Class 10 Exam  >  Class 10 Notes  >  Chapter Notes for Class 10  >  Chapter Notes: Construction

Constructions Class 10 Notes Maths

Introduction

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.

Division of a Line Segment

Construction 1: To divide a line segment in a given ratio.

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 = 2:5

Justification:

We have A2R || A7Q

Constructions Class 10 Notes Maths

Constructions Class 10 Notes Maths

Construction 2: To construct a triangle similar to a given triangle as per the given scale factor.

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. 

Case 1: When the scale factor is greater than 1.

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.

Constructions Class 10 Notes Maths

Case 2: When the scale factor is less than 1.

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.
Constructions Class 10 Notes Maths

Justification

Constructions Class 10 Notes Maths

Constructions Class 10 Notes Maths
Using BPT

Construction of tangents to a circle

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.

Construction: To construct tangents to a circle from a point outside it.

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.

Constructions Class 10 Notes Maths

The document Constructions Class 10 Notes Maths is a part of the Class 10 Course Chapter Notes for Class 10.
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FAQs on Constructions Class 10 Notes Maths

1. What is the concept of division of a line segment?
Ans. Division of a line segment refers to the process of dividing a given line segment into a specific ratio or proportion. This can be done by using various construction techniques, such as constructing perpendicular bisectors or using the concept of similar triangles.
2. How can tangents to a circle be constructed?
Ans. Tangents to a circle can be constructed using the following steps: 1. Draw the center of the circle and the radius. 2. From the center, draw a line segment perpendicular to the radius, extending it outside the circle. 3. Using a compass, measure the length of the radius. 4. Place the compass on the endpoint of the radius and draw an arc that intersects the line segment drawn in step 2. 5. From the point of intersection, draw a line that connects it to the center of the circle. This line is the tangent to the circle.
3. How is the division of a line segment related to similar triangles?
Ans. The concept of dividing a line segment into a specific ratio or proportion is related to similar triangles. When a line segment is divided into a specific ratio, the resulting segments create similar triangles with the original triangle. This is because the ratios of the corresponding sides of similar triangles are equal.
4. What is the significance of constructing tangents to a circle?
Ans. Constructing tangents to a circle is significant because it helps in determining the point of contact between a line and the circle. Tangents are used in various applications, such as calculating the angle of reflection in optics or determining the direction of motion in physics. They also play a crucial role in geometric constructions and proofs involving circles.
5. Can a line segment be divided into two unequal parts?
Ans. Yes, a line segment can be divided into two unequal parts. The division of a line segment can be done by choosing a specific ratio or proportion, which can result in one part being longer than the other. The ratio can be expressed as a fraction or a decimal, and it determines the length of each segment.
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