In quadrilateral ABCD angled `D=90^(@)`, BC=38cm and DC=25cm. A circle is inscribed in this quadrilateral which touches AB at point Q such that QB=27cm. Find the radius of the circle.

Options

A. 28cm
B. 14cm
C. 35cm
D. 19cm

Answer

To find the radius of the inscribed circle in quadrilateral ABCD where angle D is 90 degrees, BC = 38 cm, DC = 25 cm, and QB = 27 cm, we can follow these steps:

### Step 1: Identify the lengths of the tangents from points B and C.
- Since QB = 27 cm, and the tangents from a point to a circle are equal, we have:
- \( BR = BQ = 27 \, \text{cm} \)

### Step 2: Calculate the length of CR.
- Given that BC = 38 cm, we can find CR:
- \( CR = BC - BQ = 38 \, \text{cm} - 27 \, \text{cm} = 11 \, \text{cm} \)

### Step 3: Identify the lengths of the tangents from point C.
- Since CR = 11 cm, and the tangents from point C to the circle are equal:
- \( CS = CR = 11 \, \text{cm} \)

### Step 4: Calculate the length of DS.
- We know that DC = 25 cm, so we can find DS:
- \( DS = DC - CS = 25 \, \text{cm} - 11 \, \text{cm} = 14 \, \text{cm} \)

### Step 5: Identify the lengths of the tangents from point D.
- Since DS = 14 cm, and the tangents from point D to the circle are equal:
- \( DP = DS = 14 \, \text{cm} \)

### Step 6: Conclusion about the radius.
- The radius of the inscribed circle (r) is equal to the lengths of the tangents from any of the points where the circle touches the sides of the quadrilateral. Therefore, the radius of the circle is:
- \( r = DS = 14 \, \text{cm} \)

### Final Answer:
The radius of the inscribed circle is **14 cm**.

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