Question

answer i i will mark you as brainliest

Answer 1

Solution :

In given fig.

ABCD is a trapezium & CD || AB and it's diagonal intersect at O.

We know that property of trapezium that, diagonals of Isosceles trapezium are equal.

$\underline{\boldsymbol{According\:to\:the\:question\::}}}}$

$\sf{\triangle DOB\:and\:\triangle BOC \:are\:vertically\:opposite\:equal.}$

$\longrightarrow\sf{\dfrac{AO}{CO} =\dfrac{OB}{OD} }\\\\\\\longrightarrow\sf{\dfrac{(5x-7)}{(2x+1)} =\dfrac{(7x-5)}{(7x+1)} }\\\\\\\longrightarrow\sf{(7x+1)(5x-7)=(2x+1)(7x-5)}\\\\\longrightarrow\sf{35x^{2}-49x+5x-7=14x^{2}-10x+7x-5}\\\\\longrightarrow\sf{35x^{2}-44x-7=14x^{2}-3x-5}\\\\\longrightarrow\sf{35x^{2}-14x^{2}-44x+3x-7+5=0}\\\\\longrightarrow\sf{21x^{2}-41x-2=0}$

$\underline{\boldsymbol{Using\:by\:quadratic\:formula\::}}}}$

As we know that given equation compared with ax² + bx + c;

• a = 21
• b = -41
• c = -2

Now;

$\longrightarrow\sf{x=\dfrac{-b\pm\sqrt{b^{2}-4ac} }{2a} }\\\\\\\longrightarrow\sf{x=\dfrac{-(-41)\pm\sqrt{(-41)^{2}-4\times 21\times (-2)} }{2\times 21} }\\\\\\\longrightarrow\sf{x=\dfrac{41\pm\sqrt{1681-4\times (-42)} }{42} }\\\\\\\longrightarrow\sf{x=\dfrac{41\pm\sqrt{1681+168} }{42} }\\\\\\\longrightarrow\sf{x=\dfrac{41\pm\sqrt{1849} }{42} }\\\\\\\longrightarrow\sf{x=\dfrac{41\pm43}{42} }\\\\\\\longrightarrow\sf{x=\dfrac{41+43}{42} \:\:Or\:\:x=\dfrac{41-43}{42} }$

$\longrightarrow\sf{x=\cancel{\dfrac{84}{42}} \:\:Or\:\:x=\cancel{\dfrac{-2}{42} }}\\\\\\\longrightarrow\bf{x=2\:\:Or\:\:x\neq -1/21}$

We know that negative value isn't acceptable.

Thus;

The value of x will be 2 cm .