CBSE Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 10 - MCQExams.com

Area bounded by the curves satisfying the conditions $$\displaystyle \frac{x^{2}}{25}+\frac{y^{2}}{36}\leq 1\leq\frac{x}{5}+\frac{y}{6}$$ is given by
  • $$15(\displaystyle \dfrac{\pi}{2}+1)$$ sq.units
  • $$\dfrac{15}{4}(\dfrac{\pi}{2}-1)$$ sq.units
  • $$30(\pi-1)$$ sq.unit
  • $$\displaystyle \dfrac{15}{2}(\pi-2)$$ sq.unit
If $$\mathrm{A}_{1}$$ and $$\mathrm{A}_{2}$$ respectively represents the area bounded by the curves $$\mathrm{f}(\mathrm{x},\mathrm{y})$$ $$4x^{2}\leq y\leq 3x$$ and $$\mathrm{g}$$ (x,y)$$4x^{2}\leq y\leq|3x|$$ then $$\mathrm{A}_{1}$$ : $$\mathrm{A}_{2}$$ equals:
  • 2:1
  • 3:1
  • 1:2
  • 1:3
$$\sin x$$ & $$\cos x$$ meet each other at a number of points and develop symmetrical area. Area of one such region is
  • $$4\sqrt{2}$$
  • $$3\sqrt{2}$$
  • $$2\sqrt{2}$$
  • $$\sqrt{2}$$
Area bounded by the curves $$\displaystyle \frac{y}{x}=\log x$$ and $$\displaystyle \frac{y}{2}=-x^{2}+x$$ equals:
  • 7/12
  • 12/7
  • 7/6
  • 6/7
The area of the plane region bounded by the curves $$x+2y^{2}=0$$ and $$x+3y^{2}=1$$ is
  • $$\displaystyle \frac{1}{3}$$
  • $$\displaystyle \frac{2}{3}$$
  • $$\displaystyle \frac{4}{3}$$
  • $$\displaystyle \frac{5}{3}$$
The area enclosed between the curves, $$x^{2}=y$$ and $$y^{2}=x$$ is equal to:
  • $$\dfrac{1}{3}$$ sq. unit
  • $$2\displaystyle \int_{0}^{1}(x-x^{2})dx$$
  • Area enclosed by the region $$\{(x,y):x^{2}\leq y\leq\sqrt x\}$$
  • Area enclosed by the region $$\{(x, y):x^{2}\leq y\leq x\}$$
The function $$\displaystyle \mathrm{f}(\mathrm{x})=\max$$ $$\{x^{2},(1-x)^{2},2x(1-x) \forall 0\leq x \leq 1\}$$ then area of the region bounded by the curve $$\mathrm{y}=\mathrm{f}(\mathrm{x})$$ , $$\mathrm{x}$$-axis and $$\mathrm{x}= 0,\ \mathrm{x} =$$ 1 is equals
  • $$\dfrac{27}{17}$$
  • $$\dfrac{17}{27}$$
  • $$\dfrac{18}{17}$$
  • $$\dfrac{19}{17}$$

The ratio in which the area bounded by the curves $$y^2=12x $$ and $$x^2=12y$$ is divided by the line x $$=$$ 3 is
  • 15 : 16
  • 15 : 49
  • 1 : 2
  • None of these
The area bounded by the tangent and normal to the curve $$y(6-x)=x^{2}$$ at $$(3,3)$$ and the x-axis is
  • $$5$$
  • $$6$$
  • $$15$$
  • $$3$$
If $$\left| z- (4 + 4i) \right| \geq  4$$, then area of the region bounded by the locii of $$z,\; iz,\; - z$$ and $$-iz$$ is:
  • $$4(4-\pi )$$
