Jiseph
Answered

Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Get quick and reliable solutions to your questions from knowledgeable professionals on our comprehensive Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

WILL MARK BRAINLIEST

WILL MARK BRAINLIEST class=

Sagot :

[tex]\huge\mathfrak{\underline{Answer:}}[/tex]

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎[tex]\large\bf{=20 }[/tex]

__________________________________________

[tex]\large\bf{\underline{Given:}}[/tex]

  • A trapezium ABDE with sides 38 , 16 , 50 and x

[tex]\large\bf{\underline{To\: find :}}[/tex]

  • The value of x

[tex]\large\bf{\underline{Construction:}}[/tex]

  • Join C to E || AB

[tex]\setlength{\unitlength}{0.8cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(.3,2.5){\large\bf 12}\put(2.8,.3){\large\bf 16}\put(1.02,1.02){\framebox(0.3,0.3)}\put(.7,4.8){\large\bf D}\put(.8,.3){\large\bf C}\put(5.8,.3){\large\bf E}\qbezier(4.5,1)(4.3,1.25)(4.6,1.7)\put(3.8,1.3){\large\bf $\Theta$}\end{picture}[/tex]

[tex]\setlength{\unitlength}{0.75cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large 16 cm}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large 38 cm}\put(-0.5,-0.4){\bf B}\put(-0.5,3.2){\bf C}\put(5.3,-0.4){\bf A}\put(5.3,3.2){\bf E}\end{picture}[/tex]

[tex]\large\bf{\underline{Hence,}}[/tex]

  • CD = BD-BC

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎⟹CD = 50 - 38

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎⟹CD = 50 - 38

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎⟹CD = 12

[tex]\large\bf{Since,}[/tex]

  • AB || CE And ABDE is a trapezium , Therefore ABCE is a rectangle

[tex]\large\bf{\underline{Therefore}}[/tex]

  • AB = CE

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎⟹ CE = 16

[tex]\large\bf{\underline{In\: triangle\:DCE}}[/tex]

[tex]{\large\bf{Using\: Pythagoras\: theorem:}[/tex]

[tex]\boxed{\large\bf\pink{DE^2 = CD^2 + CE^2}}[/tex]

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎[tex]\large\bf{⟹x^2 = 12^2 + 16^2}[/tex]

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎[tex]\large\bf{⟹x^2 = 144 + 256}[/tex]

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎[tex]\large\bf{⟹x^2 = 400}[/tex]

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎[tex] \large \bf \:⟹ {x} = \sqrt{400} [/tex]

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎[tex]\boxed{\large\bf{⟹x= 20}}[/tex]