Constructions — Analytical & Application Based Questions

Use ruler and compass to answer this question. Construct a triangle ABC where AB = 5.5 cm, BC = 4.5 cm and angle ABC = 135°. Construct the circumcircle to the triangle ABC. Measure and write down the length of AC.

Answer

Steps of construction :

1. Draw a line segment BC = 4.5 cm

2. Construct XB such that ∠XBC = 135°.

3. Cut AB = 5.5 cm from XB.

4. Join and measure AC.

5. Draw PQ and RS the perpendicular bisectors of BC and AB.

6. Mark point O the intersection of PQ and RS.

7. With O as center and radius as OA, draw a circle touching the vertices A, B and C.

On measuring AC = 9.1 cm and radius = 6.5 cm.

Use a ruler and a compass for this question.

Construct a regular hexagon ABCDEF of side 4.3 cm and construct its circumscribed circle. Also, construct tangents to the circumscribed circle at points B and C which meets each other at point P. Measure and record ∠BPC.

Answer

We know that each angle in a regular hexagon = 120°.

1. Draw a line segment AB = 4.3 cm.

2. At A and B draw rays making an angle of 120° each and cut off AF = BC = 4.3 cm.

3. At F and C, draw rays making angle of 120° each and cut off EF = CD = 4.3 cm.

4. Join ED. Hence, ABCDEF is the required hexagon.

5. Draw the perpendicular bisector of AB and AF. Let these bisectors meet at the point O.

6. With O as center and radius equal to OA or OB draw a circle which passes through the vertices of the hexagon. This is the required circumcircle of hexagon ABCDEF.

7. Draw the radius OB and OC.

8. At point B, construct a line perpendicular to OB. This line is the tangent at B.

9. At point C, construct a line perpendicular to OC. This line is the tangent at C.

10. The two tangents will intersect at point P.

11. Measure ∠BPC.

Hence, ∠BPC = 120°.

Use a ruler and a compass for this question.

(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.

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.

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 = $\dfrac{∠C}{2}$ (As CZ is angle bisector)

In △ IRC,

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

⇒ ∠RIC = 180° - $\dfrac{∠C}{2}$ - 90°

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

In △ IQC,

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

⇒ ∠QIC = 180° - 90° - $\dfrac{∠C}{2}$

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

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

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

⇒ ∠RIQ = 180° - ∠C

⇒ ∠RIQ = 180° - ∠RCQ

⇒ ∠RIQ + ∠RCQ = 180°.

Hence, ∠RIQ + ∠QCR = 180°.