MHT CET 2025 25th April Morning Shift (Mathematics) Question Paper with Solutions

$\begin{array}{cc}(𝑓\circ 𝑔)\left(𝑥\right)=𝑓\left(𝑔\right(𝑥\left)\right)& =𝑓\left(\frac{3𝑥+{𝑥}^3}{1+3{𝑥}^2}\right)\\ & =\log \left(\frac{1+\frac{3𝑥+{𝑥}^3}{1+3{𝑥}^2}}{1-\frac{3𝑥+{𝑥}^3}{1+3{𝑥}^2}}\right)\\ & =\log \left(\frac{1+3{𝑥}^2+3𝑥+{𝑥}^3}{1+3{𝑥}^2-3𝑥-{𝑥}^3}\right)\\ & =\log \frac{(1+𝑥{)}^3}{(1-𝑥{)}^3}\\ & =\log {\left(\frac{1+𝑥}{1-𝑥}\right)}^3\\ & =3f\left(𝑥\right)\end{array}$

$\begin{array}{cc}& \sec 𝑥+\tan 𝑥=2...\left(𝑖\right)\\ & {\sec }^2𝑥-{\tan }^2𝑥=1\\ & \Rightarrow (\sec 𝑥+\tan 𝑥)(\sec 𝑥-\tan 𝑥)=1\\ & \Rightarrow \sec 𝑥-\tan 𝑥=\frac{1}{2}...\left(𝑖𝑖\right)\end{array}$

Adding (i) and (ii), we get

$\sec 𝑥=\frac{5}{4}\Rightarrow \cos 𝑥=\frac{4}{5}$

Subtracting (ii) from (i), we get

$\begin{array}{cc}& \tan 𝑥=\frac{3}{4}\\ & \begin{array}{cc}\sin 𝑥& =\tan 𝑥\cos 𝑥\\ & =\frac{3}{4}\times \frac{4}{5}=\frac{3}{5}\\ 0<𝑥& <\frac{𝜋}{2}\Rightarrow 0<\frac{𝑥}{2}<\frac{𝜋}{4}\end{array}\end{array}$

$\therefore \sin \frac{𝑥}{2}$ and $\cos \frac{𝑥}{2}$ are positive.

$\begin{array}{cc}\sin \frac{𝑥}{2}& =\sqrt{\surd \frac{1-\cos 𝑥}{2}}\\ & =\sqrt{\surd \frac{1-\frac{4}{5}}{2}}\\ & =\frac{1}{\sqrt{\surd 10}}\\ {\cos }^2\frac{𝑥}{2}& =1-{\sin }^2\frac{𝑥}{2}\\ & =1-\frac{1}{10}\\ \therefore {\cos }^2\frac{𝑥}{2}& =\frac{9}{10}\\ \Rightarrow \cos \frac{𝑥}{2}& =\frac{3}{\sqrt{\surd 10}}\\ \sin \frac{𝑥}{4}& =\sqrt{\surd \frac{1-\cos \frac{𝑥}{2}}{2}}\\ & =\sqrt{\surd \frac{1-\frac{3}{\sqrt{\surd 10}}}{2}}\\ & =\sqrt{\surd \frac{10-3\sqrt{\surd 10}}{20}}\\ & =\frac{1}{\sqrt{\surd 2(10+3\sqrt{\surd 10})}}\end{array}$

$\begin{array}{cc}& 𝑥=3(\cos t+\sin t),𝑦=4(\cos t-\sin t)\\ & \Rightarrow \frac{𝑥}{3}=\cos t+\sin t,\frac{𝑦}{4}=\cos t-\sin t\\ & \Rightarrow \frac{{𝑥}^2}{9}=1+\sin 2t,\frac{{𝑦}^2}{16}=1-\sin 2t\end{array}$

$\Rightarrow \frac{{𝑥}^2}{9}+\frac{{𝑦}^2}{16}=2$

$\Rightarrow \frac{{𝑥}^2}{18}+\frac{{𝑦}^2}{32}=1$, which is an ellipse.

Here, $a^2=18,b^2=32$

$𝑒=\sqrt{\surd 1-\frac{{𝑎}^2}{{𝑏}^2}}=\sqrt{\surd 1-\frac{9}{16}}=\frac{\sqrt{\surd 7}}{4}$

Let P be the population at time t years.

