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ICSE Class 10 Mathematics Question 26 of 28

Short Answer Questions 2 — Question 26

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Question 103

(a) Construct a triangle ABC such that BC = 8 cm, AC = 10 cm and ∠ABC = 90°.

(b) Construct an incircle to this triangle. Mark the centre as I.

(c) Measure and write the length of the in-radius.

(d) Measure and write the length of the tangents from vertex C to the incircle.

(e) Mark points P, Q and R where the incircle touches the sides AB, BC, and AC of the triangle respectively. Write the relationship between ∠RIQ and ∠QCR.

(Use a ruler and a compass for this question.)

Answer

Steps of construction :

  1. Draw a line segment BC = 8 cm.

  2. Draw BX perpendicular to BC.

  3. With C as center and radius = 10 cm, draw an arc cutting BX at A.

  4. Join AB and AC.

  5. Draw AW, BY and CZ the angle bisectors of A, B and C respectively.

  6. Mark the point of intersection as I.

  7. Draw IR perpendicular to side AC.

  8. With I as center and radius IR draw a circle, which is the required incircle.

  9. Mark points P, Q and R where the incircle touches the sides AB, BC, and AC of the triangle respectively.

  10. Measure CQ and CR.

Construct a triangle ABC such that BC = 8 cm, AC = 10 cm and ∠ABC = 90°. Maths Competency Focused Practice Questions Class 10 Solutions.

From figure,

⇒ ∠IRC = ∠IQC = 90° (The radius from the center of the circle to the point of tangency is perpendicular to the tangent line.)

⇒ ∠RCI = ∠QCI = ∠C2\dfrac{∠C}{2} (As CZ is angle bisector)

In △ IRC,

⇒ ∠RIC = 180° - ∠RCI - ∠IRC [∵ Sum of ∠'s in a Δ = 180°]

⇒ ∠RIC = 180° - ∠C2\dfrac{∠C}{2} - 90°

⇒ ∠RIC = 90° - ∠C2\dfrac{∠C}{2} ............(1)

In △ IQC,

⇒ ∠QIC = 180° - ∠IQC - ∠ICQ [∵ Sum of ∠'s in a Δ = 180°]

⇒ ∠QIC = 180° - 90° - ∠C2\dfrac{∠C}{2}

⇒ ∠QIC = 90° - ∠C2\dfrac{∠C}{2} ............(2)

Adding equations (1) and (2), we get :

⇒ ∠RIC + ∠QIC = 90° - ∠C2\dfrac{∠C}{2} + 90° - ∠C2\dfrac{∠C}{2}

⇒ ∠RIQ = 180° - ∠C

⇒ ∠RIQ = 180° - ∠RCQ

⇒ ∠RIQ + ∠RCQ = 180°.

Hence, ∠RIQ + ∠QCR = 180°.