Question

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Answer 1

$\Large\bf{\color{indigo}GiVeN,}$

The length of diagonals of a rhombus are in ratio 3 : 4.

It's perimeter is 80 cm.

$\bf\pink{Let,}$

The length of one diagonal is 3X.

=》 d₁ = 3X

And length of other diagonal is 4X.

=》 d₂ = 4X

$\bf\blue{We\:have,}$

$\red\bigstar\:\:\bf{\color{peru}Perimeter\:=\:4\times{Side}\:}$

$:\implies\:\:\bf{80\:=\:4\times{Side}\:}$

$:\implies\:\:\bf{Side\:=\:\dfrac{80}{4}\:}$

$:\implies\:\:\bf\green{Side\:=\:20\:cm\:}$

$\bf\purple{We\:know\:that,}$

❶ In rhombus all sides are equal and two diagonals are intersect at the mid point of each other.

❷ At the intersection point two diagonals makes 90° angle between them.

$\bf\red{So,}$

$\pink\bigstar\:\:\bf{\color{coral}(Side)^2\:=\:\Big(\dfrac{d_1}{2}\Big)^2\:+\:\Big(\dfrac{d_2}{2}\Big)^2\:}$

$:\implies\:\:\bf{(20)^2\:=\:\Big(\dfrac{3X}{2}\Big)^2\:+\:\Big(\dfrac{4X}{2}\Big)^2\:}$

$:\implies\:\:\bf{400\:=\:\dfrac{9X^2}{4}\:+\:\dfrac{16X^2}{4}\:}$

$:\implies\:\:\bf{400\:=\:\dfrac{9X^2\:+\:16X^2}{4}\:}$

$:\implies\:\:\bf{400\times{4}\:=\:25X^2\:}$

$:\implies\:\:\bf{1600\:=\:25X^2\:}$

$:\implies\:\:\bf{X^2\:=\:\dfrac{1600}{25}}$

$:\implies\:\:\bf{X\:=\:\sqrt{\dfrac{1600}{25}}\:}$

$:\implies\:\:\bf{X\:=\:\dfrac{40}{5}\:}$

$:\implies\:\:\bf{\color{peru}X\:=\:8\:cm\:}$

________________

$\bf\orange{Hence,}$

=》 d₁ = 3 × 8 = 24 cm

=》 d₂ = 4 × 8 = 32 cm

$\Large\bold\therefore$ The length of the diagonals are 24 cm & 32 cm.

Answer 2

Step-by-step explanation:

Step-by-step explanation:

Given Equation:-

⠀⠀⠀⠀ $\sf{\bigg[ \dfrac{5 {x}^{2} - 10}{12 } \bigg] }$

To find:-

value of x

Solution:-

use factor theorem

take the value of equation =0

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow \left [\dfrac {5x^2-10}{12}\right]=0$

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow \dfrac {5x^2-10}{12}=0$

using cross multiplication

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow 5x^2-10=0$

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow 5x^2=10$

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow x^2=\dfrac {10}{5}$

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow x^2=2$

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow x=\sqrt {2}$

$\\\\\therefore\sf x=\sqrt {2}.$