A circle is inscribed in a quadrilateral ABCD, touching sides AB, BC CD and DA at P, Q, R and S, respectively. If AS = 6 cm, BC = 12 cm, and CR = 5 cm, then the length of AB (in cm) is:
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Circle Inscribed in Quadrilateral: Applying Tangent Properties
This problem involves a circle inscribed within a quadrilateral. The key property to solve this type of geometry problem is that tangents drawn from an external point to a circle are equal in length. In this scenario, the vertices of the quadrilateral ABCD are the external points, and the points where the circle touches the sides (P, Q, R, S) are the tangent points.
Let's use the given information and the tangent property to find the required length AB.
Understanding the Given Information
- The circle touches side AB at P, BC at Q, CD at R, and DA at S.
- Length of segment AS = 6 cm.
- Length of side BC = 12 cm.
- Length of segment CR = 5 cm.
Applying the Tangent Property
According to the property that tangents from an external point to a circle are equal:
- From vertex A: Tangents are AS and AP. So, AS = AP.
- From vertex B: Tangents are BP and BQ. So, BP = BQ.
- From vertex C: Tangents are CQ and CR. So, CQ = CR.
- From vertex D: Tangents are DR and DS. So, DR = DS.
Step-by-Step Calculation of Length AB
We want to find the length of side AB, which is the sum of the lengths of segments AP and BP (since P is the point of tangency on AB). AB = AP + BP.
Step 1: Find AP using AS.
We are given that AS = 6 cm. Since tangents from A are equal, AP = AS.
\(\text{AP} = 6 \text{ cm}\)
Step 2: Find CQ using CR.
We are given that CR = 5 cm. Since tangents from C are equal, CQ = CR.
\(\text{CQ} = 5 \text{ cm}\)
Step 3: Find BQ using BC and CQ.
The side BC is made up of segments BQ and CQ. So, BC = BQ + CQ. We know BC = 12 cm and CQ = 5 cm.
\(12 \text{ cm} = \text{BQ} + 5 \text{ cm}\)
Subtract 5 cm from both sides:
\(\text{BQ} = 12 \text{ cm} - 5 \text{ cm}\)
\(\text{BQ} = 7 \text{ cm}\)
Step 4: Find BP using BQ.
Since tangents from B are equal, BP = BQ. We found BQ = 7 cm.
\(\text{BP} = 7 \text{ cm}\)
Step 5: Calculate AB using AP and BP.
Finally, the length of side AB is the sum of AP and BP.
\(\text{AB} = \text{AP} + \text{BP}\)
Substitute the values we found:
\(\text{AB} = 6 \text{ cm} + 7 \text{ cm}\)
\(\text{AB} = 13 \text{ cm}\)
Thus, the length of side AB is 13 cm.
| Segment | Length (cm) | Reason (Tangent Property) |
|---|---|---|
| AS | 6 | Given |
| AP | 6 | AP = AS (from A) |
| CR | 5 | Given |
| CQ | 5 | CQ = CR (from C) |
| BC | 12 | Given |
| BQ | 7 | BQ = BC - CQ = 12 - 5 |
| BP | 7 | BP = BQ (from B) |
| AB | 13 | AB = AP + BP = 6 + 7 |
Revision Table: Circle and Quadrilateral Tangents
| Concept | Description | Application in this problem |
|---|---|---|
| Circle inscribed in a quadrilateral | A circle that is tangent to all four sides of the quadrilateral. | The problem setup involves this specific geometric configuration. |
| Tangent segments from external point | Tangents drawn from an external point to a circle have equal length from the point to the tangent point. | Used to establish AP=AS, BP=BQ, CQ=CR, DR=DS. |
| Side length of quadrilateral | The side is formed by the sum of two tangent segments from the vertices at the ends of the side. | AB = AP + BP, BC = BQ + CQ, CD = CR + DR, DA = DS + AS. |
Additional Information: Properties of Tangents
Understanding tangents is crucial in circle geometry. Here are a few more points about tangents:
- A tangent line touches the circle at exactly one point, called the point of tangency.
- The radius drawn to the point of tangency is always perpendicular to the tangent line at that point. This forms a 90-degree angle.
- In the case of a quadrilateral circumscribing a circle (which is the same as a circle inscribed in a quadrilateral), the sum of opposite sides are equal. That is,
AB + CD = BC + DA. This property can sometimes be used to solve related problems.
In this specific problem, we used the property of equal tangent segments from external points to find the length of AB.
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Important Questions from Quadrilaterals
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