Question

Please solve. I need it fast.

Answer 1

$\setlength{\unitlength}{1.6mm}\begin{picture}(30,20)\linethickness{0.1mm}\put(0,0){\line(0,1){37.5}}\put(0,0){\line(1,0){25}}\put(25,0){\line(-2,3){25}}\end{picture}\put(-32,18){5}\put(-18,-1.5){}\put(-18,20){6}$

$\:\:$

Using Pythagoras theorem

$\:\:$

$\red{\tt{{Hypotenuse}^{2} = {Height}^{2} + {Base}^{2} }}$

$\red{\tt{\implies {(6unit)}^{2} = {(5unit)}^{2} + {(Base)}^{2} }}$

$\red{\tt{\implies {36unit}^{2} = {25unit}^{2} + {Base}^{2} }}$

$\red{\tt{\implies {36unit}^{2} - {25unit}^{2} = {Base}^{2} }}$

$\red{\tt{\implies {11unit}^{2} = {Base}^{2} }}$

$\red{\tt{\implies \sqrt{{11unit}^{2}} = Base}}$

$\red{\tt{\implies 3.31unit = Base }}$

$\:\:$

$\green{\tt{Area\: of\: triangle = \frac{1}{2} \times base \times height }}$

$\green{\tt{\implies Area\: of\: triangle = \frac{1}{2} \times 3.31unit \times 5unit }}$

$\green{\tt{\implies Area\: of\: triangle = \frac{1}{2} \times {16.55unit}^{2} }}$

$\green{\tt{\implies Area\: of\: triangle = {8.27unit}^{2} }}$

$\green{\tt{\implies Area\: of\: triangle = {8.3unit}^{2} }}$

$\:\:$

$\blue{\bf{ Area\: of\: triangle = {8.3unit}^{2} }}$

Answer 2

Question -

$\sf{To \ the \nearest \ tenth, \ find \ the \ area \ of \ the \ triangle \ in \ the \diagram \ given \ below \ - }$

Options -

$\sf{ Option \ 1 \ - \ 16.6 \ square \ units . } \\ \\ \sf{ Option \ 2 \ - \ 15.0 \ square \ units . } \\ \\ \sf{ Option \ 3 \ - \ 10.0 \ square \ units . } \\ \\ \sf{ Option \ 4 \ - \ 8.3 \ square \ units . }$

Solution -

$\sf{ According \ To \ The \ Pythagoras \ Theorem \ - } \\ \\ \sf{ {a}^2 + {b}^2 = {c}^2 } \\ \\ \sf{\bold{ Where \ - }} \\ \\ \sf{ a \ and \b \ are \: two \ sides . } \\ \\ \sf{c \ is \ the \ hypotenuse . } \\ \\ \sf{ Substituting \ the \ given \ values \ - } \\ \\ \sf{ => a^2 + 5^2 = 6^2 } \\ \\ \sf{ => a^2 + 25 = 36 } \\ \\ \sf{ => a^2 = 36 - 25 } \\ \\ \sf{ a^2 = 11 } \\ \\ \sf{ a = + \sqrt{11} \approx \3.3 \ as \ it \ can't \ be \ negative . } \\ \\ \sf{ Now, \ we \ know \ that \ - } \\ \\ \sf{ Area \ Of \ A \ right \ Angled \ Triangle = \dfrac{1}{2} \ \times \ base \ \times \height } \\ \\ \sf{ Substituting \ the \ given \ values \ - } \\ \\ \sf{ Area \ Of \ Given \Triangle \ - } \\ \\ \sf{ => \dfrac{1}{2} \times 3.3 \times 5 } \\ \\ \sf{ => 8.3 \ {unit}^2 } \\ \\ \sf{ Hence \ The \ Correct \ Option \ Is \ D }$