Question

if area of triangle is 384m² and its height is 30m. then its base?

a.30.5m
b.25.6m
c.29.2m
d.none of these ​

Answer 1

he required value is x^3-y^3-6xy=8x

3

−y

3

−6xy=8 .

Step-by-step explanation:

Given : x-y=2x−y=2

To find : What is x^3-y^3-6xy=?x

3

−y

3

−6xy=? ?

Solution :

x-y=2x−y=2

Cubing both side,

(x-y)^3=(2)^3(x−y)

3

=(2)

3

Using algebraic identity, (a-b)^3=a^3-b^3-3ab(a-b)(a−b)

3

=a

3

−b

3

−3ab(a−b)

x^3-y^3-3xy(x-y)=8x

3

−y

3

−3xy(x−y)=8

Substitute the value,

x^3-y^3-3xy(2)=8x

3

−y

3

−3xy(2)=8

x^3-y^3-6xy=8x

3

−y

3

−6xy=8

Therefore, the required value is x^3-y^3-6xy=8x

3

−y

3

−6xy=8 .

#Learn more

- Simplify: (x - y)^2 + 2(x - y)(x + y) + (x + y)^2.

https://brainly.in/question/15806151

Answer 2

Given : The area of triangle is 384m² and its height is 30m.

Need To Find : The Base of Triangle.

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❍ Let's Consider the Base of Triangle be x cm .

$\dag\underline {\frak{ As,\:We\:know\::}}$

$\star\boxed {\pink{\sf{ Area_{(Triangle)} = \dfrac{1}{2} \times b \times h }}}$

Where ,

• b is the Base of Triangle in m and h is the Height of Triangle in m .

$\underline {\frak{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

$:\implies \sf { 384 = \dfrac{1}{2} \times 30 \times x } \\\\ :\implies \sf { 384 = \dfrac{1}{\cancel{2}} \times \cancel {30} \times x } \\\\ :\implies \sf { 384 = 15 \times x } \\\\ :\implies \sf { \cancel{\dfrac{384}{15} }= x } \\\\ \underline {\boxed{\pink{ \mathrm {  x = 25.6\: m}}}}\:\bf{\bigstar}$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {  Base \:of\:Triangle \:is\:\bf{25.6\: m}}}}$

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$\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}$

$\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}$

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