Question

1. Using suitable identity find

$\\ \bigg( \frac{3p}{2} - \frac{2}{3} q \bigg) ^{2}$

2. Find the volume of the rectangular box whose length, breadth and height are ( 2x + 1 ), ( x - 2 ) and 3x.

3. Using suitable identity, evaluate ( 103² ).

4. If p + q = 25 and p² + q² = 225 then find pq.

5. Show that

$\\ \bigg( \: \frac{4}{3} m \: - \: \frac{3}{4} n \: \bigg)^{2} \: + \: 2mn \: = \: \frac{16}{9} m^{2} + \frac{9}{16} n^{2}$
​

Answer 1

$\large\underline{\sf{Solution-1}}$

Given expression is

$\rm \: \bigg( \dfrac{3p}{2} - \dfrac{2q}{3} \bigg) ^{2}$

We know,

$\boxed{\sf{  \:\rm \: {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} \: \: }}$

So, using this identity, we get

$\rm \: = \: {\bigg(\dfrac{3p}{2} \bigg) }^{2} + {\bigg(\dfrac{2q}{3} \bigg) }^{2} - 2 \times \dfrac{3p}{2} \times \dfrac{2q}{3}$

$\rm \: = \:\dfrac{9 {p}^{2} }{4} + \dfrac{ {4q}^{2} }{9} - 2pq$

Hence,

$\rm\implies\boxed{\sf{  \:\rm \: \bigg( \frac{3p}{2} - \frac{2}{3} q \bigg) ^{2}= \:\dfrac{9 {p}^{2} }{4} + \dfrac{ {4q}^{2} }{9} - 2pq \: \: }}$

$\rule{190pt}{2pt}$

$\large\underline{\sf{Solution-2}}$

Length of rectangular box = 2x + 1

Breadth of rectangular box = x - 2

Height of rectangular box = 3x

So,

$\rm \: Volume_{(rectangular\:box)} = length \times breadth \times height$

$\rm \: = \:(2x + 1)(x - 2)(3x)$

$\rm \: = \:3x( {2x}^{2} - 4x + x - 2)$

$\rm \: = \:3x( {2x}^{2} - 3x - 2)$

$\rm \: = \: {6x}^{3} - {9x}^{2} - 6x$

Hence,

$\boxed{\sf{  \:\rm \:Volume_{(rectangular\:box)} = \: {6x}^{3} - {9x}^{2} - 6x \: \: }}$

$\rule{190pt}{2pt}$

$\large\underline{\sf{Solution-3}}$

$\rm \: {(103)}^{2}$

$\rm \: = \: {(100 + 3)}^{2}$

We know,

$\boxed{\sf{  \:\rm \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \: \: }}$

So, using this identity, we get

$\rm \: = \: {(100)}^{2} + {3}^{2} + 2 \times 100 \times 3$

$\rm \: = \:10000 + 9 + 600$

$\rm \: = \:10609$

$\rule{190pt}{2pt}$

$\large\underline{\sf{Solution-4}}$

Given that,

$\rm \: p + q = 25$

On squaring both sides, we get

$\rm \: {(p + q)}^{2} = {25}^{2}$

$\rm \: {p}^{2} + {q}^{2} + 2pq \: = \: 625$

$\rm \: 225 + 2pq \: = \: 625$

$\rm \: 2pq \: = \: 625 - 225$

$\rm \: 2pq \: = \: 400$

$\rm\implies \:pq \: = \: 200$

$\rule{190pt}{2pt}$

$\large\underline{\sf{Solution-5}}$

Consider, LHS

$\rm \: \bigg( \: \dfrac{4}{3} m \: - \: \dfrac{3}{4} n \: \bigg)^{2} \: + \: 2mn \:$

$\rm \: = \:{\bigg(\dfrac{4m}{3} \bigg) }^{2} + {\bigg(\dfrac{3n}{4} \bigg) }^{2} - 2 \times \dfrac{4m}{3} \times \dfrac{3n}{4} + 2mn$

$\rm \: = \:\dfrac{16 {m}^{2} }{9} + \dfrac{9 {n}^{2} }{16} - 2mn + 2mn$

$\rm \: = \:\dfrac{16 {m}^{2} }{9} + \dfrac{9 {n}^{2} }{16}$

Hence,

$\boxed{\sf{  \:\rm \: \bigg( \: \frac{4}{3} m \: - \: \frac{3}{4} n \: \bigg)^{2} \: + \: 2mn \:= \:\dfrac{16 {m}^{2} }{9} + \dfrac{9 {n}^{2} }{16} \: \: }}$

Answer 2

Answer:

Solution−1

Given expression is

\begin{gathered}\rm \: \bigg( \dfrac{3p}{2} - \dfrac{2q}{3} \bigg) ^{2} \\ \end{gathered}

(

2

3p

−

3

2q

)

2

We know,

\begin{gathered}\boxed{\sf{ \:\rm \: {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} \: \: }} \\ \end{gathered}

(x−y)

2

=x

2

−2xy+y

2

So, using this identity, we get

\begin{gathered}\rm \: = \: {\bigg(\dfrac{3p}{2} \bigg) }^{2} + {\bigg(\dfrac{2q}{3} \bigg) }^{2} - 2 \times \dfrac{3p}{2} \times \dfrac{2q}{3} \\ \end{gathered}

=(

2

3p

)

2

+(

3

2q

)

