Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# Let ABCD be a quadrilateral with area 18, with side AB parallel to the side CD and AB=2CD.Let AD be perpendicular to AB and CD. If a circle is drawn inside the quadrilateral ABCD touching all the side, then its radius isA) 3B) $2$C) $\dfrac{3}{2}$D) 1

Last updated date: 20th Jun 2024
Total views: 374.7k
Views today: 3.74k
Verified
374.7k+ views
Hint: The above question is based on the concept of mensuration. The main approach towards solving this is by identifying which quadrilateral it is and by using the area of the quadrilateral we can find the distance between the parallel sides from which we can find the radius of the circle. But before that we have to find the relation between the radius of the circle and the sides of the quadrilateral.

Complete step by step solution:
In the above question, given is the quadrilateral with parallel sides AB and CD. So, this is a property of trapezium. Quadrilateral is basically a four sided 2D figure whose sum of internal angles is 360.So there are different types of quadrilateral so since the property mentioned is having parallel sides then this property is of trapezium.
So in the question given is AB=2CD that means one side is twice the other and the side ad is perpendicular to AB and CD which means it has angle 90 to the side.
Let CD=l
So AB = 2

In the above figure, we have constructed the $CP$ line which is perpendicular to the $AB$.
Let, the radius of the circle is r.
So AD = diameter of the circle = r+r = 2r
From the figure, we get that,
$AM = r$
Let’s find the value of CN.
$CN = CD-DN = l-r$
but CN = MP = l-r

Now let’s find the value of PB.
$PB = AB-AP = 2l-l=l$

We know that, if two tangents are drawn from one point to a circle then, the length of two tangents will be the same.
So $BM = BG$
But $BM = BA-AM$
$BG = BM = 2l-r$

As per the same theorem, $CG = CN = l-r$

Now let's find the side $BC= BG+CG$
$BC = 2l-r + l-r= 3l-2r$

BPC is a right angle triangle. So we can apply the pythagorean theorem here.
$BP^2+PC^2 =BC^2$
$\Rightarrow l^2 + (2r)^2= (3l-2r)^2$
$\Rightarrow l^2+4r^2=9l^2+ 4r^2-12lr$
$4r^2$ will get cancelled. By simplifying this, we get
$\Rightarrow 8l^2 = 12lr$
Now here divide l on both the side,
$\Rightarrow l= \dfrac{3}{2}r$
Area of the trapezium = 18
$\dfrac{1}{2} (AB+CD) AD= 18$
Multiplying 2 in both the sides and putting the value of AB,CD and AD, we get
$\Rightarrow (2l+l) \times 2r= 36$
$\Rightarrow 6l \times r = 36$
$\Rightarrow lr=6$
putting the value of l in terms of r.
$\Rightarrow \dfrac{3}{2}r \times r = 6$
$\Rightarrow r^2=4$
taking square root on both the sides, we get
$\Rightarrow r=2$
So the radius of the circle is 2cm.
Hence, the correct answer is option B.

Note: An important thing to note is that the side AD is perpendicular to the side AB and DC. So the length becomes the distance between the two sides and also becomes the diameter of the circle and dividing it by 2 we get the radius.