Then, $\frac{\mathrm{dP}}{\mathrm{dt}}=\mathrm{kP}$

$\Rightarrow \frac{\mathrm{dP}}{P}=\mathrm{kdt}$

Integrating on both sides, we get

$\log 𝑃=𝑘𝑡+𝑐$

When $t=0,P=400000$

$\begin{array}{cc}\therefore & \log 400000=0+c\\ & \Rightarrow c=\log 400000\\ \therefore & \log P=\mathrm{kt}+\log 400000\\ & \Rightarrow \log \left(\frac{P}{400000}\right)=\mathrm{kt}...\left(𝑖\right)\end{array}$

When $𝑡=20$ years, $𝑃=600000$

$\begin{array}{cc}\therefore & \log \left(\frac{600000}{400000}\right)=20k\\ & \Rightarrow k=\frac{1}{20}\log \left(\frac{3}{2}\right)\\ \therefore & \log \left(\frac{P}{400000}\right)=\frac{t}{20}\log \frac{3}{2}...\left[𝐹𝑟𝑜𝑚\right(𝑖\left)\right]\end{array}$

$\therefore$ Population after another 20 years i.e. $t=40$ years,

$\begin{array}{cc}& \log \left(\frac{P}{400000}\right)=\frac{40}{20}\log \frac{3}{2}\\ & \Rightarrow \log \left(\frac{P}{400000}\right)=2\log \frac{3}{2}\\ & \Rightarrow \frac{P}{400000}=\frac{9}{4}\\ & \Rightarrow P=9\text{ lakhs }\end{array}$

$\begin{array}{cc}& \text{ Let }\overline{𝑐}=𝑝\widehat{𝑖}+𝑞\widehat{𝑗}+𝑟\widehat{𝑘}\\ & \begin{array}{cc}& \overline{𝑎}\times \overline{𝑐}=\mid \begin{array}{ccc}\widehat{𝑖}& \widehat{𝑗}& \widehat{𝑘}\\ 1& 1& 1\\ 𝑝& 𝑞& 𝑟\end{array}\mid \\ & =\widehat{𝑖}(𝑟-𝑞)-\widehat{𝑗}(𝑟-𝑝)+\widehat{𝑘}(𝑞-𝑝)\\ & \overline{𝑎}\times \overline{𝑐}=\overline{𝑏}\\ & \Rightarrow \widehat{𝑖}(𝑟-𝑞)-\widehat{𝑗}(𝑟-𝑝)+\widehat{𝑘}(𝑞-𝑝)=\widehat{𝑗}-\widehat{𝑘}\\ & \Rightarrow 𝑟-𝑞=0,𝑝-𝑟=1,𝑝-𝑞=1...\left(𝑖\right)\end{array}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \overline{𝑎}\cdot \overline{𝑐}=3\\ & \Rightarrow (\widehat{𝑖}+\widehat{𝑗}+\widehat{𝑘})\cdot (\widehat{𝑝}\widehat{𝑖}+𝑞\widehat{𝑗}+𝑟\widehat{𝑘})=3\\ & \Rightarrow 𝑝+𝑞+𝑟=3...\left(𝑖𝑖\right)\end{array}\\ & \text{ From (i) and (ii), we get }\\ & \begin{array}{cc}& 𝑝=\frac{5}{3},𝑞=\frac{2}{3},𝑟=\frac{2}{3}\\ \therefore & \overline{𝑐}=\frac{5}{3}\widehat{𝑖}+\frac{2}{3}\widehat{𝑗}+\frac{2}{3}\widehat{𝑘}\end{array}\end{array}$

The feasible region lies on the origin side of $|𝑥-𝑦|=1$ and is in the first quadrant.

$\therefore$ The common region is unbounded.