2

−2×

2

3p

×

3

2q

\begin{gathered}\rm \: = \:\dfrac{9 {p}^{2} }{4} + \dfrac{ {4q}^{2} }{9} - 2pq \\ \end{gathered}

=

4

9p

2

+

9

4q

2

−2pq

Hence,

\begin{gathered}\rm\implies\boxed{\sf{ \:\rm \: \bigg( \frac{3p}{2} - \frac{2}{3} q \bigg) ^{2}= \:\dfrac{9 {p}^{2} }{4} + \dfrac{ {4q}^{2} }{9} - 2pq \: \: }} \\ \end{gathered}

⟹

(

2

3p

−

3

2

q)

2

=

4

9p

2

+

9

4q

2

−2pq

\rule{190pt}{2pt}

\large\underline{\sf{Solution-2}}

Solution−2

Length of rectangular box = 2x + 1

Breadth of rectangular box = x - 2

Height of rectangular box = 3x

So,

\begin{gathered}\rm \: Volume_{(rectangular\:box)} = length \times breadth \times height \\ \end{gathered}

Volume

(rectangularbox)

=length×breadth×height

\begin{gathered}\rm \: = \:(2x + 1)(x - 2)(3x) \\ \end{gathered}

=(2x+1)(x−2)(3x)

\begin{gathered}\rm \: = \:3x( {2x}^{2} - 4x + x - 2) \\ \end{gathered}

=3x(2x

2

−4x+x−2)

\begin{gathered}\rm \: = \:3x( {2x}^{2} - 3x - 2) \\ \end{gathered}

=3x(2x

2

−3x−2)

\begin{gathered}\rm \: = \: {6x}^{3} - {9x}^{2} - 6x \\ \end{gathered}

=6x

3

−9x

2

−6x

Hence,

\begin{gathered}\boxed{\sf{ \:\rm \:Volume_{(rectangular\:box)} = \: {6x}^{3} - {9x}^{2} - 6x \: \: }}\\ \end{gathered}

Volume

(rectangularbox)

=6x

3

−9x

2

−6x

\rule{190pt}{2pt}

\large\underline{\sf{Solution-3}}

Solution−3

\rm \: {(103)}^{2}(103)

2

\begin{gathered}\rm \: = \: {(100 + 3)}^{2} \\ \end{gathered}

=(100+3)

2

We know,

\begin{gathered}\boxed{\sf{ \:\rm \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \: \: }} \\ \end{gathered}

(x+y)

2

=x

2

+y

2

+2xy

So, using this identity, we get

\begin{gathered}\rm \: = \: {(100)}^{2} + {3}^{2} + 2 \times 100 \times 3 \\ \end{gathered}

=(100)

2

+3

2

+2×100×3

\begin{gathered}\rm \: = \:10000 + 9 + 600 \\ \end{gathered}

=10000+9+600

\begin{gathered}\rm \: = \:10609 \\ \end{gathered}

=10609

\rule{190pt}{2pt}

\large\underline{\sf{Solution-4}}

Solution−4

Given that,

\rm \: p + q = 25p+q=25

On squaring both sides, we get

\begin{gathered}\rm \: {(p + q)}^{2} = {25}^{2} \\ \end{gathered}

(p+q)

2

=25

2

\begin{gathered}\rm \: {p}^{2} + {q}^{2} + 2pq \: = \: 625 \\ \end{gathered}

p

2

+q

2

+2pq=625

\begin{gathered}\rm \: 225 + 2pq \: = \: 625 \\ \end{gathered}

225+2pq=625

\begin{gathered}\rm \: 2pq \: = \: 625 - 225 \\ \end{gathered}

2pq=625−225

\begin{gathered}\rm \: 2pq \: = \: 400 \\ \end{gathered}

2pq=400

\begin{gathered}\rm\implies \:pq \: = \: 200 \\ \end{gathered}

⟹pq=200

\rule{190pt}{2pt}

\large\underline{\sf{Solution-5}}

Solution−5

Consider, LHS

\rm \: \bigg( \: \dfrac{4}{3} m \: - \: \dfrac{3}{4} n \: \bigg)^{2} \: + \: 2mn \:(

3

4

m−

4

3

n)

2

+2mn

\begin{gathered}\rm \: = \:{\bigg(\dfrac{4m}{3} \bigg) }^{2} + {\bigg(\dfrac{3n}{4} \bigg) }^{2} - 2 \times \dfrac{4m}{3} \times \dfrac{3n}{4} + 2mn \\ \end{gathered}

=(

3

4m

)

2

+(

4

3n

)

2

−2×

3

4m

×

4

3n

+2mn

\begin{gathered}\rm \: = \:\dfrac{16 {m}^{2} }{9} + \dfrac{9 {n}^{2} }{16} - 2mn + 2mn \\ \end{gathered}

=

9

16m

2

+

16

9n

2

−2mn+2mn

\begin{gathered}\rm \: = \:\dfrac{16 {m}^{2} }{9} + \dfrac{9 {n}^{2} }{16} \\ \end{gathered}

=

9

16m

2

+

16

9n

2

Hence,

\begin{gathered}\boxed{\sf{ \:\rm \: \bigg( \: \frac{4}{3} m \: - \: \frac{3}{4} n \: \bigg)^{2} \: + \: 2mn \:= \:\dfrac{16 {m}^{2} }{9} + \dfrac{9 {n}^{2} }{16} \: \: }} \\ \end{gathered}

(

3

4

m−

4

3

n)

2

+2mn=

9

16m

2

+

16

9n

2