  • $$16(4-\pi )$$
  • $$16(\pi -1)$$
  • $$4(\pi -1)$$
The area of the region bounded by the curve y $$\displaystyle =\frac{16-x^{2}}{4}$$ and $$\displaystyle y=sec^{-1}[-sin^{2}x],$$ where [.] stands for the greatest integer function is:
  • $$(4-\pi )^{3/2}$$
  • $$\dfrac{8}{3}(4-\pi )^{3/2}$$
  • $$\dfrac{4}{3}(4-\pi )^{3/2}$$
  • $$\dfrac{8}{3}(4-\pi )$$
The parabola $$y^{2} = 4x$$ and $$x^{2}  = 4y$$ divide the square region bounded by the lines $$x = 4, y = 4$$ and the coordinate axes. If $$S_{1}, S_{2}, S_{3}$$ are the areas of these parts numbered from top to bottom respectively, then
  • $$S_{1} : S_{2}\equiv 1 : 1$$
  • $$S_{2} : S_{3}\equiv 1 : 2$$
  • $$S_{1} : S_{3}\equiv 1 : 1$$
  • $$S_{1} : (S_{1} + S_{2})\equiv 1 : 2$$
The area of the smaller region in which the curve $$y=\left [ \frac{x^{3}}{100}+\frac{x}{50} \right ],$$ where[.] denotes the greatest integer function, divides the circle $$\left ( x-2 \right )^{2}+\left ( y+1 \right )^{2}=4,$$ is equal to
  • $$\frac{2\pi-3\sqrt{3}}{3}sq. units$$
  • $$\frac{3\sqrt{3}-\pi}{3}sq. units$$
  • $$\frac{4\pi-3\sqrt{3}}{3}sq. units$$
  • $$\frac{5\pi-3\sqrt{3}}{3}sq. units$$
  • $$\frac{4\pi-3\sqrt{3}}{6}sq. units$$
Area of the region bounded by the curve $$y = x^{2}$$ and $$y = sec^{-1} [sin^{2}x]$$ (where [ . ] denotes the greatest integer function) is
  • $$\frac{\pi}{3}\sqrt{\pi}$$
  • $$\frac{2\pi\sqrt{\pi}}{3}$$
  • $$\frac{4\pi\sqrt{\pi}}{3}$$
  • $$\frac{6\pi\sqrt{\pi}}{3}$$
  • $$\frac{3\pi\sqrt{\pi}}{2}$$
If $$A_1$$ is the area bounded by $$y= \cos x, y = \sin x$$ &  $$x=0$$  and $$A_2$$ the area bounded by $$y = \cos x , y = \sin x , y = 0$$ in $$(0,\frac{\pi}{2})$$ then $$\displaystyle  \dfrac{A_1}{A_2}$$ equals to :
  • $$\dfrac{1}{2}$$
  • $$\dfrac{1}{\sqrt2}$$
  • $$1$$
  • None of these
The area bounded by the curves $$y = sin^{-1} |sin  x|$$ and $$y = (sin^{-1} | sin  x|)^{2},$$ where $$0\leq x\leq 2\pi ,$$ is:
  • $$\dfrac{1}{3}+\dfrac{\pi ^{2}}{4}$$ sq. units
  • $$\dfrac{1}{6}+\dfrac{\pi ^{3}}{8}$$ sq. units
  • $$2$$ sq. units
  • $$\displaystyle \frac{4}{3}+\pi ^{2}\frac{\pi -3}{6}  sq.units$$
Area lying in the first quadrant and bounded by the circle $$x^2+y^2=4$$ and the line $$x=y\sqrt{3}$$ is:
  • $$\pi$$
  • $$\displaystyle\frac{\pi}{2}$$
  • $$\displaystyle\frac{\pi}{3}$$
  • None of these
Area bounded by the region $$R\equiv \left \{ (x,y):y^{2}\leq x\leq |y| \right \}$$ is
