Q. Construct an isosceles triangle with base 5 cm and equal sides of 6 cm. Then, construct another triangle whose sides are of the corresponding sides of the (4/3)th of first triangle .
Steps of construction:
(i) Draw BC = 5 cm
(ii) With B and C as the centre and radius 6 cm, draw arcs on the same side of BC, intersecting at A.
(iii) Join AB and AC to get the required ΔABC.
(iv) Draw a ray , making an acute angle with BC on the side opposite to the vertex A.
(v) Mark 4 points B_{1}, B_{2}, B_{3}, B_{4}, along BX such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}
(vi) Join B_{3}C. Draw a line through B_{4} parallel to B_{3}C, making an angle equal to ∠BB_{3}C intersecting the extended line segment BC at C´.
(vii) Through point C´, draw a line parallel to CA, intersecting extended BA at A´.
(viii) The resulting ΔA´BC´ is the required triangle.
Q. Draw Line segment PQ=9cm and divide it in the ratio 2:5. Justify your construction.
Steps of construction:
(i) Draw Line Segment PQ=9cm
(ii) Draw a Ray PX, making an acute angle with PQ.
(iii) Mark 7 points A_{1} , A_{2} , A_{3} …A_{7} along PX such that PA_{1} = A_{3}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5} = A_{5}A_{6} = A_{6}A_{7}
(iv) Join QA_{7}
(v) Through the point A_{2} , draw a line parallel to A7Q by making an angle equal to ∠PA_{7}Q at A_{2}, intersecting PQ at point R. PR:RQ = 2:5Justification:
We have A_{2}R  A_{7}Q
Q. Draw a circle of radius 3 cm. From a point 5 cm away from its centre, construct a pair
Steps of construction:
(i) Draw a circle with centre O and radius 3 cm. Take a point P such that OP = 5 cm, and then join OP.
(ii) Draw the perpendicular bisector of OP. Let M be the mid point of OP.
(iii) With M as the centre and OM as the radius, draw a circle. Let it intersect the previously drawn circle at A and B.
(iv) Joint PA and PB. Therefore, PA and PB are the required tangents. It can be observed that PA=PB=4cm.
Q. Draw a ΔABC with sides BC = 8 cm, AC = 7 cm, and ÐB = 70°. Then, construct a similar triangle whose sides are (3/5)th of the corresponding sides of the ΔABC.
Steps of construction:
(i) Draw BC = 8 cm
(ii) At B, draw ∠XBC = 70°
(iii) With C as centre and radius 7 cm, draw an arc intersecting BX at A.
(iv) Join AB, and DABC is thus obtained.
(v) Draw a ray , making an acute angle with BC.
(vi) Mark 5 points, B_{1}, B_{2}, B_{3}, B_{4}, B_{5}, along BY such that
(vii) BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4} = B_{4}B_{5}
(viii) Join CB_{5}
(ix) Through the point B_{3}, draw a line parallel to B_{5}C by making an angle equal to ∠BB_{5}C, intersecting BC at C´.
(x) Through the point C´, draw a line parallel to AC, intersecting BA at A´. Thus, ΔA´BC´ is Required Triangle.Justification
Using BPT
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