Question

ABCD is a parallelogram in which AB = 20 cm, BC = 13 cm, AC = 21 cm. Find the area of parallelogram ABCD

Answer 1

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• ABCD is a //gm
• in which

AB = 20cm and BC =13cm

• AC=21 cm

${\sf{\green{\underline{\large{To\:find}}}}}$

• Area of //gm

${\sf{\pink{\underline{\Large{Explanation}}}}}$

Diagram

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(8.6,3){\large{A}}\put(7.7,0.9){\large{B}}\put(9.5,0.7){\sf{\large{Base}}}\put(11.1,0.9){\large{C}}\put(8,1){\line(1,0){3}}\put(11,1){\line(1,2){1}}\put(9,3){\line(3,0){3}}\put(7.7,2){\large{ \sf\:side }}\put(8,1){\line(1,2){1}}\put(12.1,3){\large{D}}\put(8,1){\line(2,1){4}}\put(11,1){\line(-1,1){2}}\end{picture}$

Taking Small ∆s

∆ABD

$\setlength{\unitlength}{5mm}\begin{picture}(5,5)\put(0,0){\line(1,0){8}}\put(0,0){\line(2,1){6}}\put(8,0){\line(-2,3){2}}\end{picture}$

Once Again

∆ABC

$\setlength{\unitlength}{5mm}\begin{picture}(5,5)\put(0,0){\line(1,0){8}}\put(0,0){\line(2,1){6}}\put(8,0){\line(-2,3){2}}\end{picture}$

Solution

∆ABD

From Heron's Formula

$\tt\sqrt{S(S-A)(S-B)(S-C)}$

$\rightarrow\sf\dfrac{A+B+C}{2}$

¶utting the value

$\rightarrow\sf\dfrac{A+B+C}{2}$

$\rightarrow\sf\dfrac{13+21+20}{2}$

$\rightarrow\sf\dfrac{54}{2}$

$\rightarrow\sf{27}$

$\tt\sqrt{S(S-A)(S-B)(S-C)}$

$\rightarrow\tt\sqrt{27(27-13)(27-21)(27-20)}$

$\rightarrow\tt\sqrt{27(14)(6)(7)}$

=>√3×3×3×7×2×3×2×7

=>3×3×2×7

=>126cm²

Properties of //gm

• Opposite sides equal
• Diagonal bisect the //gm in two equal area

Therefore

Area of ∆ABC =. Area of ∆ABD

Required area of //gm ABCD is

=>Area of ∆ABC + Area of ∆ABD=Area of //gm

=>126+126

=>252cm²

Answer 2

✯✯ QUESTION ✯✯

ABCD is a parallelogram in which AB = 20 cm, BC = 13 cm, AC = 21 cm. Find the area of parallelogram ABCD..

━━━━━━━━━━━━━━━━━━━━

✰✰ ANSWER ✰✰

➙Given : -

AB = 20cm , BC = 13cm , AC = 21 cm

➙Now : -

• Area ΔABC = Area ΔCDA
• Area (ABCD) = Area ΔACD + Area ΔABC

➜By Using Heron's Formula : -

Consider Δ ABC : -

$\longmapsto\tt{a=20cm , b= 21cm , c=13cm}$

$\longmapsto\tt{S=\dfrac{a+b+c}{2}}$

$\longmapsto\tt{S=\dfrac{20+21+13}{2}}$

$\longmapsto\tt{S=\cancel\dfrac{54}{2}}$

$\longmapsto\tt{27cm}$

➙Now : -

$\longmapsto\tt{Area=\sqrt{s(s-a)(s-b)(s-c)}}$

$\longmapsto\tt{\sqrt{27(27-20)(27-21)(27-13)}}$

$\longmapsto\tt{\sqrt{27(7)(6)(14)}}$

$\longmapsto\tt{\sqrt{3}\times{3}\times{3}\times{7}\times{{3}\times{2}\times{7}\times{2}}}$

$\longmapsto\tt{3\times{3}\times{7}\times{2}}$

$\longmapsto\tt{126{cm}^{2}}$

☛Area of Δ ABC = 126 cm²

$\longmapsto\tt{Area\:of\:Quad.\:ABCD = 126\times{2}}$

$\longmapsto\tt{\large{\boxed{\boxed{\bold{\orange{\sf{252{cm}^{2}}}}}}}}$

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✔Properties of Parallelogram : -

• Opposite Sides of parallelogram are equal..
• Opposite Angles of parallelogram are equal.
• The diagonals of parallelogram bisects each other..

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