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

Q: 29. (A) In the given figure, $\triangle ABC$ is a right triangle in which $\angle B = 90^\circ$, $AB = 4\text{ cm}$ and $BC = 3\text{ cm}$. Find the radius of the circle inscribed in the triangle ABC.

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

Circles

Radius of incircle of a right triangle

Official Answer

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

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

Marking Scheme

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

Hint

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

Quick Oral Answer

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

Analysis & 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: $r = (\text{leg}_1 + \text{leg}_2 - \text{hypotenuse})/2$ .
Alternative general formula: $r = \text{Area}/\text{semiperimeter}$ , 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 $\text{hypotenuse}/2 = 2.5\text{ cm}$ .
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.

Common Mistakes

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

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 $r = (a + b - c)/2$ 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.

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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 $r = (a + b - c)/2$ . Here $r = (4 + 3 - 5)/2 = 1\text{ cm}$ . 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 $\angle OPB = \angle OQB = 90^\circ$ , and $\angle B = 90^\circ$ . With $OP = OQ = r$ , quadrilateral OPBQ has four right angles and two equal adjacent sides, making it a square, hence $BP = BQ = r$ .

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

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