Question

If a trapezoid has a height of 10cm and sides of 7 and 8 cm, what is its area?

Answer 1

$\huge\underline{\sf{given}}$

$\star \: \bold{height = 10 \: cm} \\ \star \: \bold{base} \: \tiny{1} \: \small\bold{= 8 \: cm }\\ \star \: \bold{base \: \tiny{2} \: \small\bold{ = 7 \: cm}}$

$\huge \underline{\sf{find}}$

$\bigstar \: \red{\underline\bold{area \: of \: trapezoid}}$

$\huge\mathfrak{steps}$

$\underline{\dagger \: \sf{first \: know \: the \: formulae - }}$

$\purple{\leadsto\bold{area = \frac{a + b}{2}h}}$

$\sf \green{\mapsto \: area = \frac{8 + 7}{2} \: 10} \\ \\ \sf \green{\mapsto \: area = \frac{15}{2} \times 10 = \frac{150}{2} = 75}$

$\blacktriangleright \bold\pink{\: so \: area \: of \: trapezoid = \fbox{75}}$

_______________________________

$\bigstar \: \underline\bold{some \: more \: formulae}$

$\sf \star \: area \: of \: rectangle = l \times b$

$\sf \star \: area \: of \: square = {a}^{2}$

$\sf \star \: area \: of \: triangle = \sqrt{s(s - a)(s - b)(s - c)}$

$\sf \star \: area \: of \: triangle = \frac{1}{2} \times b \times h$

$\sf\star \: area \: of \: parallelogram = b \times h$

________________________________

Answer 2

Answer:

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{green}{Given:}}}}}}}\end{gathered}$

• $\dashrightarrow{\sf{{B_1} \: of \: trapezoid = 7m}}$
• $\dashrightarrow{\sf{{B_2} \: of \: trapezoid = 8 \: cm}}$
• $\dashrightarrow{\sf{Height \: of \: trapezoid = 10 cm}}$

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{green}{To Find :}}}}}}}\end{gathered}$

• $\dashrightarrow{\sf{Area \: of \: Trapezoid}}$

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{green}{Using Formula:}}}}}}}\end{gathered}$

${\dag{\underline{\boxed{\sf{Area \: of \: Trapezoid = \bigg\{\dfrac{1}{2} \times h({B_1} + {B_2})\bigg\}}}}}}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{green}{Solution:}}}}}}}\end{gathered}$

$\bigstar{\underline{\underline{\frak{\red{\pmb{Finding \: the \: area \: by \: using \: formula}}}}}}$

$\quad{:\implies{\sf{Area \: of \: Trapezoid ={\bf\dfrac{1}{2} \times h({B_1} + {B_2})}}}}$

• Substituting the values

$\quad{:\implies{\sf{Area \: of \: Trapezoid ={\bf\dfrac{1}{2} \times 10({7} + {8})}}}}$

$\quad{:\implies{\sf{Area \: of \: Trapezoid ={\bf\dfrac{10}{2}({15})}}}}$

$\quad{:\implies{\sf{Area \: of \: Trapezoid ={\bf\dfrac{10}{2} \times {15}}}}}$

$\quad{:\implies{\sf{Area \: of \: Trapezoid ={\bf{\cancel{\dfrac{10}{2}} \times {15}}}}}}$

$\quad{:\implies{\sf{Area \: of \: Trapezoid ={\bf{5}\times {15}}}}}$

$\quad{:\implies{\sf{Area \: of \: Trapezoid ={\bf{75 \: {cm}^{2} }}}}}$

${\dag{\underline{\boxed{\rm{\color{purple}{Area \: of \: Trapezoid = 75 \: {cm}^{2}}}}}}}$

• Henceforth,The Area of Trapezoid is 75 cm²

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\Large{\textsf{\textbf{\underline{\underline{\color{green}{Learn More :}}}}}}}\end{gathered}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto CSA \: of \: sphere \: = \: 2 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto SA \: of \: sphere \: = \: 4 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto TSA \: of \: sphere \: = \: 3 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Radius \: of \: circle \: = \: \dfrac{d}{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: sphere \: = \: \dfrac{4}{3} \pi r^{3}}}}$

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

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Circumference \: of \: circle \: = \: 2 \pi r}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}}$

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

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

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

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

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

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

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

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

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

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: cylinder \: = \: \pi r^{2}h}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Surface \: area \: of \: cylinder \: = \: 2 \pi rh + 2 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Lateral \: area \: of \: cylinder \: = \: 2 \pi rh}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Base \: area \: of \: cylinder \: = \: \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Height \: of \: cylinder \: = \: \dfrac{v}{\pi r^{2}}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Radius \: of \: cylinder \: = \:\sqrt \frac{v}{\pi h}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto TSA \: of \: cuboid \: = \: 2(l \times b + b \times h + l \times h}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto LSA \: of \: cuboid \: = \: 2h(l+b)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: cuboid \: = \: L \times B \times H}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diagonal \: of \: cuboid \: = \: \sqrt 3l}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: cuboid \: = \: 12 \times Sides}}}$