JEE Advanced 2007

LEVELJEE Advanced

Animated Solution for Mathematics - Circles: Let $A B C D$ be a quadrilateral with area 18, with side $A B$ parallel to the side $C D$ and $2 A B = C D$ . Let $A D$ be perpendicular to $A B$ and $C D$ . If a circle is drawn inside the quadrilateral $A B C D$ touching all the sides, then its radius is

Q: Why do we reject the negative root for $r$?

In geometry, the radius $r$ represents a physical length. Lengths must always be positive. Therefore, when we solve the quadratic equation $r^2 = 4$, we only consider the positive solution, $r = 2$.

Select Answer:

Visualized Solution

Visualizing the Geometry

Given: Quadrilateral $A B C D$ with Area $= 18$ .
$A B ∥ C D$ and $2 A B = C D$ .
$A D ⊥ A B$ and $A D ⊥ C D$ .
A circle is inscribed, touching all sides.

Defining Variables $a$ and $r$

Let $A B = a$ . Then $C D = 2 a$ .
Let the radius of the circle be $r$ .
Since the circle touches $A B$ and $C D$ , the height $A D = 2 r$ .

Using the Area Constraint

Area of Trapezoid $= \frac{1}{2} \times (A B + C D) \times A D$
Substitute the known variables:
$18 = \frac{1}{2} \times (a + 2 a) \times 2 r$

Simplifying the Area Equation

Simplify the expression:
$18 = 3 a \times r$
$a r = 6$ --- (Equation 1)

Constructing Right Triangle $△ B E C$

Draw $B E ⊥ C D$ .
In $△ B E C$ :
$B E = A D = 2 r$
$E C = C D - A B = 2 a - a = a$

Applying Pythagoras Theorem

Using Pythagoras in $△ B E C$ :
$B C^{2} = B E^{2} + E C^{2}$
$B C^{2} = (2 r)^{2} + a^{2}$
$B C^{2} = 4 r^{2} + a^{2}$ --- (Equation 2)

The Tangential Quadrilateral Property

For a tangential quadrilateral, sums of opposite sides are equal:
$A B + C D = A D + B C$

Expressing $B C$ via Tangential Property

Substitute the variables:
$a + 2 a = 2 r + B C$
$3 a = 2 r + B C$
$B C = 3 a - 2 r$ --- (Equation 3)

Equating the Expressions for $B C$

Equating $B C^{2}$ from Equation 2 and Equation 3:
$(3 a - 2 r)^{2} = 4 r^{2} + a^{2}$

Expanding the Equation

Expand the left side:
$9 a^{2} + 4 r^{2} - 12 a r = 4 r^{2} + a^{2}$

Simplifying to Relate $a$ and $r$

Cancel $4 r^{2}$ and rearrange:
$8 a^{2} = 12 a r$
$2 a = 3 r ⟹ a = \frac{3}{2} r$

Substituting back into Area Equation

Substitute $a = \frac{3}{2} r$ into Equation 1 ( $a r = 6$ ):
$(\frac{3}{2} r) \times r = 6$

Solving for the Radius

$\frac{3}{2} r^{2} = 6$
$r^{2} = 4$
Taking the positive root: $r = 2$

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

Imagine a quadrilateral

A B C D

where

A B ∥ C D

and

2 A B = C D

. We are given that

A D

is perpendicular to both

A B

and

C D

, defining this as a right trapezoid.

Let

A B = a

. Consequently,

C D = 2 a

Since a circle is inscribed within the trapezoid touching all four sides, the height of the trapezoid

A D

must be equal to the diameter of the circle. If

r

is the radius, then

A D = 2 r

The Area Constraint

The area of a trapezoid is given by the formula:

Area = \frac{1}{2} \times (A B + C D) \times A D

Substituting our defined variables into the area equation of

18

18 = \frac{1}{2} \times (a + 2 a) \times 2 r

Simplifying this expression, the

2

in the numerator and denominator cancels out:

18 = 3 a \times r

a r = 6

This is our first crucial relationship.

