}\put(20,-4.6){\line(-1,-6){2.6}}\put(20,-4.6){\line(1,-6){2.6}}\put(14.4,-20){\line(-5,-6){3}}\put(25.6,-20){\line(5,-6){3}}\put(17.3,-20){\line(0,-1){3.7}}\put(17.3,-20){\line(-1,0){3}}\put(17.3,-23.6){\line(-1,0){6}}\put(22.6,-20){\line(0,-1){3.7}}\put(22.6,-20){\line(1,0){3}}\put(22.6,-23.6){\line(1,0){6}}\end{picture}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\put(1,1.2){\line(1,0){6.5}}\end{picture}$