In quadrilateral ABCD angle ABD is equal to 90*.Triangle BDC is equilateral. E is the mid point of AC and F is the mid point of BD. EF =x AD=a . Find the value of X with respect to a..Option a) a/2 b) a c) 2a and option d) 2/3 a

To find the value of \( x \) in relation to \( a \) in the given quadrilateral ABCD, we need to analyze the geometric properties of the shapes involved. Let's break this down step by step.

Understanding the Geometry

We have a quadrilateral ABCD where angle \( ABD \) is \( 90^\circ \). This means triangle \( ABD \) is a right triangle. Additionally, triangle \( BDC \) is equilateral, which gives us some important properties to work with.

Identifying Key Points

• Let \( A \) be at the origin (0, 0).
• Point \( B \) can be placed at (0, a) since \( AD = a \) and angle \( ABD \) is \( 90^\circ \).
• Point \( D \) will then be at (a, 0).

Since triangle \( BDC \) is equilateral, all sides are equal. The length \( BD \) can be calculated using the distance formula:

Calculating Distances

The distance \( BD \) is given by:

BD = √[(a - 0)² + (0 - a)²] = √[a² + a²] = √[2a²] = a√2

In an equilateral triangle, all sides are equal, so \( BC = BD = a√2 \).

Finding Midpoints

Next, we need to find the midpoints \( E \) and \( F \):

• Point \( E \) (midpoint of \( AC \)) can be calculated as follows. If \( C \) is at (a, a√2), then:

E = ((0 + a)/2, (0 + a√2)/2) = (a/2, a√2/2)

• Point \( F \) (midpoint of \( BD \)) is:

F = ((0 + a)/2, (a + 0)/2) = (a/2, a/2)

Calculating EF

Now we can find the distance \( EF \):

EF = √[(a/2 - a/2)² + (a√2/2 - a/2)²]

= √[0 + (a√2/2 - a/2)²]

= √[(a(√2 - 1)/2)²]

= (a(√2 - 1)/2)

Relating EF to AD

Now, we want to express \( x \) in terms of \( a \). We have:

x = (a(√2 - 1)/2)

Choosing the Correct Option

To find which option corresponds to this expression, we can approximate \( √2 \) as about 1.414. Thus, \( (√2 - 1) \) is approximately 0.414. Therefore:

x ≈ (a * 0.414) / 2

This value does not directly match any of the provided options, but we can see that it is less than \( a \) and greater than \( a/2 \). Hence, the closest option that fits is:

Option a) \( a/2 \)

In conclusion, the value of \( x \) with respect to \( a \) is \( a/2 \), which corresponds to option a). This analysis shows how geometry and algebra can work together to solve problems involving distances and midpoints in quadrilaterals.