Question

solve 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

$\huge\mathtt{\fbox{\orange{★Answer★}}}$

$\large\underline\red{GIVEN,}$

$\sf\dashrightarrow \blue{THE\:GIVEN\:FRACTION\:IS\:A\:RATIONAL\:NUMBER.}$

$\sf\dashrightarrow {\blue{\mathbb{\text{ denominator is less than its numerator by 5.}}}}$

$\sf\therefore \blue{let\:the\:numerator\:be\:x}$

$\sf\dashrightarrow \blue{denominator= x-5}$

$\sf\dashrightarrow \blue{\dfrac{x}{x-5}}$

$\sf\dashrightarrow \bold\pink{if \:5 \:is \:added\: to \:the \:numerator, \:numerator \:becomes\: ; \dfrac{11}{6}}$

$\sf\dashrightarrow \red{numerator= x+5}$

THE EQUATION FORM IS,

$\rm{\boxed{\sf{\green{ \circ\:\: \dfrac{x+5}{x-5} = \dfrac{11}{6}\:\: \circ}}}}$

$\large\underline\purple{TO\:FIND,}$

$\sf\dashrightarrow \red{\:THE\:ORIGINAL\:RATIONAL\:NUMBER}$

$\sf\implies \green{\dfrac{x + 5}{x - 5} = \dfrac{11}{6}}$

$\sf\implies \green{6 \times (x+5)= 11 \times (x-5)}$

$\sf\implies \green{6x+30= 11x-55}$

$\sf\implies \green{30+55=11x-6x}$

$\sf\implies \green{85= 5x}$

$\sf\implies \green{x= \dfrac{85}{5}}$

$\sf\implies \green{x= \cancel \dfrac{85}{5}}$

$\sf\implies \orange{x = 17}$

$\rm{\boxed{\sf{ \circ\:\: x= 17 \:\: \circ}}}$

THE ORIGINAL NUMBERS ARE,

$\sf\implies \red{numerator=x =17}$

$\sf\implies \red{d enominator= x-5}$

$\sf\implies \red{denominator=17-5}$

$\sf\implies \pink{denominator=12}$

$\large\underline\orange{FRACTION,}$

$\sf\dashrightarrow \purple{\dfrac{x}{x-5}= \dfrac{17}{12}}$

$\sf\dashrightarrow \purple{\dfrac{NUMERATOR}{DENOMINATOR}= \dfrac{17}{12}}$

$\rm\underline\blue{NUMERATOR\:IS\:17\:AND\: DENOMINATOR\:IS\:12}$

$\rm{\boxed{\sf{ \circ\:\: \dfrac{NUMERATOR}{DENOMINATOR}= \dfrac{17}{12} \:\: \circ}}}$

Answer 2

Step-by-step explanation:

Given :

Area of rhombus = 120 cm²

Length of diagonal = 8 cm

To find :

Length of another diagonal

According to the question,

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

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

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

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

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

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

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