Question

If the perpendicular sides of a
right angled triangle are 8 cm
and 12 cm long, then what is the
area of that right angled
triangle?​

Answer 1

${\large{\pmb{\sf{\underline{About \; question}}}}}$

☀ Correct question: If the perpendicular sides of a right angled triangle are 8 cm and 12 cm long, then what is the area of that right angled triangle?

☀ Understanding the question: This question means that we have to find out thebarea of that right angled triangle whose length is 12 cm and 8 cm is its base and both are given as perpendicular sides of that right angled triangle.

${\large{\pmb{\sf{\underline{Given \; that}}}}}$

✠ Base of right angled triangle = 8 cm

✠ Perpendicular or height of right angled triangle = 12 cm

${\large{\pmb{\sf{\underline{To \; find}}}}}$

✠ Area of right angled triangle

${\large{\pmb{\sf{\underline{Solution}}}}}$

✠ Area of right angled triangle = 48 cm²

${\large{\pmb{\sf{\underline{Using \; concept}}}}}$

✠ Formula to find area of traingle

${\large{\pmb{\sf{\underline{Using \; formula}}}}}$

✠ Area of triangle = ½ × Base × Height

${\large{\pmb{\sf{\underline{Full \; Solution}}}}}$

$:\implies \sf Area \: of \: triangle \: = \dfrac{1}{2} \times Base \times Height \\ \\ :\implies \sf Area \: of \: triangle \: = \dfrac{1}{2} \times 8 \times 12 \\ \\ :\implies \sf Area \: of \: triangle \: = \dfrac{1}{2} \times 96 \\ \\ :\implies \sf Area \: of \: triangle \: = 48 cm^{2}$

${\large{\pmb{\sf{\underline{Additional \; knowledge}}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: rectangle \: = \: Length \times Breadth}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: rectangle \: = \:2(length+breadth)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: square \: = \: 4 \times sides}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: square \: = \: Side \times Side}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: triangle \: = \: \dfrac{1}{2} \times breadth \times height}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: paralloelogram \: = \: Breadth \times Height}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: circle \: = \: \pi b^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: triangle \: = \: (1st \: + \: 2nd \: + 3rd) \: side}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: paralloelogram \: = \: 2(a+b)}}}$

Answer 2

Answer:

$\Large\underline\frak\red{\pmb{Given}}$

• ➥ If the perpendicular sides of a right angled triangle are 8 cm and 12 cm long.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\Large\underline\frak\red{\pmb{To \: Find }}$

• ➥ Area of that right angled triangle.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\Large \underline \frak\red{\pmb{Using \: Formula }}$

$\displaystyle \sf{ A_{\Delta }} = \big( \dfrac{1}{2} \big) \times (base) \times (height)$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\Large\underline\frak\red{\pmb{Solution }}$

${: \implies {\displaystyle \sf\pink{ A_{\Delta }}} = \sf\purple {\big( \dfrac{1}{2} \big) \times (base) \times (height)}}$

• Substituting the values

${: \implies {\displaystyle \sf\pink{ A_{\Delta }}} = \sf{\purple {\dfrac{1}{2} \times 8 \times 12}}}$

${: \implies {\displaystyle \sf\pink{ A_{\Delta }}}} = \sf \cancel{ \purple { \dfrac{96}{2} }}$

${: \implies {\displaystyle \sf\pink{ A_{\Delta }}}} = \sf \purple{48 \: {cm}^{2}}$

$\: \: \: \: \: \large \underline{\boxed{\frak \pink{Area \: of \: Triangle } = \sf \purple{48 \: {cm}^{2} }}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\Large\underline\frak\red{\pmb{Therefore}}$

• ➠ The area of right angled triangle is 48 cm²

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\Large\underline\frak\red{\pmb{Know \: More }}$

$\begin{gathered}\small\begin{gathered}\begin{gathered} \bigstar \: \bf \underline{More \: Useful \: formulae} \: \bigstar\\ \begin{gathered} \\ {\boxed{\begin{array}{cccc} \\ {\displaystyle \star\sf{Perimeter \: of \: Triangle }} = \sf{Length +Breadth + Height} \\ \\ \star\sf{Area \: of \: Rectangle = Lenght×Breadth} \\ \\ {\star\small\sf{Perimeter \: of\: Rectangle = 2(Length + Breadth)}} \\ \\ {\star\small \sf{Diagonal \: of \: Rectangle = \sqrt{ {Length }^{2} + { Breadth}^{2} }}} \\ \\ \star\small\sf{Area \: of \: Square } = {a}^{2} \\ \\ \star\small\sf{Perimeter \: of \: Square = 4a } \\ \\ \star\small\sf{Diagonal \: of \: Square = a \sqrt{2} }\end{array}}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$