Question 323 marksShort AnswerSection C

  1. (A) In the given figure, △ABC\triangle ABC is a right triangle in which ∠B=90∘\angle B = 90^\circ, AB=4 cmAB = 4\text{ cm} and BC=3 cmBC = 3\text{ cm}. Find the radius of the circle inscribed in the triangle ABC.

Circles
Radius of incircle of a right triangle

Official answer

The radius of the inscribed circle is r=1 cmr = 1\text{ cm}. Since ∠B=90∘\angle B = 90^\circ, AC(hypotenuse)=42+32=5 cmAC (\text{hypotenuse}) = \sqrt{4^2 + 3^2} = 5\text{ cm} by Pythagoras theorem. Using the tangent-length relation for a right triangle, r=(AB+BC−AC)/2=(4+3−5)/2=1 cmr = (AB + BC - AC)/2 = (4 + 3 - 5)/2 = 1\text{ cm}. This is confirmed by the area method: r=Area/semiperimeter=6/6=1 cmr = \text{Area}/\text{semiperimeter} = 6/6 = 1\text{ cm}.

Answer keywords:incircleinradiusright trianglePythagoras theoremtangents from external pointAC = 5 cmr = (a + b - c)/2Area by semiperimeter

Marking scheme

  1. 1.1 mark: Correctly applying Pythagoras theorem to obtain AC=5 cmAC = 5\text{ cm}.
  2. 2.1 mark: Setting up the tangent relations (BP=BQ=rBP = BQ = r, AR=4−rAR = 4 - r, CR=3−rCR = 3 - r) and forming the equation 5=(4−r)+(3−r)5 = (4 - r) + (3 - r); or correctly stating r=Area/sr = \text{Area}/s.
  3. 3.1 mark: Solving to get r=1 cmr = 1\text{ cm}. Full marks for the equivalent formula r=(AB+BC−AC)/2=1 cmr = (AB + BC - AC)/2 = 1\text{ cm} or the area method r=6/6=1 cmr = 6/6 = 1\text{ cm}.

Hint

First find the hypotenuse (AC=5 cmAC = 5\text{ cm}). For a right triangle the inradius is r=(leg1+leg2−hypotenuse)/2r = (\text{leg1} + \text{leg2} - \text{hypotenuse})/2, or use r=Area/semiperimeterr = \text{Area}/\text{semiperimeter}.

Quick oral answer

The hypotenuse AC is 5 cm by Pythagoras. Using equal tangents, the tangent lengths give 5=(4−r)+(3−r)5 = (4 - r) + (3 - r), so 2r=22r = 2 and r=1 cmr = 1\text{ cm}; the area method r=6/6r = 6/6 confirms it.

Common mistakes

  1. 1.Confusing the incircle with the circumcircle and giving R=hypotenuse/2=2.5 cmR = \text{hypotenuse}/2 = 2.5\text{ cm} instead of the inradius 1 cm.
  2. 2.Forgetting to compute the hypotenuse first, or using AB and BC as if one of them were the hypotenuse.
  3. 3.Sign or setup error in the tangent equation, e.g. writing AC=(4+r)+(3+r)AC = (4 + r) + (3 + r) instead of (4−r)+(3−r)(4 - r) + (3 - r).

Analysis and explanation

Combines the Pythagoras theorem with the equal-tangents property of circles to find the incircle radius of a right triangle.


Concept & key formula

  • Shortcut for any right triangle: r=(leg1+leg2−hypotenuse)/2r = (\text{leg}_1 + \text{leg}_2 - \text{hypotenuse})/2.
  • Alternative general formula: r=Area/semiperimeterr = \text{Area}/\text{semiperimeter}, useful as a cross-check.
  • The 3-4-5 triangle is the smallest Pythagorean triple, chosen so the incircle radius comes out as the whole number 1 cm.

Common mistakes

  • Confusing the incircle with the circumcircle — the circumradius here would wrongly be taken as hypotenuse/2=2.5 cm\text{hypotenuse}/2 = 2.5\text{ cm}.
  • Assuming the incircle radius equals half the shortest side, which is not a valid general rule.

Real-world relevance

  • The incircle radius equals the largest circular shaft, pipe, or cutter that fits inside a triangular frame — used in mechanical design and tiling layouts.

Interesting facts

The 3-4-5 right triangle is the smallest Pythagorean triple and was used by ancient Egyptian 'rope-stretchers' (harpedonaptae) to lay out perfect right angles when building the pyramids.

For every right triangle the inradius satisfies the neat relation r=(a+b−c)/2r = (a + b - c)/2 where c is the hypotenuse — a formula that follows purely from equal tangent lengths.

The incircle is the unique largest circle that fits inside a triangle; in manufacturing, it determines the biggest round rod that can pass through a triangular opening.

Question details

Q32 of 53 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Basic)
Year
2026
Section
Section C
Marks
3
Type
Short Answer
Chapter
Circles
Topic
Radius of incircle of a right triangle
Bloom's level
apply
Syllabus topic
Unit 4: Geometry — Circles: Tangent to a Circle and the Incircle of a Right Triangle (CBSE Class 10 Mathematics)
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

What is the quick formula for the inradius of a right triangle?

For a right triangle with legs a and b and hypotenuse c, the inradius is r=(a+b−c)/2r = (a + b - c)/2. Here r=(4+3−5)/2=1 cmr = (4 + 3 - 5)/2 = 1\text{ cm}. It follows directly from the equal-tangent property.

Why is OPBQ a square in this problem?

O is the incentre and OP, OQ are radii drawn to the points where the circle touches AB and BC. A radius is perpendicular to the tangent at the contact point, so ∠OPB=∠OQB=90∘\angle OPB = \angle OQB = 90^\circ, and ∠B=90∘\angle B = 90^\circ. With OP=OQ=rOP = OQ = r, quadrilateral OPBQ has four right angles and two equal adjacent sides, making it a square, hence BP=BQ=rBP = BQ = r.

Can I use r = Area/semiperimeter for any triangle?

Yes. r=Area/sr = \text{Area}/s holds for every triangle, not just right triangles. Here Area=6 cm2\text{Area} = 6\text{ cm}^2 and s=6 cms = 6\text{ cm}, giving r=1 cmr = 1\text{ cm} — the same answer as the tangent method, which is a good way to verify your result.