  • $$\displaystyle \frac{4}{3}$$
  • $$\displaystyle \frac{3}{4}$$
  • $$\displaystyle \frac{1}{3}$$
  • $$\displaystyle \frac{1}{4}$$
Area bounded by curves $$y^2 = x$$ and $$y = | x |$$ is given by
  • $$\displaystyle \frac{1}{6}$$
  • $$\displaystyle \frac{2}{3}$$
  • $$-\displaystyle \frac{1}{6}$$
  • $$\displaystyle \frac{1}{3}$$
If the area bounded by the curve $$|y|=sin^{-1}|x|$$ and  $$x=1$$ is $$a(\pi+b)$$, then the value $$a-b$$ is:
  • 1
  • 2
  • 3
  • 4
The area bounded by $$y=3-|3-x|$$ and $$\displaystyle y=\frac{6}{|x+1|}$$ is:
  • $$\dfrac{15}{2} -6$$ ln 2 sq. units
  • $$\dfrac{13}{2} -3$$ ln 2 sq. units
  • $$\dfrac{13}{2} -6$$ ln 2 sq. units
  • None of these
The area bounded by $$y=\sec^ {-1}{x}, y= \text{cosec}^{-1}{x}$$ and the line $$x-1=0$$ is:
  • $$\ln(3+2\sqrt{2})-\displaystyle \frac{\pi}{2}$$
  • $$\displaystyle \frac{\pi}{2}+\ln(3+2\sqrt{2})$$
  • $$\pi - \ln3$$
  • $$\pi + \ln3$$
The area of the smaller part bounded by the semi-circle $$\displaystyle y=\sqrt { 4-{ x }^{ 2 } } ,y=x\sqrt { 3 } $$ and x-axis is
  • $$\displaystyle \frac { \pi  }{ 3 } $$
  • $$\displaystyle \frac { 2\pi  }{ 3 } $$
  • $$\displaystyle \frac { 4\pi  }{ 3 } $$
  • none of these
The area enclosed by $$x^2 + y^2 = 4, y = x^2 + x + 1,$$  $$y=\left [ \sin^{2}\displaystyle \frac{x}{4}+\cos\displaystyle \frac{x}{4} \right ]$$ and $$x$$-axis (where $$[.]$$ denotes the greatest integer function) is:
  • $$\displaystyle \frac{2\pi}{3}+\sqrt{3}-\displaystyle \frac{1}{6}$$
  • $$\displaystyle \frac{2\pi}{3}+2\sqrt{3}-\displaystyle \frac{1}{6}$$
  • $$2\sqrt3-\frac{1}{3}$$
  • $$\displaystyle \frac{\pi}{3}+\sqrt3$$
The area bounded by the function $$f(x)=x^{2}:R^{+}\rightarrow R^{+}$$ and its inverse function is:
  • $$\dfrac{1}{2}$$sq.units
  • $$\dfrac{1}{3}$$sq.units
  • $$\dfrac{2}{3}$$sq.units
  • $$\dfrac{1}{6}$$sq.units
Find the area of the region bounded by the curves $$y= log_{e}x $$, $$ y=\sin ^{4} \pi x $$, $$x=0 $$
  • $$\displaystyle\ \frac{11}{8}sq.units$$
  • $$\displaystyle\ \frac{9}{8}sq.units$$
  • $$\displaystyle\ \frac{13}{8}sq.units$$
  • $$\displaystyle\ \frac{15}{8}sq.units$$
State the following statement is True or False
The area bounded by the circle $$x^2 + y^2 = 1, x^2 + y^2 = 4$$ and the pair of lines $$\sqrt{3} (x^2 + y^2) = 4xy$$, is equal to $$\dfrac{\pi}{2}$$. The statement is true or false.
