In Class IX, we performed geometric constructions using a ruler and compass, like angle bisecting and drawing perpendicular bisectors. Now, we'll delve into more constructions, building on the previous knowledge, with a focus on understanding the mathematical reasoning behind each construction.
Q: Draw Line segment PQ=9cm and divide it in the ratio 2:5. Justify your construction.
Sol:
Steps of construction:
(i) Draw Line Segment PQ=9cm
(ii) Draw a Ray PX, making an acute angle with PQ.
(iii) Mark 7 points A1 , A2 , A3 …A7 along PX such that PA1 = A3A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7
(iv) Join QA7
(v) Through the point A2 , draw a line parallel to A7Q by making an angle equal to ∠PA7Q at A2, intersecting PQ at point R. PR:RQ = 2:5Justification:
We have A2R || A7Q
Scale Factor means the ratio of the sides of the triangle to be constructed with the corresponding sides of the given triangle.
This construction involves two cases:(i) Where the new triangle is smaller than the given triangle.
(ii) Where the new triangle is larger than the given triangle.
Q: Construct an isosceles triangle with a base of 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 the first triangle.
Sol:
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 B1, B2, B3, B4, along BX such that BB1 = B1B2 = B2B3 = B3B4
(vi) Join B3C. Draw a line through B4 parallel to B3C, making an angle equal to ∠BB3C 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 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.
Sol:
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, B1, B2, B3, B4, B5, along BY such that
(vii) BB1 = B1B2 = B2B3 = B3B4 = B4B5
(viii) Join CB5
(ix) Through the point B3, draw a line parallel to B5C by making an angle equal to ∠BB5C, 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
We know that a point inside a circle has no tangent. On the circle, there's a unique tangent perpendicular to the radius at that point. Extend the radius and draw a line perpendicular to it to create the tangent. For a point outside the circle, two tangents can be drawn. We'll now see how to construct these tangents.
Q: Draw a circle of radius 3 cm. From a point 5 cm away from its center, construct a pair of tangents to the circle and measure their lengths.
Sol:
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.
1. What is the concept of division of a line segment? |
2. How can tangents to a circle be constructed? |
3. How is the division of a line segment related to similar triangles? |
4. What is the significance of constructing tangents to a circle? |
5. Can a line segment be divided into two unequal parts? |
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