The Hidden Triangle

Construct a perpendicular from point

B

to the side

C D

, meeting at point

E

. This creates a right-angled triangle

△ B E C

The height

B E

is equal to

A D

, which is

2 r

. The base

E C

is the difference between the parallel sides:

E C = C D - A B = 2 a - a = a

Applying the Pythagorean theorem to

△ B E C

B C^{2} = B E^{2} + E C^{2}

B C^{2} = (2 r)^{2} + a^{2} = 4 r^{2} + a^{2}

The Tangential Magic

For any tangential quadrilateral, the sum of opposite sides must be equal. Therefore:

A B + C D = A D + B C

Substituting our variables:

a + 2 a = 2 r + B C

3 a = 2 r + B C \Rightarrow B C = 3 a - 2 r

Now, equate the two expressions for

B C^{2}

(3 a - 2 r)^{2} = 4 r^{2} + a^{2}

Expanding the left side:

9 a^{2} + 4 r^{2} - 12 a r = 4 r^{2} + a^{2}

The

4 r^{2}

terms cancel out, leaving:

9 a^{2} - 12 a r = a^{2}

8 a^{2} = 12 a r

Since $a eq 0$, we divide by

a

8 a = 12 r \Rightarrow a = \frac{3}{2} r

The Final Victory

Substitute

a = \frac{3}{2} r

back into our first relationship,

a r = 6

(\frac{3}{2} r) \times r = 6

\frac{3}{2} r^{2} = 6

Solving for

r^{2}

r^{2} = 6 \times \frac{2}{3} = 4

Since the radius must be positive, we find the final result: $r = 2$

Common Traps & FAQs

Wait, why is the height

A D

equal to

2 r

Great question! Because the circle is inscribed in the quadrilateral, it touches both parallel sides

A B

and

C D

. The distance between these two parallel lines is the diameter of the circle, which is

2 r

. Since

A D

is perpendicular to both, it represents this distance perfectly.

Why can we use the tangential quadrilateral property

A B + C D = A D + B C

This is a fundamental property for any quadrilateral that circumscribes a circle. The tangents drawn from a point to a circle are equal in length. If you label the points of tangency on the four sides, you will see that the sum of the opposite sides naturally equals the sum of the four tangent segments, proving the property.

Why do we reject the negative root for

r

In geometry, the radius

r

represents a physical length. Lengths must always be positive. Therefore, when we solve the quadratic equation

r^{2} = 4

, we only consider the positive solution,

r = 2

Select Answer:

Visualized Solution

Visualizing the Geometry

Defining Variables $a$ and $r$

Using the Area Constraint

Simplifying the Area Equation

Constructing Right Triangle $△ B E C$

Applying Pythagoras Theorem

The Tangential Quadrilateral Property

Expressing $B C$ via Tangential Property

Equating the Expressions for $B C$

Expanding the Equation

Simplifying to Relate $a$ and $r$

Substituting back into Area Equation

Solving for the Radius

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Area Constraint

The Hidden Triangle

The Tangential Magic

The Final Victory

Similar Questions

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Let $A B$ be a chord of length 12 of the circle $(x - 2)^{2} + (y + 1)^{2} = \frac{169}{4}$ . If tangents drawn to the circle at points $A$ and $B$ intersect at the point $P$ , then five times the distance of point $P$ from chord $A B$ is equal to ________

Select Answer:

Visualized Solution

Visualizing the Geometry

Defining Variables .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Using the Area Constraint

Simplifying the Area Equation

Constructing Right Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △BEC

Applying Pythagoras Theorem

The Tangential Quadrilateral Property

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BC via Tangential Property

Equating the Expressions for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BC

Expanding the Equation

Simplifying to Relate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Substituting back into Area Equation

Solving for the Radius

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Area Constraint

The Hidden Triangle

The Tangential Magic

The Final Victory

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the tangents at two points A and B on the circle x2+y2−4x+3=0 meet at origin O(0,0). Then the area of the triangle of OAB is

Defining Variables $a$ and $r$

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Expressing $B C$ via Tangential Property

Equating the Expressions for $B C$

Simplifying to Relate $a$ and $r$

Let the tangents at two points $A$ and $B$ on the circle $x^{2} + y^{2} - 4 x + 3 = 0$ meet at origin $O (0, 0)$ . Then the area of the triangle of $O A B$ is