If the lines $\frac{𝑥-{𝑥}_1}{{𝑎}_1}=\frac{𝑦-{𝑦}_1}{{𝑏}_1}=\frac{𝑧-{𝑧}_1}{{𝑐}_1}$ and $\frac{𝑥-{𝑥}_2}{a_2}=\frac{𝑦-{𝑦}_2}{b_2}=\frac{z-z_2}{c_2}$ are coplanar, then $\mid \begin{array}{ccc}{𝑥}_2-{𝑥}_1& {𝑦}_2-{𝑦}_1& {𝑧}_2-{𝑧}_1\\ {𝑎}_1& {𝑏}_1& {𝑐}_1\\ {𝑎}_2& {𝑏}_2& {𝑐}_2\end{array}\mid =0$ and equation of plane containing them is

$\mid \begin{array}{ccc}𝑥-{𝑥}_1& 𝑦-{𝑦}_1& 𝑧-{𝑧}_1\\ {𝑎}_1& {𝑏}_1& {𝑐}_1\\ {𝑎}_2& {𝑏}_2& {𝑐}_2\end{array}\mid =0$

Since the given lines are coplanar.

$\begin{array}{cc}& \therefore \mid \begin{array}{ccc}2-0& 6-2& 3+3\\ 1& 2& 𝜆\\ 2& 3& 𝜆\end{array}\mid =0\\ & \Rightarrow 2(-𝜆)-4(-𝜆)+6(-1)=0\\ & \Rightarrow 2𝜆=6\\ & \Rightarrow 𝜆=3\end{array}$

Equation of the plane containing the given lines is

$\mid \begin{array}{ccc}𝑥-0& 𝑦-2& 𝑧+3\\ 1& 2& 𝜆\\ 2& 3& 𝜆\end{array}\mid =0$

$\begin{array}{cc}& \begin{array}{cc}& \Rightarrow 𝑥(-𝜆)-(𝑦-2)(-𝜆)+(𝑧+3)(-1)=0\\ & \Rightarrow -3𝑥+3𝑦-6-𝑧-3=0\\ & \Rightarrow 3𝑥-3𝑦+𝑧+9=0\end{array}\\ & \text{ Only option (D) satisfies this equation. }\end{array}$

$\begin{array}{cc}𝐿𝑒𝑡𝐼& ={\int }_3^1\frac{\log {𝑥}^2}{\log (16{𝑥}^2-8{𝑥}^3+{𝑥}^4)}𝑑𝑥\\ & ={\int }_3^1\frac{2\log 𝑥}{\log \left[{𝑥}^2(16-8𝑥+{𝑥}^2)\right]}𝑑𝑥\\ & ={\int }_3^1\frac{2\log 𝑥}{\log {𝑥}^2+\log (16-8𝑥+{𝑥}^2)}𝑑𝑥\\ & ={\int }_3^1\frac{2\log 𝑥}{\log {𝑥}^2+\log (4-𝑥{)}^2}𝑑𝑥\\ & ={\int }_3^1\frac{2\log 𝑥}{2\log 𝑥+2\log (4-𝑥)}𝑑𝑥\\ & ={\int }_3^1\frac{\log 𝑥}{\log 𝑥+\log (4-𝑥)}𝑑𝑥\\ & =\frac{1}{2}(3-1)=1\\ & \cdots [\because {\int }_{𝑏}^{𝑎}\frac{𝑓\left(𝑥\right)}{𝑓\left(𝑥\right)+𝑓(𝑎+𝑏-𝑥)}𝑑𝑥=\frac{1}{2}(𝑏-𝑎\left)\right]\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \frac{{𝑥}^2}{{𝑎}^2}+\frac{{𝑦}^2}{{𝑏}^2}=1\\ & \Rightarrow {𝑏}^2{𝑥}^2+{𝑎}^2{𝑦}^2={𝑎}^2{𝑏}^2...\left(𝑖\right)\end{array}\\ & \text{ Differentiating w.r.t. }𝑥\text{, we get }\\ & \begin{array}{cc}& 2{𝑏}^2𝑥+2{𝑎}^2𝑦\frac{𝑑𝑦}{𝑑𝑥}=0\\ & \Rightarrow 2{𝑎}^2𝑦\frac{𝑑𝑦}{𝑑𝑥}=-2{𝑏}^2𝑥\\ & \Rightarrow \frac{𝑑𝑦}{𝑑𝑥}=\frac{-{𝑏}^2}{{𝑎}^2}\left(\frac{𝑥}{𝑦}\right)...\left(𝑖𝑖\right)\\ & \Rightarrow \frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}=\frac{-{𝑏}^2}{{𝑎}^2}\left[\frac{𝑦-𝑥\frac{𝑑𝑦}{𝑑𝑥}}{{𝑦}^2}\right]\\ & =\frac{-{𝑏}^2}{{𝑎}^2{𝑦}^2}[𝑦-𝑥\left(\frac{-{𝑏}^2𝑥}{{𝑎}^2𝑦}\right)]\dots [\text{ From (ii)] }\\ & =\frac{-{𝑏}^2}{{𝑎}^2{𝑦}^2}\left[\frac{{𝑎}^2{𝑦}^2+{𝑏}^2{𝑥}^2}{{𝑎}^2𝑦}\right]\\ & =\frac{-{𝑏}^2}{{𝑎}^2{𝑦}^2}\times \frac{{𝑎}^2{𝑏}^2}{{𝑎}^2𝑦}\dots [\text{ From (i)] }\\ & =\frac{-{𝑏}^4}{{𝑎}^2{𝑦}^3}\end{array}\end{array}$

