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den301095 [7]
3 years ago
5

WILL MARK BRAINLIEST

Mathematics
2 answers:
Anit [1.1K]3 years ago
8 0

Answer:

x = 20

Step-by-step explanation:

50 - 38 = 12

a^2 + b^2 = c^2

16^2 + 12^2 = x^2

256 + 144 = x^2

400 = x^2

x = \sqrt{400}

x = 20

Rus_ich [418]3 years ago
5 0

\huge\mathfrak{\underline{Answer:}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{=20 }

__________________________________________

\large\bf{\underline{Given:}}

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

\large\bf{\underline{To\: find :}}

  • The value of x

\large\bf{\underline{Construction:}}

  • Join C to E || AB

\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}

\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}

\large\bf{\underline{Hence,}}

  • CD = BD-BC

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎⟹CD = 50 - 38

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎⟹CD = 50 - 38

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎⟹CD = 12

\large\bf{Since,}

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

\large\bf{\underline{Therefore}}

  • AB = CE

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎⟹ CE = 16

\large\bf{\underline{In\: triangle\:DCE}}

{\large\bf{Using\: Pythagoras\: theorem:}

\boxed{\large\bf\pink{DE^2 = CD^2 + CE^2}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹x^2 = 12^2 + 16^2}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹x^2 = 144 + 256}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹x^2 = 400}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large \bf \:⟹ {x} =   \sqrt{400}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\boxed{\large\bf{⟹x= 20}}

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