Q32
3 marksShort AnswerSection C

In the given figure, Δ ABC is a right triangle in which ∠B=90∘\angle B = 90^\circ, AB = 4 cm and BC = 3 cm. Find the radius of the circle inscribed in the triangle ABC.

Right triangle ABC with the right angle at B (marked with a square), A at the top apex, C at bottom-left, B at bottom-ri
Fig. for Q32
Circles
Incircle of a Right Triangle
Official Answer

Radius of inscribed circle r=1 cmr = 1\text{ cm}. Using Pythagoras, AC=5 cmAC = 5\text{ cm} (3-4-5 triplet); since a right triangle's incircle radius is r=AB+BC−AC2=4+3−52=1 cmr = \frac{AB+BC-AC}{2} = \frac{4+3-5}{2} = 1\text{ cm}, confirmed by r=Areas=66=1 cmr = \frac{\text{Area}}{s} = \frac{6}{6} = 1\text{ cm}.

incircleinscribed circleright trianglePythagoras theoremtangent lengthradius3-4-5 triplet

Marking Scheme

  • 11 mark: correctly finding AC=5 cmAC = 5\text{ cm} using Pythagoras theorem (AC2=AB2+BC2AC^2 = AB^2 + BC^2).
  • 21 mark: setting up correct relation for the incircle — either r=AB+BC−AC2r = \frac{AB+BC-AC}{2} or Area=r×s\text{Area} = r \times s with justification (tangent lengths from a vertex are equal).
  • 31 mark: correct final computation giving r=1 cmr = 1\text{ cm} with proper units.

Hint

First find the hypotenuse using Pythagoras theorem, then use r=AB+BC−AC2r = \frac{AB + BC - AC}{2} for the incircle of a right triangle, or r=Areasemi-perimeterr = \frac{\text{Area}}{\text{semi-perimeter}}.

Quick Oral Answer

First find AC=5 cmAC = 5\text{ cm} by Pythagoras theorem, then use the shortcut for a right triangle's incircle: r=AB+BC−AC2=4+3−52=1 cmr = \frac{AB+BC-AC}{2} = \frac{4+3-5}{2} = 1\text{ cm}.

Analysis & Explanation

A right-triangle mensuration problem linking Pythagoras' theorem with the incircle tangent-length property.


Concept

  • AC (hypotenuse) is found via Pythagoras: AC2=AB2+BC2AC^2 = AB^2 + BC^2; recognise the 3-4-5 triplet instantly.
  • For a right triangle with legs a, b and hypotenuse c, the incircle radius r=a+b−c2r = \frac{a+b-c}{2}.

Key points

  • Alternative method: r=Areasemi-perimeterr = \frac{\text{Area}}{\text{semi-perimeter}}, using Area=12×AB×BC\text{Area} = \frac{1}{2}\times AB\times BC.
  • Both methods must give the same answer — a strong self-check.

Common mistakes

  • Forgetting that tangents from the same external point are equal, so BP = BQ = r only because ∠B = 90° makes OPBQ a square.
  • Skipping the cross-check with Area=r×s\text{Area} = r\times s.

Real-world

  • Incircle radius calculations are used in design and fabrication where a circular fitting must be inscribed within a triangular frame.

Common Mistakes

  1. 1Forgetting to first calculate the hypotenuse AC and mistakenly using AB or BC as the hypotenuse in the radius formula.
  2. 2Using the circumradius formula (hypotenuse2=2.5 cm\frac{\text{hypotenuse}}{2} = 2.5\text{ cm}) instead of the incircle radius formula by confusing inscribed and circumscribed circles.
  3. 3Errors in the semi-perimeter method: forgetting to divide the sum of all three sides by 2 before dividing area by it.

Interesting Facts

The 3-4-5 triangle is the smallest and most famous Pythagorean triple, known to ancient Egyptian surveyors ('rope stretchers') who used knotted ropes of these proportions to construct right angles for pyramids and land boundaries.

For any right triangle, the incircle radius always equals r=sum of legs−hypotenuse2r = \frac{\text{sum of legs} - \text{hypotenuse}}{2}, a fast shortcut that avoids computing area and semi-perimeter separately.

The point where the incircle touches the two legs of a right triangle, together with the right-angle vertex, always forms a perfect square of side r — this is why the tangent length from the right-angle vertex equals r exactly.

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Frequently Asked Questions

What is the direct formula for the incircle radius of a right triangle?

For a right triangle with legs a, b and hypotenuse c, the incircle radius is r=a+b−c2r = \frac{a + b - c}{2}. This comes from equal tangent lengths from each vertex to the incircle.

Can this problem be solved using area instead of the direct formula?

Yes. Compute Area=12×AB×BC=6 cm2\text{Area} = \frac{1}{2}\times AB\times BC = 6\text{ cm}^2 and semi-perimeter s=AB+BC+AC2=6 cms = \frac{AB+BC+AC}{2} = 6\text{ cm}, then r=Areas=1 cmr = \frac{\text{Area}}{s} = 1\text{ cm} — the same answer as the direct formula.