  • True
  • False
Find the area bounded by the curves $$\displaystyle\ y=\sqrt{1-x^{2}}$$ and $$\displaystyle\ y=x^{3}-x $$. Also find the ratio in which the y-axis divide this area
  • $$\displaystyle\ \frac{\pi }{2}$$ , $$\displaystyle\ \frac{\pi -1}{\pi +1} $$
  • $$\displaystyle\ \frac{\pi }{4}$$ , $$\displaystyle\ \frac{\pi -1}{\pi +1} $$
  • $$\displaystyle\ \frac{\pi }{2}$$ , $$\displaystyle\ \frac{\pi +1}{\pi -1} $$
  • None of these
Area bounded by the curve $$y = \sin^{-1} (\sin x), y = x^2 -\pi x$$ is
  • $$\dfrac{\pi ^2}{4}+\pi ^3$$
  • $$\dfrac{\pi ^2}{4}+\dfrac{\pi ^3}{2}$$
  • $$\dfrac{\pi ^3}{6}$$
  • $$\dfrac{\pi ^2}{4}+\dfrac{\pi ^3}{6}$$
Find the area enclosed the curves : $$y=ex\log { x } $$ and $$\displaystyle y=\frac { \log { x }  }{ ex } $$ where $$\log { e } =1$$
  • $$\displaystyle\frac { { e }^{ 2 }-5 }{ 4e } $$
  • $$\displaystyle\frac { { e }^{ 2 }+5 }{ 4e } $$
  • $$\displaystyle\frac { { e }^{ 2 }-3 }{ 2e } $$
  • $$\displaystyle\frac { { e }^{ 2 }+3 }{ 2e } $$
Sketch the region bounded by the curves $$ \displaystyle y=x^{2}$$ & $$ \displaystyle\ y= 2/(1+x^{2})$$. Find the area:
  • $$\displaystyle\ \pi -\frac{2}{3}$$
  • $$\displaystyle\ \pi -\frac{1}{3}$$
  • $$\displaystyle\ \pi -\frac{5}{3}$$
  • $$\displaystyle\ \pi -\frac{7}{3}$$
The ratio in which the area bounded by the curves $$y^{2}=4x$$ and $$x^{2}=4y$$ is divided by the line $$x=1$$ is
  • $$64 : 49$$
  • $$15 : 34$$
  • $$15 : 49$$
  • none of these
Find the equation of the line passing through the origin and dividing the curvilinear triangle with vertex at the origin, bounded by the curves $$y=2x-x^2, y=0$$ and $$x=1$$ into two parts of equal area.
  • $$y=2x/3$$
  • $$y=x/3$$
  • $$y=2x/5$$
  • $$y=5x/3$$
For the curve $$\displaystyle f(x)=\frac{1}{1+x^2}$$, let two points on it be $$A(\alpha ,f(\alpha )), B \left ( -\dfrac{1}{\alpha }, f\left (- \dfrac{1}{\alpha } \right )\right) (\alpha > 0)$$. Find the minimum area bounded by the line segments OA, OB and $$f(x)$$, where 'O' is the origin.
  • $$\displaystyle \frac{(\pi -1)}{2}$$
  • $$\dfrac{\pi}{2}$$
  • $$\dfrac{(\pi-2)}{2}$$
  • Maximum area is always infinite
Find the area of the region enclosed between the two circles $$\displaystyle\ x^{2}+y^{2}=1$$ & $$(x-1)^{2}+y^{2}=1$$
  • $$\displaystyle\ \frac{\pi }{6}-\frac{\sqrt{3}}{2}$$ sq.units
  • $$\displaystyle\ \frac{\pi }{3}-\frac{\sqrt{3}}{2}$$ sq.units
  • $$\displaystyle\ \frac{\pi }{6}-\frac{\sqrt{3}}{4}$$ sq.units
  • $$\displaystyle\ \frac{\pi }{3}-\frac{\sqrt{3}}{4}$$ sq.units
The area bounded by the region by the curves $$ \displaystyle \begin{vmatrix}x\end{vmatrix} =1-y^{2} $$ and $$ \displaystyle \begin{vmatrix}x\end{vmatrix}+\begin{vmatrix}y\end{vmatrix}=1 $$ is
  • $$ \displaystyle \frac{1}{3} $$
  • $$1$$
  • $$ \displaystyle \frac{1}{2} $$
  • $$ \displaystyle \frac{2}{3} $$
Compute the area of the curvilinear triangle bounded by the y-axis & the curve, $$\displaystyle\ y=\tan x$$ & $$ \displaystyle\ y=(2/3) \cos x$$
  • $$\displaystyle\ \frac{1}{3}+ ln\left [ \frac{\sqrt{3}}{2} \right ]sq.units$$
  • $$\displaystyle\ \frac{1}{3}- ln\left [ \frac{\sqrt{3}}{2} \right ]sq.units$$
  • $$\displaystyle\ \frac{2}{3}+ ln\left [ \frac{\sqrt{3}}{2} \right ]sq.units$$
  • $$\displaystyle\ \frac{1}{3}+ ln\left [ \frac{\sqrt{1}}{2} \right ]sq.units$$
The area lying in the first quadrant inside the circle $${ x }^{ 2 }+{ y }^{ 2 }=12$$ and bounded by the parabolas $${ y }^{ 2 }=4x,{ x }^{ 2 }=4y$$ is:
  • $$\displaystyle 2\left( \frac { \sqrt { 2 }  }{ 3 } +\frac { 3 }{ 2 } \sin ^{ -1 }{ \frac { 1 }{ 3 }  }  \right) $$
  • $$\displaystyle 4\left( \frac { \sqrt { 2 }  }{ 3 } +\frac { 3 }{ 2 } \sin ^{ -1 }{ \frac { 1 }{ 3 }  }  \right) $$
  • $$\displaystyle \left( \frac { \sqrt { 2 }  }{ 3 } +\frac { 3 }{ 2 } \sin ^{ -1 }{ \frac { 1 }{ 3 }  }  \right) $$
  • none of these
A polynomial function f(x) satisfies the condition $$f(x+1) =f(x)+2x+1$$. Find $$f(x)$$ if $$f(0)=1$$. Find also the equations of the pair of tangents from the origin on the curve $$y=f(x)$$ and compute the area enclosed by the curve and the pair of tangents.
