Question

find the area of the shaded region .​

Answer 1

Given

• Side PC = 5cm
• Side PC = 6cm
• Side DC = 9cm

To Find

• Area of Shaded region

Solution

In ∆PDC , we have

• Side PC = 5cm
• Side PC = 6cm
• Side DC = 9cm

Any triangle has 3 sides. We represent the length of the 3 sides as ‘a’, ‘b’, ‘c’ units respectively. Therefore the sum of lengths of all the 3 sides (perimeter) is P = a + b+ c. Hence, the semi perimeter of the triangle is

$\sf{ ~~~~\implies ~~ s = \dfrac{P}{2} = \dfrac{ (a+b+c)}{2}}$

$\sf{ ~~~~\implies ~~ s = \dfrac{ (5 + 6 + 9)}{2}}$

$\sf{ ~~~~\implies ~~ s = \dfrac{ 20}{2}}$

$\sf{ ~~~~\implies ~~ s =10}$

By Heron’s formula, the area of the triangle is

$\\\\{ ~~~~\implies ~~ \sf Area ~~of~~ ∆PDC=\sqrt{s(s-a)(s-b)(s-c)} }$

${ ~~~~\implies ~~ \sf Area ~~of~~ ∆PDC=\sqrt{10(10-5)(10-6)(10-9)} }$

${ ~~~~\implies ~~ \sf Area ~~of~~ ∆PDC=\sqrt{10 \times 5 \times 4 \times 1} }$

${ ~~~~\implies ~~ \sf Area ~~of~~ ∆PDC=\sqrt{10 \times 5 \times 2 \times 2} }$

${ ~~~~\implies ~~ \sf Area ~~of~~ ∆PDC=\sqrt{10 \times 10 \times 2} }$

${ ~~~~\implies ~~ \sf Area ~~of~~ ∆PDC=\sqrt{ {10}^{2} \times 2} }$

${ ~~~~\implies ~~ \sf Area ~~of~~ ∆PDC=10\sqrt{ 2} }$

Height of ∆PDC = H

Base of ∆PDH = 9cm

$\\\\{ ~~~~\implies ~~ \sf Area ~~of~~ ∆PDC= \dfrac{1}{2} \times Base × Height }$

${ ~~~~\implies ~~ \sf 10 \sqrt{2} = \dfrac{1}{2} \times 9 × Height }$

${ ~~~~\implies ~~ \sf \dfrac{2 \times 10 \sqrt{2}}{9} = Height }$

${ ~~~~\implies ~~ \sf Height = \dfrac{20 \sqrt{2}}{9} }$

In Rectangle ABCD, we have

• Length = 9
• ${\sf Height = \dfrac{20 \sqrt{2}}{9} }$

$\\\\ \sf{ ~~~~\implies ~~ Area \: \: of \: \: Rectangle ABCD = Length × Breadth}$

$\sf{ ~~~~\implies ~~ Area \: \: of \: \: Rectangle ABCD = \cancel9 ×\dfrac{20 \sqrt{2}}{\cancel9} }$

$\sf{ ~~~~\implies ~~ Area \: \: of \: \: Rectangle ABCD = 20 \sqrt{2} }$

Now, We have

Area of Rectangle ABCD = 20√2cm²

Area of Traingle PDC = 10√2cm²

$\\\\\sf{ ~~~~\implies ~~ Area \: of \: shaded \: region = Area \: of \: Rectangle \: ABCD - Area \: of \: Traingle \: PDC}$

$\sf{ ~~~~\implies ~~ Area \: of \: shaded \: region = 20 \sqrt{2} - 10 \sqrt{2} }$

$~~~~~~\underbrace{\bf \pink{ \: \: Area \: of \: shaded \: region = 10 \sqrt{2} \: \: \: }}$

Answer 2

$\huge \mathbb{ \red {★᭄ꦿ᭄S} \pink{ᴏ}\purple{ʟᴜ} \blue {ᴛ} \orange{ɪ} \green{ᴏɴ★᭄ꦿ᭄}}$

