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 AD, draw arcs on both sides of the line segment AD (to intersect each other).

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.

639 Views

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 QR, 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 AC. draw arcs on both sides of the line segment AC (to intersect each other).

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

309 Views

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.

1313 Views

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

Steps of Construction

1. Draw a line segment AB = 10 cm.

2. Taking A and B as centres and radius more than 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 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 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

639 Views

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.

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 PQ, draw arcs to intersect each other, say at R. 7. Draw the ray MR. This ray MR is the required bisector of the ∠M.

1519 Views