Construction 6. Construct a triangle ABC in which BC = 6cm, and the attitude through A is 4.5cm. Measure the length of median through A. Write the steps of construction.
Procedure:-
- BC = 6cm is drawn and is made downwards with BC of any length.is drawn Perpendicular bisector RQ of BC is drawn which cut BC at M. and intersect BE at O.Taking O as centre and OB as radius, a circle is drawn. ML = 4.5cm is cut from RQ.A line XY, parallel to BC is drawn through L to intersect the circle at A and A'.
AB, AC, A’B and A’C are joined.
ABC and A’BC are the required triangle
Medium AM = A'M = 5.5cm (app.)
Construction 7. Construct a triangle ABC in which BC = 5cm, and median AD through A is of length 3.5cm. Also, determine the length of the altitude drawn from A on the side BC (Write the steps of construction also).
Procedure:-
- BC = 5cm is drawn and is constructed downwards.
- BX is drawn perpendicular to BY.
- PQ is drawn perpendicular bisector if BC intersecting BX at O and cutting BC at E.
- Taking O as a centre and OB as radius, a circle is drawn.
- Taking E as centre and radius equal to 3.5cm, arc is drawn to cut the circle at A.
- AC and AB are joined
- AD is drawn perpendicular to BC from A to cut BC at D.
- By measuring we find that AD = 3cm.
Construction 8. Construction a to a equilateral with side 5cm such that each its sides is 6/7th of the corresponding side of Also draw the circumcircle of .
Procedure:-
- A ray QX is drawn making any angle with QR and opposite to P.
- Starting from Q, seven equal line segments QQ1, Q1R2, Q2Q3, Q3Q4, Q4Q5, Q5Q6, Q6Q7 are cut of from QX.
- RQ7 is joined and a line CQ6 is drawn parallel to RQ4 to intersect QR at C.
- Line CA is drawn parallel to PR.
ABC is the required triangle.
Construction 9. Construct a triangle ABC in which BC = 6cm, and median AD = 5cm. Also construct another triangle BPQ similar to triangle BCA such that the side BP = 3/2BC.
Procedure:-
- A line segment BC of length 6cm is drawn.
- At B, is drawn on downwards.At B, is drawnPerpendicular bisector of BC is drawn which intersect BY at O and BC at D.Taking O as a center and OB as a radius a circle passing through B and C is drawn.Taking D as a centre and radius 5cm an arc is drawn to intersect the circle at A. AB and AC are joined. The required triangle is ABC.Taking C as centre and CD as radius an arc is drawn to intersect BC produced at P such that BP = 3/2BC.Through P, PQ is drawn parallel to CA meeting BA produced at Q.BPQ is the required triangle similar to triangle BCA.
Consyruction 10. Construct a quadrilateral ABCD in which AB = 2.5cm, BC = 3.5cm, AC = 4.2cm, CD = 3.5cm and AD = 2.5cm. Construct another quadrilateral AB’C’D’ with diagonal AC’ = 6.3cm such that it is similar to quadrilateral ABCD.
Procedure:-
- A line segment Ac = 4.2cm is drawn.
- With A as a centre and radius 2.5cm, two arcs, one above AC and one below AC are drawn.
- With C as centre and radius 3.5cm, two arcs arc drawn intersecting previous arcs at B and D.
- AB, AD, BC and CD are joined ABCD is the required quadrilateral.
- Taking A as a centre and radius 6.3cm an arc is drawn to intersect AC produced at C’.
- Through C’, C’B’ and C’D’ are drawn parallel to CB and CD respectively.
AB’C’D’ is the required quadrilateral similar to ABCD.
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