• ⚘Side PC = 5cm
• ⚘Side PC = 6cm
• ⚘Side DC = 9cm

⚘To Find

⚘Area of Shaded region

⚘Solution

⚘In ∆PDC , we have

• ⚘Side PC = 5cm
• ⚘Side PC = 6cm
• ⚘Side DC = 9cm

⚘Any triangle has 3 sides. We represent the length of the 3 sides as ‘a’, ‘b’, ‘c’ units respectively. Therefore the sum of lengths of all the 3 sides (perimeter) is P = a + b+ c. Hence, the semi perimeter of the triangle is

$\tt{ ~~~~\implies ~~ s = \dfrac{P}{2} = \dfrac{ (a+b+c)}{2}}$

$\tt{ ~~~~\implies ~~ s = \dfrac{ (5 + 6 + 9)}{2}}$

$\tt{ ~~~~\implies ~~ s = \dfrac{ 20}{2}}$

$\begin{gathered} \tt{ ~~~~\implies ~~ s =10}\\\\\end{gathered}$

$\begin{gathered}\\\\{ ~~~~\implies ~~ \tt Area ~~of~~ ∆PDC=\sqrt{s(s-a)(s-b)(s-c)} }\end{gathered}$

${ ~~~~\implies ~~ \tt Area ~~of~~ ∆PDC=\sqrt{10(10-5)(10-6)(10-9)} }$

${ ~~~~\implies ~~ \tt \: Area ~~of~~ ∆PDC=\sqrt{10 \times 5 \times 4 \times 1} }$

${ ~~~~\implies ~~ \tt \: Area ~~of~~ ∆PDC=\sqrt{10 \times 5 \times 2 \times 2} }$

${ ~~~~\implies ~~ \tt Area ~~of~~ ∆PDC=\sqrt{10 \times 10 \times 2} }$

${ ~~~~\implies ~~ \tt Area ~~of~~ ∆PDC=\sqrt{ {10}^{2} \times 2} }$

$\begin{gathered}{ ~~~~\implies ~~ \tt \: Area ~~of~~ ∆PDC=10\sqrt{ 2} } \\ \\ \end{gathered}$

• ⚘By Heron’s formula, the area of the triangle is

$\begin{gathered}\\\\{ ~~~~\implies ~~ \tt \: Area ~~of~~ ∆PDC= \dfrac{1}{2} \times Base × Height }\end{gathered}$

${ ~~~~\implies ~~ \tt 10 \sqrt{2} = \dfrac{1}{2} \times 9 × Height }$

${ ~~~~\implies ~~ \tt \dfrac{2 \times 10 \sqrt{2}}{9} = Height }$

$\begin{gathered}{ ~~~~\implies ~~ \tt \: Height = \dfrac{20 \sqrt{2}}{9} } \\ \\ \end{gathered}$

• ⚘Height of ∆PDC = H
• ⚘Base of ∆PDH = 9cm

${\tt Height = \dfrac{20 \sqrt{2}}{9} }$

$\begin{gathered}\\\\ \tt{ ~~~~\implies ~~ Area \: \: of \: \: Rectangle \: ABCD = Length × Breadth}\end{gathered}$

$\tt{ ~~~~\implies ~~ Area \: \: of \: \: Rectangle \: ABCD = \cancel9 ×\dfrac{20 \sqrt{2}}{\cancel9} }$

$\tt{ ~~~~\implies ~~ Area \: \: of \: \: Rectangle \: ABCD = 20 \sqrt{2} }$

• ⚘Now, We have
• ⚘Area of Rectangle ABCD = 20√2cm²
• ⚘Area of Traingle PDC = 10√2cm²

$\begin{gathered}\\\\\tt{ ~~~~\implies ~~ Area \: of \: shaded \: region = Area \: of \: Rectangle \: ABCD - Area \: of \: Traingle \: PDC}\end{gathered}$

$\tt{ ~~~~\implies ~~ Area \: of \: shaded \: region = 20 \sqrt{2} - 10 \sqrt{2} }$

$~~~~~~\underbrace{\mathbb \pink{ \: \: Area \: of \: shaded \: region = 10 \sqrt{2} \: \: \: }}$

$\huge\pink{\mid{\fbox{\tt{thanks}}\mid}}$

$\huge \color{red} \boxed{\colorbox{lime}{------------×-----------}}$