Let $f\left(𝑥\right)={𝑥}^3+a{𝑥}^2+b𝑥$

Since $f\left(𝑥\right)$ satisfies the Rolle's theorem,

$\begin{array}{cc}& f\left(1\right)=f\left(2\right)\\ & \Rightarrow (1{)}^3+a(1{)}^2+b\left(1\right)=(2{)}^3+a(2{)}^2+b\left(2\right)\\ & \Rightarrow 1+a+b=8+4a+2b\\ & \Rightarrow 3a+b=-7...\left(𝑖\right)\\ & f\left(𝑥\right)={𝑥}^3+{\mathrm{ax}}^2+\mathrm{bx}\\ \therefore & f^{\prime }\left(𝑥\right)=3{𝑥}^2+2\mathrm{ax}+b\end{array}$

Now, $f^{\prime }\left(c\right)=0$

$\begin{array}{cc}& \Rightarrow {𝑓}^{\prime }\left(\frac{4}{3}\right)=0\\ & \Rightarrow 3{\left(\frac{4}{3}\right)}^2+2𝑎\left(\frac{4}{3}\right)+𝑏=0\\ & \Rightarrow 8𝑎+3𝑏=-16...\left(𝑖𝑖\right)\end{array}$

Solving (i) and (ii), we get

$𝑎=-5,𝑏=8$

$\begin{array}{cc}𝐿𝑒𝑡𝐼& =\int \frac{1}{e^{𝑥}+1}d𝑥\\ & =\int \frac{1+e^{𝑥}-e^{𝑥}}{e^{𝑥}+1}d𝑥=\int (1-\frac{e^{𝑥}}{e^{𝑥}+1})d𝑥\\ & =𝑥-\log |e^{𝑥}+1|+c\end{array}$

$\begin{array}{cc}𝑦& ={\tan }^{-1}\left(\frac{4𝑥}{1+5{𝑥}^2}\right)+{\cot }^{-1}\left(\frac{3-2𝑥}{2+3𝑥}\right)\\ & ={\tan }^{-1}\left(\frac{4𝑥}{1+5{𝑥}^2}\right)+{\tan }^{-1}\left(\frac{2+3𝑥}{3-2𝑥}\right)\\ & ={\tan }^{-1}\left(\frac{5𝑥-𝑥}{1+5𝑥\left(𝑥\right)}\right)+{\tan }^{-1}\left(\frac{\frac{2}{3}+𝑥}{1-\frac{2}{3}𝑥}\right)\\ & ={\tan }^{-1}\left(5𝑥\right)-{\tan }^{-1}𝑥+{\tan }^{-1}\left(\frac{2}{3}\right)+{\tan }^{-1}𝑥\\ & ={\tan }^{-1}\left(5𝑥\right)+{\tan }^{-1}\left(\frac{2}{3}\right)\\ \therefore \frac{𝑑𝑦}{𝑑𝑥}& =\frac{1}{1+(5𝑥{)}^2}\cdot \frac{𝑑}{𝑑𝑥}\left(5𝑥\right)=\frac{5}{1+25{𝑥}^2}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& 𝑥\frac{d𝑦}{d𝑥}=2𝑦\\ & \Rightarrow \frac{d𝑦}{𝑦}=\frac{2d𝑥}{𝑥}\end{array}\\ & \text{ Integrating on both sides, we get }\\ & \begin{array}{cc}& \log 𝑦=2\log 𝑥+\log 𝑐\\ & \Rightarrow \log 𝑦=\log {𝑥}^2+\log 𝑐\end{array}\end{array}$

