Question

Find the area of quadrilateral ABCD. Here, AC = 22 cm , BM = 3 cm , DN = 3 cm and ⊥

, ⊥ ?​

Answer 1

Answer:

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Answer 2

Answer:

Rɪɢʜᴛ Qᴜᴇsᴛɪᴏɴ :

↝ Find the area of the quadrilateral ABCD.Here, AC = 22 cm, BM = 3 cm,DN = 3 cm, and BM⊥AC, DN⊥AC.

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

Gɪᴠᴇɴ :

• ➛ AC = 22 cm
• ➛ BM = 3 cm
• ➛ DN = 3
• ➛ BM ⊥ AC
• ➛ DN ⊥ AC

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

Tᴏ Fɪɴᴅ :

• ➛ Area of ΔADC
• ➛ Area of ΔABC
• ➛ Area of quadrilateral ABCD.

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

Cᴏɴᴄᴇᴘᴛ :

↝ Here the concept of Area of quadrilateral has been used. We need to calculate the area of quadrilateral ABCD,

↝ So,First we'll calculate the area of ΔABC, and ΔADC and after finding area of ΔABC, and ΔADC, we'll find the area of quadrilateral ABCD.

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

Usɪɴɢ Fᴏʀᴍᴜʟᴀs :

${\longrightarrow{\small{\underline{\boxed{\pmb{\sf{Area \: of \: quadrilateral \: ABCD = Area \: of \Delta ABC + Area \: of \: \Delta ADC}}}}}}}$

${\longrightarrow{\small{\underline{\boxed{\pmb{\sf{Area \: of \Delta ABC = \frac{1}{2} \times b \times h}}}}}}}$

${\longrightarrow{\small{\underline{\boxed{\pmb{\sf{Area \: of \: \Delta ADC = \frac{1}{2} \times b \times h}}}}}}}$

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

Sᴏʟᴜᴛɪᴏɴ :

☼ Finding the area of ΔABC

${\longrightarrow{\small{\underline{\boxed{\pmb{\sf{Area \: of \Delta ABC = \dfrac{1}{2} \times b \times h}}}}}}}$

★ Here :-

• ➛ b = AC (22 cm)
• ➛ h = BM (3 cm)

★ Now :-

${\longrightarrow{\small{\sf{Area \: of \Delta ABC = \dfrac{1}{2} \times b \times h}}}}$

${\longrightarrow{\small{\sf{Area \: of \Delta ABC = \dfrac{1}{2} \times 22\times 3}}}}$

${\longrightarrow{\small{\sf{Area \: of \Delta ABC = \dfrac{1 \times 22 \times 3}{2}}}}}$

${\longrightarrow{\small{\sf{Area \: of \Delta ABC = \dfrac{22 \times 3}{2}}}}}$

${\longrightarrow{\small{\sf{Area \: of \Delta ABC = \dfrac{66}{2}}}}}$

${\longrightarrow{\small{\sf{Area \: of \Delta ABC = \cancel{\dfrac{66}{2}}}}}}$

${\longrightarrow{\small{\sf{Area \: of \Delta ABC = 33 \: cm}}}}$

${\longrightarrow{\small{\red{\underline{\boxed{\sf{Area \: of \Delta ABC = 33 \: {cm}^{2}}}}}}}}$

∴ The area of ΔABC is 33 cm².

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☼ Finding the area of ΔADC :-

${\longrightarrow{\small{\underline{\boxed{\pmb{\sf{Area \: of \: \Delta ADC = \frac{1}{2} \times b \times h}}}}}}}$

★ Here :-

• ➛ b = 22 cm
• ➛ h = 3 cm

★ Now :-

${\longrightarrow{\small{\sf{Area \: of \: \Delta ADC = \dfrac{1}{2} \times b \times h}}}}$

${\longrightarrow{\small{\sf{Area \: of \: \Delta ADC = \dfrac{1}{2} \times 22 \times 3}}}}$

${\longrightarrow{\small{\sf{Area \: of \: \Delta ADC = \dfrac{1 \times 22 \times 3}{2}}}}}$

${\longrightarrow{\small{\sf{Area \: of \: \Delta ADC = \dfrac{22 \times 3}{2}}}}}$

${\longrightarrow{\small{\sf{Area \: of \: \Delta ADC = \dfrac{66}{2}}}}}$

${\longrightarrow{\small{\sf{Area \: of \: \Delta ADC = \cancel{\dfrac{66}{2}}}}}}$

${\longrightarrow{\small{\sf{Area \: of \: \Delta ADC = 33 \: cm}}}}$

${\longrightarrow{\small{\red{\underline{\boxed{\sf{Area \: of \: \Delta ADC = 33 \: {cm}^{2}}}}}}}}$

∴ The area of ΔADC is 33 cm².

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☼ Now, finding the area of quadrilateral ABCD :-

${\longrightarrow{\small{\underline{\boxed{\pmb{\sf{Area \: of \: ABCD = Area \: of \Delta ABC + Area \: of \: \Delta ADC}}}}}}}$

★ Here :-

• ➛ Area of ΔABC = 33 cm
• ➛ Area of ΔACD = 33 cm

★ Now :-

${\longrightarrow{\small{\sf{Area \: of \: ABCD = Area \: of \Delta ABC + Area \: of \: \Delta ADC}}}}$

${\longrightarrow{\small{\sf{Area \: of \: ABCD = 33 \: {cm}^{2} + 33 \: {cm}^{2}}}}}$

${\longrightarrow{\small{\sf{Area \: of \: ABCD = 66 \: {cm}^{2}}}}}$

${\longrightarrow{\small{\red{\underline{\boxed{\sf{Area \: of \: quadrilateral \: ABCD = 66 \: {cm}^{2}}}}}}}}$

∴ The area of quadrilateral ABCD is 66 cm².

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

Lᴇᴀʀɴ Mᴏʀᴇ :

$\boxed{\begin {minipage}{9cm}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {minipage}}$

$\underline{\rule{100mm}{0.9mm}}$