Question

solve it !Solve this............................solve it solve it solve it solve it solve it solve it solve it solve it solve it solve it solve it solve it solve it solve it solve it solve it​​

Answer 1

Given,

$\longrightarrow\left|\begin{array}{ccc}x+2&x+6&x-1\\x+6&x-1&x+2\\x-1&x+2&x+6\end{array}\right|=0$

Performing $R_1\to R_1-R_3$ and $R_2\to R_2-R_3,$

$\longrightarrow\left|\begin{array}{ccc}3&4&-7\\7&-3& -4\\x-1&x+2&x+6\end{array}\right|=0$

Performing $C_1\to C_1+C_2+C_3,$

$\longrightarrow\left|\begin{array}{ccc}0&4&-7\\0&-3&-4\\3x+7&x+2&x+6 \end{array}\right|=0$

Expanding along $C_1,$

$\longrightarrow(3x+7)(-16-21)=0$

Anyways,

$\longrightarrow3x+7=0$

$\longrightarrow\underline{\underline{x=-\dfrac{7}{3}}}$

Answer 2

Given :

Area of rhombus = 120 cm²

Length of diagonal = 8 cm

To find :

Length of another diagonal

According to the question,

$\sf{ : \implies Area \: of \: rhombus = \dfrac{1}{2} \times d _{1} \times d_{2} }$

$\sf : \implies{ {120 \: cm}^{2} = \dfrac{1}{2} \times 8 \: cm \times x}$

$\sf : \implies{ {120 \: cm}^{2} \times 2 = 8 \: cm \times x }$

$\sf : \implies{ {240 \: cm}^{2} = 8 \: cm \times x}$

$\sf : \implies{ \dfrac{240}{8} \: cm = x}$

${ \underline{ \boxed{ \sf \pink{ : \implies{ \bm3 \bm0 \: c m =x}}}}}$

${ \therefore{ \underline{\sf{So \:,the \: length \: of \: other \: diagonal \: is \: 3 0 \: cm}}}}$