$\begin{array}{cc}& \Rightarrow \log 𝑦=\log 𝑐{𝑥}^2\\ & \Rightarrow 𝑦=𝑐{𝑥}^2\end{array}$

The above equation represents a family of parabolas with vertex at the origin and axis along the Y -axis.

$\begin{array}{cc}& \int e^{𝑥}\left(\frac{𝑥+5}{(𝑥+6{)}^2}\right)d𝑥\\ & =\int e^{𝑥}\left(\frac{𝑥+6-1}{(𝑥+6{)}^2}\right)d𝑥\\ & =\int e^{𝑥}[\frac{1}{𝑥+6}-\frac{1}{(𝑥+6{)}^2}]d𝑥\\ & =\frac{e^{𝑥}}{𝑥+6}+𝑐\\ & \dots [\because \int e^{𝑥}\left[f\right(𝑥)+f^{\prime }(𝑥\left)\right]d𝑥=e^{𝑥}f(𝑥)+c]\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& (5+3\sin 𝜃)(2\cos 𝜃+1)=0\\ & \Rightarrow \sin 𝜃=\frac{-5}{3},\text{ which is not possible }\\ & \text{ or }\cos 𝜃=\frac{-1}{2}\\ & \Rightarrow \cos 𝜃=\cos \frac{2𝜋}{3}\text{ and }\cos 𝜃=\cos \frac{4𝜋}{3}\end{array}\\ & \therefore \text{ The principal solutions are }\frac{2𝜋}{3}\text{ and }\frac{4𝜋}{3}\text{. }\end{array}$

$\text{ The probability distribution of }X\text{ is }$

$\begin{array}{cccccc}X=𝑥& 0& 1& 2& 3& 4\\ P(X=𝑥)& 0.1& K& 2k& 2k& k\end{array}$

$\begin{array}{cc}& \text{ Since }\sum _{𝑥=0}^{4}P(X=𝑥)=1\\ & 0.1+k+2k+2k+k=1\\ & \Rightarrow 6k=0.9\\ & \begin{array}{cc}& \Rightarrow k=0.15\\ & P(X\ge 2)=1-P(X\le 2)\\ & =1-(0.1+k)\\ & =1-(0.1+0.15)\\ & =0.75\end{array}\end{array}$

$\begin{array}{cc}& \text{ Principal value of }{\cos }^{-1}𝑥\text{ is }[0,𝜋]\text{. }\\ & \begin{array}{cc}& {\cos }^{-1}\left[\frac{1}{\sqrt{\surd 2}}(\cos \frac{9𝜋}{10}-\sin \frac{9𝜋}{10})\right]\\ & ={\cos }^{-1}(\frac{1}{\sqrt{\surd 2}}\cos \frac{9𝜋}{10}-\frac{1}{\sqrt{\surd 2}}\sin \frac{9𝜋}{10})\\ & ={\cos }^{-1}[\cos (\frac{9𝜋}{10}+\frac{𝜋}{4})]\end{array}\end{array}$

$\begin{array}{cc}& ={\cos }^{-1}[\cos \left(\frac{23𝜋}{20}\right)]\\ & ={\cos }^{-1}[\cos (2𝜋-\frac{23𝜋}{20})]\\ & ={\cos }^{-1}(\cos \frac{17𝜋}{20})\\ & =\frac{17𝜋}{20}\end{array}$

The length of the perpendicular from a point $({𝑥}_1,{𝑦}_1,{𝑧}_1)$ to the plane $a𝑥+b𝑦+\mathrm{cz}+d=0$ is

