Question

In the given figure if AB || DC, then find the value of x.

Answer 1

Hint: To solve this problem we need to know that the if the two sides of any quadrilateral are parallel and other two are non-parallel then it forms a trapezium and the diagonals of the trapezium divides each other proportionally so we get $\dfrac{AO}{OC}=\dfrac{BO}{OD}$ and we are given all the values of AO, BO, OC and OD in terms of x, so we will put the required values and solve it to find the value of x.

Complete step-by-step answer:
We are given that in the figure,

AB || DC,
And we know that when two sides of the quadrilateral are parallel and other two are not then it forms a trapezium,
And the diagonals of the trapezium divides each other proportionally, i.e. we get
$\dfrac{AO}{OC}=\dfrac{BO}{OD}\,\,...\left( 1 \right)$
And we are given that,
AO = x + 5
BO = x – 1
OC = x + 3
OD = x – 2
Now putting the above values in the equation 1 we get,
$\dfrac{x+5}{x+3}=\dfrac{x-1}{x-2}$
Now cross multiplying we get,
$\begin{align}
  & \left( x+5 \right)\left( x-2 \right)=\left( x-1 \right)\left( x+3 \right) \\
 & {{x}^{2}}+3x-10={{x}^{2}}+2x-3 \\
\end{align}$
Cancelling out ${{x}^{2}}$ from both the sides we get,
$\begin{align}
  & 3x-10=2x-3 \\
 & x=10-3 \\
 & x=7 \\
\end{align}$
Hence we get the value of x as 7.

Note: To solve problems related to the standard geometric figures you should know about all of their properties else you cannot solve its problems easily like in this problem we used the property that the diagonals of the trapezium divides each other proportionally. Try to solve more problems of this kind and also remember the property of trapezium we used in this problem.