Draw a line segment AB = 5 cm. From the point A draw a line segment AD = 6 cm making an angle of 60°. Draw perpendicular bisector of AD.

Steps of Construction
1. Draw a line segment AB = 5 cm.
2. Taking A as centre and some radius, draw an arc of a circle, which intersects AB, say at a point P.
3. Taking P as centre and with the same radius as before, drawn an arc intersecting the previously draw arc, say at a point E.
4. Draw the ray AC passing through E.
5. From ray AC, cut off AD = 6 cm.
Then, ∠DAB is the required angle of 60° such that AD = 6 cm.
6. Now, taking A and D as centres and radius more than 1 half AD, draw arcs on both sides of the line segment AD (to intersect each other).


Steps of Construction1. Draw a line segment AB = 5 cm.2. Taking A a

7. Let these arcs intersect each other at M and N. Join MN.
8. Let MN intersect AD at the point L. Then line MLN is the required perpendicular bisect or of AD.


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Draw a line segment of 10 cm. Divide it into four equal parts. Write the measure of each part. 


Steps of Construction


Steps of Construction
1. Draw a line segment AB = 10 cm.2. Taking A

1. Draw a line segment AB = 10 cm.
2. Taking A and B as centres and radius more than 1 half AB, draw arcs on both sides of the line segment AB (to intersect each other).
3. Let these arcs intersect each other at P and Q. Join PQ.
4. Let PQ intersect AB at the point M.
5. Taking A and M as centres and radius more than 1 half AM, draw arcs on both sides of the line segment AM (to intersect each other).

6. Let these arcs intersect each other at C and D. Join CD.

7. Let CD intersect AM at the point E.
8. Taking M and B as centres and radius more than 1 half MB, draw arcs on both sides of the line segment MB (to intersect each other).
9. Let these arcs intersect each other at F and G Join FG.
10. Let FG intersect MB at H.

Then, AE = EM = MH = HB, i.e., AE, EM, MH and HB are the four equal parts. By measurement,
AE = EM = MH = HB = 2.5 cm

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Construct an equilateral triangle with one side 6 cm. 

Steps of Construction
1. Draw BC = 6 cm.
2. With B as centre and 6 cm as radius, draw an arc on one side of BC.
3. With C as ccntre and 6 cm as radius, draw another arc on the same side of BC to intersect the former arc at A.
4. Joint AB and AC.
Then, ∆ABC is the required equilateral triangle.


Steps of Construction1. Draw BC = 6 cm.2. With B as centre and 6 cm

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Construct an angle ∠BAC = 90° on a line segment AB = 6 cm. Cut AC = 5 cm. Bisect AC at D. Measure AD and DC.


Steps of Construction

1. Draw a line segment AB = 6 cm.
2. Taking A as centre and some radius, draw an arc of a circle, which intersects AB, say at a point P.
3. Taking P as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point Q.
4. Taking Q as centre and with the same radius as before, drawn an arc intersecting the arc drawn in step 1, say at R.
5. Next, taking Q and R as centres and with the radius more than 1 halfQR, draw arcs to intersect each other, say at S.
6. Draw the ray AS.
7. From ray AS, cut off AC = 5 cm.

Then. ∠BAC = 90° is such that AB = 6 cm and AC = 5 cm.
8. Taking A and C as centres and radius more than 1 half AC. draw arcs on both sides of the line segment AC (to intersect each other).


Steps of Construction
1. Draw a line segment AB = 6 cm.2. Taking A

9. Let these arcs intersect each other at M and N. Join MN.
10. Let MN intersect AC at the point D.
Then, AD = DC
By measurement,
AD = DC = 2.5 cm



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Construct an equilateral triangle LMN, one of whose sides is 5 cm. Bisect ∠M of the triangle.

Steps of Construction
1. Draw a line segment MN = 5 cm.
2. With M as centre and 5 cm as radius, draw an arc on one side of MN.
3. With N as centre and 5 cm as radius, draw another arc on the same side of MN to intersect the former arc at L.
4. Join LM and LN.
Then, ∆LMN is the required equilateral triangle.


Steps of Construction1. Draw a line segment MN = 5 cm.2. With M as

5. Taking M as centre and any radius, draw an arc to intersect the line segments MN and ML at P and Q respectively.
6. Next, taking P and Q as centres and with the radius more than 1 halfPQ, draw arcs to intersect each other, say at R. 7. Draw the ray MR. This ray MR is the required bisector of the ∠M.

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