Question

If ABCD is a triangle find the area of the shaded portion.​

Answer 1

↝ To Find :

The Area of the Shaded Region.

↝ Given :

• Length of AB = 12 cm

• Length of BC = 16 cm

• Length of EH = 4 cm

• Length of FG = 8 cm

• Length of HG = 6 cm

↝ We Know :

Area of a Rectangle :

$\sf{\underline{\boxed{A = length \times breadth}}}$

Area of a Trapezium :

$\sf{\underline{\boxed{A = \dfrac{1}{2} \times (P_{1} + P_{2}) \times h}}}$

Where ,

• P = Parallel side of the Trapezium

• h = height of the Trapezium

↝ Concept :

According to the question , to find the area of the Shaded Region , first we have to find the area of the rectangle and the Trapezium.

Then by subtracting the Area of Trapezium from the Area of Rectangle .i.e,

$\sf{A_{Rectangle} - Area_{Trapezium} = Area_{shaded\:Region}}$

↝ Solution :

Area of the Rectangle :

Given ,

• Length = 16 cm

• Breadth = 12 cm

Using the formula, and Substituting the values in it we get :

$\sf{\underline{\boxed{A = length \times breadth}}}$

$\sf{\Rightarrow A = 16 \times 12}$

$\sf{\Rightarrow A = 192 cm^{2}}$

Hence ,the area of the Rectangle is 192 cm².

Area of the Trapezium :

Given ,

• Parallel side ➝ P¹ = 8 cm

• Parallel side ➝ P² = 4 cm

• Height ➝ h = 6 cm

Using the formula and substituting the values in it ,we get :

$\sf{\underline{\boxed{A = \dfrac{1}{2} \times (P_{1} + P_{2}) \times h}}}$

$\sf{\Rightarrow A = \dfrac{1}{2} \times (8 + 4) \times 6}$

$\sf{\Rightarrow A = \dfrac{1}{2} \times 12 \times 6}$

$\sf{\Rightarrow A = \dfrac{1}{2} \times 72}$

$\sf{\Rightarrow A = \dfrac{1}{\cancel{2}} \times \cancel{72}}$

$\sf{\Rightarrow A = 36 cm^{2}}$

Hence ,the area of the Trapezium is 36 cm².

Area of the figure :

According to the Concept used ,

$\sf{A_{Rectangle} - Area_{Trapezium} = Area_{Shaded\:region}}$

Putting the value in it ,we get :

$\sf{192 - 36 = Area}$

$\sf{156 = Area}$

Hence ,the area of the Shaded Region is 156 cm ².

» Additional information :

• Area of a Parallelogram = Base × height

• Area of a Sector = lr/2

• Area of a Rhombus = ½ × product of it's diagonals

• Volume of a Cylinder = πr²h

Answer 2

A NSWER✔

$\large\underline\bold{GIVEN,}$

$\sf\dashrightarrow AB = 12 cm$

$\sf\dashrightarrow AD = 16 cm$

$\sf\dashrightarrow FG = 4 cm$

$\sf\dashrightarrow EH = 8 cm$

$\sf\dashrightarrow HG = 6 cm$

$\sf\dashrightarrow IF=6cm$

$\large\underline\bold{TO\:FIND}$

$\sf\dashrightarrow \:AREA\:OF\:SHADED\:REGION.$

✯FORMULA IN USE,

$\rm{\boxed{\bf{ \star\: AREA\:OF\:RECTANGLE= LENGTH \times BREADTH}}}$

$\rm{\boxed{\bf{\star\: AREA\:OF\: TRAPEZIUM= \dfrac{1}{2} \times (sum\:of\:parallel\:sides) \times height }}}$

$\sf\large\therefore \:taking\:three\:cases,$

$\sf\dashrightarrow in\:case\:1,we\:will\:find\:the\:area\:of\:rectangle$

$\sf\dashrightarrow in\:case\:2,we\:will\:find\:the\:area\:of\:trapezium$

$\sf\dashrightarrow in\:case\:1,we\:will\:subtract\:area\:of\: trapezium\:from\:rectangle$

✯. ACCORDING TO THE QUESTION,

$\sf\large\therefore AREA\:OF\:SHADED\:REGION = AREA\:OF\:RECTANGLE-\:AREA\:OF\:TRAPEZIUM.$

$\large\underline\bold{SOLUTION,}$

$\large{\boxed{\bf{ CASE:-1}}}$

$\sf\dashrightarrow LENGTH=16CM$

$\sf\dashrightarrow BREADTH =12CM$

$\sf\therefore area\:of\:rectangle= length \times breadth$

$\sf\implies 16 \times 12$

$\sf\implies 192cm^2$

$\rm{\boxed{\bf{ \star\:\: area\:of\:rectangle=192cm^2\:\: \star }}}$

$\large{\boxed{\bf{ CASE:-2}}}$

$\sf\dashrightarrow HEIGHT=6CM$

$\sf\dashrightarrow a= 8cm$

$\sf\dashrightarrow b=4cm$

$\sf\therefore AREA\:OF\: TRAPEZIUM= \dfrac{1}{2} \times (sum\:of\:parallel\:sides) \times height$

$\sf\implies \dfrac{1}{2} \times (8 + 4) \times 6$

$\sf\implies \dfrac{1}{2} \times (12) \times 6$

$\sf\implies \dfrac{1}{2} \times (72)$

$\sf\implies \dfrac{72}{2}$

$\sf\implies \cancel \dfrac{72}{2}$

$\sf\implies 36cm^2$

$\rm{\boxed{\bf{ \star \:\:36cm^2 \:\: \star}}}$

$\large{\boxed{\bf{CASE:-3 }}}$

$\sf\therefore AREA\:OF\:SHADED\:REGION = AREA\:OF\:RECTANGLE-\:AREA\:OF\:TRAPEZIUM$

$\sf\implies AREA\:OF\:SHADED\:REGION= 192-36$

$\sf\implies AREA\:OF\:SHADED\:REGION=156cm^2$

$\large{\boxed{\bf{\star\:\: area\:of\:shaded\:region\:= 156cm^2}}}$

$\large\underline\bold{AREA\:OF\:SHADED\:REGION\:=156cm^2}$

_________________________

NOTE:-

✯. FOR DIAGRAM PLEASE REFER THE ATTACHMENT ✔