  • $$f(x)=x^2+1;$$, $$y=\pm 2x;$$ , $$\displaystyle A=\frac{2}{3}$$ sq.units
  • $$f(x)=x^2-1;$$, $$y=\pm 2x;$$ , $$\displaystyle A=\frac{2}{3}$$ sq.units
  • $$f(x)=x^2+1;$$, $$y=\pm 2x;$$ , $$\displaystyle A=\frac{3}{2}$$ sq.units
  • $$f(x)=x^2-1;$$, $$y=\pm 2x;$$ , $$\displaystyle A=\frac{3}{2}$$ sq.units
If f(x) be an increasing function defined on [a, b] then
max {f(t) such that $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(x)  & min {f(t), $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(a) and if f(x) be decreasing function defined on [a, b] then
max {f(t), $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(a),
min {f(t), $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(x).
On the basis of above information answer the following questions.
Let $$\displaystyle f\left ( x \right )=min \left \{ 1, 1-\cos x, 2\sin x \right \}$$ then $$\displaystyle \int_{0}^{\pi}f\left ( x \right )dx$$ is
  • $$\displaystyle \frac{\pi }{3}+1-\sqrt{3}$$
  • $$\displaystyle \frac{2\pi }{3}-1+\sqrt{3}$$
  • $$\displaystyle \frac{5\pi }{6}+1-\sqrt{3}$$
  • $$\displaystyle \frac{\pi }{6}+1-\sqrt{3}$$
The area included between the curve $${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$$ and $$\displaystyle \sqrt { \left| x \right|  } +\sqrt { \left| y \right|  } =\sqrt { a } \left( a>0 \right) $$ is:
  • $$ \displaystyle \left( \pi +\frac { 2 }{ 3 }  \right) { a }^{ 2 }$$
  • $$ \displaystyle \left( \pi -\frac { 2 }{ 3 }  \right) { a }^{ 2 }$$
  • $$ \displaystyle \frac { 2 }{ 3 } { a }^{ 2 }$$
  • $$\displaystyle \frac { 2\pi }{ 3 } { a }^{ 2 }$$
Let y=f(x) be the given curve and $$x=a$$, $$x=b$$ be two ordinates then area bounded by the curve $$y=f(x)$$, the axis of x between the ordinates $$x=a$$ & $$x=b$$, is given by definite integral
$$\int_{a}^{b}ydx$$ or $$\int_{a}^{b}f\left ( x \right )dx$$ and the area bounded by the curve $$x=f(y)$$, the axis of y & two abscissae $$y=c$$ & $$y=d$$ is given by $$\int_{c}^{d}xdy$$ or $$\int_{c}^{d}f\left ( x \right )dy$$. Again if we consider two curves $$y=f(x)$$, $$y=g(x)$$ where $$f\left ( x \right )\geq g\left ( x \right )$$ in the interval [a, b] where $$x=a$$ & $$x=b$$ are the points of intersection of these two curves Shown by the graph given
Then area bounded by these two curves is given by
$$\int_{a}^{b}\left [ f\left ( x \right )-g\left ( x \right ) \right ]dx$$
On the basis of above information answer the following questions.