$\begin{array}{cc}𝑑& =\mid \frac{𝑎{𝑥}_1+𝑏{𝑦}_1+𝑐{𝑧}_1+𝑑}{\sqrt{\surd {𝑎}^2+{𝑏}^2+{𝑐}^2}}\mid \\ & =\mid \frac{2\left(1\right)-2\left(\frac{3}{2}\right)+4\left(2\right)+17}{\sqrt{\surd 4+4+16}}\mid \\ & =\frac{24}{\sqrt{\surd 24}}\\ & =\sqrt{\surd 24}\\ & =2\sqrt{\surd 6}\text{ units }\end{array}$

$\begin{array}{cc}& \begin{array}{cc}\overline{\mathrm{OA}}\times \overline{\mathrm{OB}}& =\mid \begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 1\\ 2& 1& 3\end{array}\mid \\ & =5\hat{i}-\hat{j}-3\hat{k}\\ \overline{\mathrm{AB}}\times \overline{\mathrm{AC}}& =\mid \begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 2\\ -2& -1& 1\end{array}\mid \\ & =\hat{i}-5\hat{j}-3\hat{k}\end{array}\\ & \text{ Angle between the faces }\mathrm{OAB}\text{ and }\mathrm{ABC}\text{ is }\\ & \begin{array}{cc}& \cos 𝜃=\frac{(5\hat{i}-\hat{j}-3\hat{k})\cdot (\hat{i}-5\hat{j}-3\hat{k})}{\sqrt{\surd (5{)}^2+(-1{)}^2+(-3{)}^2}\sqrt{\surd (1{)}^2+(-5{)}^2+(-3{)}^2}}\\ & =\frac{5+5+9}{\sqrt{\surd 35}\sqrt{\surd 35}}\\ & \therefore \cos 𝜃=\frac{19}{35}\\ & \therefore 𝜃={\cos }^{-1}\left(\frac{19}{35}\right)\end{array}\end{array}$

$\left|𝐴\right|=\mid \begin{array}{ccc}3& -3& 4\\ 2& -3& 4\\ 0& -1& 1\end{array}\mid =1\ne 0$

$\begin{array}{cc}\mathrm{adj}𝐴& ={⎡⎣⎢\begin{array}{ccc}1& -2& -2\\ -1& 3& 3\\ 0& -4& -3\end{array}⎤⎦⎥}^{\top }\\ & =⎡⎣⎢\begin{array}{ccc}1& -1& 0\\ -2& 3& -4\\ -2& 3& -3\end{array}⎤⎦⎥\\ \therefore {𝐴}^{-1}& =⎡⎣⎢\begin{array}{ccc}1& -1& 0\\ -2& 3& -4\\ -2& 3& -3\end{array}⎤⎦⎥\\ {𝐴}^2& =⎡⎣⎢\begin{array}{ccc}3& -3& 4\\ 2& -3& 4\\ 0& -1& 1\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}3& -3& 4\\ 2& -3& 4\\ 0& -1& 1\end{array}⎤⎦⎥\\ & =⎡⎣⎢\begin{array}{ccc}3& -4& 4\\ 0& -1& 0\\ -2& 2& -3\end{array}⎤⎦⎥\\ {𝐴}^3& ={𝐴}^2\cdot 𝐴=⎡⎣⎢\begin{array}{ccc}3& -4& 4\\ 0& -1& 0\\ -2& 2& -3\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}3& -3& 4\\ 2& -3& 4\\ 0& -1& 1\end{array}⎤⎦⎥\\ & =⎡⎣⎢\begin{array}{ccc}1& -1& 0\\ -2& 3& -4\\ -2& 3& -3\end{array}⎤⎦⎥\\ & ={𝐴}^{-1}\end{array}$

$\begin{array}{cc}\text{ Required area }& =2{\int }_1^0(4\sqrt{\surd 𝑦}-\frac{\sqrt{\surd 𝑦}}{3})d𝑦\\ & =2{[4\left(\frac{{𝑦}^{\frac{3}{2}}}{\frac{3}{2}}\right)-\frac{1}{3}\left(\frac{{𝑦}^{\frac{3}{2}}}{\frac{3}{2}}\right)]}_1^0\\ & =2(\frac{8}{3}-\frac{2}{9})\\ & =2\left(\frac{22}{9}\right)\\ & =\frac{44}{9}\text{ sq. units }\end{array}$