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Question
Use ruler and compasses only for this question. Draw a circle of radius 4 cm and mark two chords, AB and AC, of the circle of length 6 cm and 5 cm respectively.
- Construct the locus of points, inside the circle, that are equidistant from A and C. Prove your construction.
- Construct the locus of points, inside the circle, that are equidistant from AB and AC.
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Solution 1
- Draw PQ, the perpendicular bisector of chord AC. PQ is the required locus, which is the diameter of the circle.
Reason: We know each point on the perpendicular bisector of AB is equidistant from A and B. Also, the perpendicular bisector of a chord passes through the centre of the circle, and any chord passing through the centre of the circle is its diameter.
∴ PQ is the diameter of the circle. - Chords AB and AC intersect at M and N is a moving point such that LM = LN, where LM ⊥ AB and LN ⊥ AC
In right ΔALN and ΔALB
∠ANL = ∠ABL ...(90° each)
AL = AL ...(Common)
NL = BL ...[Given]
∴ ΔALN = ΔALB ...[R.H.S.]
Hence ∠MAL = ∠BAL ...c.p.c.t.
Thus, L lies on the bisector of ∠BAC.
Hence proved.
Solution 2
Draw a circle of radius 4 cm whose center is O. Take a point A on the circumference of this circle.
With A as the center and a radius of 6 cm, draw an arc to cut the circumference at B. Join AB.
Then AB is the chord of the circle of length 6 cm.
With A as the center and a radius of 5 cm, draw another arc to cut the circumference at C. Join AC; then AC is the chord of the circle of length 5 cm.
With A as the center and a suitable radius, draw two arcs on opposite sides of AC.
With C as the center and the same radius, draw two arcs on opposite sides of AC to intersect the former arcs at P and Q.
Join PQ and produce to cut the circle at D and E.
Join DE. Then chord DE is the locus of points inside the circle that Ls equidistant from A and C.
As chord DE passes through the center O of the circle, it is a diameter. To prove the construction, take any point S inside the circle on DE.
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