The area bounded by parabolas $$y=x^{2}+2x+1$$ & $$y=x^{2}-2x+1$$ and the line $$\displaystyle y=\frac{1}{4}$$ is equal to

161838_6c80fc7958864f1f961bdcd5221bb036.png
  • $$\displaystyle \frac{2}{3}$$ square unit
  • $$\displaystyle \frac{1}{3}$$ square unit
  • $$\displaystyle \frac{3}{2}$$ square unit
  • $$\displaystyle \frac{1}{2}$$ square unit
Let $$ \displaystyle f\left ( x \right )=\min \left \{ x+1,\sqrt{1-x} \right \} $$ then area bounded by $$ \displaystyle y=f\left ( x \right ) $$ and x-axis is:
  • $$ \displaystyle \frac{7}{6} $$
  • $$ \displaystyle \frac{5}{6} $$
  • $$ \displaystyle \frac{1}{6} $$
  • $$ \displaystyle \frac{11}{6} $$
If f(x) be an increasing function defined on [a, b] then
max {f(t) such that $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(x)  & min {f(t), $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(a) and if f(x) be decreasing function defined on [a, b] then
max {f(t), $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(a),
min {f(t), $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(x).
On the basis of above information answer the following questions.
$$\int_{0}^{\pi /2}min\left \{ \sin x, \cos x \right \}dx$$ equals
  • $$2\left ( \sqrt{3}-1 \right )$$
  • $$\sqrt{2}\left ( \sqrt{2}-1 \right )$$
  • $$\left ( \sqrt{3}-1 \right )$$
  • $$\sqrt{2}\left ( \sqrt{2}+1 \right )$$
If f(x) be an increasing function defined on [a, b] then
max {f(t) such that $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(x)  & min {f(t), $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(a) and if f(x) be decreasing function defined on [a, b] then
max {f(t), $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(a),
min {f(t), $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(x).
On the basis of above information answer the following questions.
$$\int_{0}^{\pi }max\left \{ \sin x, \cos x \right \}dx$$ is equal to
  • $$\sqrt{2}-1$$
  • $$\displaystyle 1-\frac{1}{\sqrt{2}}$$
  • $$\displaystyle 1+\frac{1}{\sqrt{2}}$$
  • None of these
The ratio of the area's bounded by the curves $$ \displaystyle y^{2}=12x $$ and $$ \displaystyle x^{2}=12y $$ is divided by the line $$ \displaystyle x=3 $$ is
  • 15 : 49
  • 9 : 15
  • 7 : 15
  • 7 : 5
The function $$ \displaystyle f\left ( x \right )=\max \left \{ x^{2},\left ( 1-x \right )^{2},2x\left ( 1-x \right ) \forall 0\leq x\leq 1\right \} $$ then area of the region bounded by the curve $$ \displaystyle y=f\left ( x \right ) $$, x-axis and $$ \displaystyle x=0,x=1 $$ is equals,
  • $$ \displaystyle \frac{27}{17} $$
  • $$ \displaystyle \frac{9}{17} $$
  • $$ \displaystyle \frac{18}{17} $$
  • None of these
The area bounded by $$ \displaystyle x=a\cos ^{3}\theta,y=a\sin ^{3}\theta $$ is:
  • $$ \displaystyle \frac{3\pi a^{2}}{16} $$
  • $$ \displaystyle \frac{3\pi a^{2}}{8} $$
  • $$ \displaystyle \frac{3\pi a^{2}}{32} $$
  • $$ \displaystyle 3\pi a^{2} $$
The area bounded by $$ \displaystyle y=\frac{3x^{2}}{4} $$ and the line $$ \displaystyle 3x-2y+12=0 $$ is:
  • $$9$$
  • $$18$$
  • $$27$$
  • None of these
Find the area bounded by the curves $$\displaystyle x = y^{2}$$ and $$\displaystyle x = 3-2y^{2}$$
  • $$2$$ sq. units
  • $$4$$ sq. units
  • $$6$$ sq. units
  • $$8$$ sq. units
0:0:1


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