Determining a Point Dividing a given Line Segment, Internally in the given Ratio M : N
Let AB be the given line segment of length x cm.
Steps of Construction:
- Draw a line segment AB = x cm.
- Make an acute ∠BAX at the end A of AB.
- Use a compass of any radius and mark off arcs. Take (m + n) points A1, A2, … Am, Am+1, …, Am+n along AX such that AA1 = A1A2 = … = Am+n-1 , Am+n
- Join Am+nB.
- Passing through Am, draw a line AmP || Am+nB to intersect AB at P.
The point P so obtained is the A required point which divides AB internally in the ratio m : n.
Construction of a Tangent at a Point on a Circle to the Circle when its Centre is Known
Steps of Construction:
- Draw a circle with centre O of the given radius.
- Take a given point P on the circle.
- Join OP.
- Construct ∠OPT = 90°.
- Produce TP to T’ to get TPT’ as the required tangent.
The topics discussed in this chapter are as follows:
- Dividing a Line Segment
- Constructing Similar Triangles
- Drawing Tangents to a Circle
- Number of tangents to a circle from a given point
- Drawing tangents to a circle from a point outside the circle
- Drawing tangents to a circle from a point on the circle
Students can view and download the chapter from the link given below.
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NCERT Solutions for Chapter 11: Constructions
Also Check
Chapter 1: Real Numbers
Chapter 2: Polynomials
Chapter 3: Pair of Linear Equations in Two Variables
Chapter 4: Quadratic Equations
Chapter 5: Arithmetic Progression
Chapter 6: Triangles
Chapter 7: Coordinate Geometry
Chapter 8: Introduction to Trigonometry
Chapter 9: Some Applications of Trigonometry
Chapter 10: Circle
Chapter 12: Areas Related to Circles
Chapter 13: Surface Areas and Volume
Chapter 14: Statistics
Chapter 15: Probability
Check out Frequently Asked Questions (FAQs) for Chapter 11: Constructions
Draw a line segment of length 6 cm. Using compasses and ruler, find a point P on it which divides it in the ratio 3 : 4.
Hence, PA : PB = 3 : 4
Draw a triangle ABC in which AB = 5 cm, BC = 6 cm and ∠ABC = 60°. Then construct a triangle whose sides are 5/7 times the corresponding sides of ∆ABC
In ∆ABC
AB = 5 cm
BC = 6 cm
∠ABC = 60°
Hence, ∆A’BC’ is the required ∆.
AB = 5 cm
BC = 6 cm
∠ABC = 60°
Hence, ∆A’BC’ is the required ∆.
Construct a triangle with sides 5 cm, 5.5 cm and 6.5 cm. Now construct another triangle, whose sides are 3/5 times the corresponding sides of the given triangle.
∴ ∆AB’C’ is the required ∆.
Draw a triangle PQR such that PQ = 5 cm, ∠P = 120° and PR = 6 cm. Construct another triangle whose sides are 3/4 times the corresponding sides of ∆PQR.
In ∆PQR,
PQ = 5 cm, PR = 6 cm, ∠P = 120°
∴ ∆POʻR’ is the required ∆.
PQ = 5 cm, PR = 6 cm, ∠P = 120°
∴ ∆POʻR’ is the required ∆.
Construct a triangle with sides 5 cm, 4 cm and 6 cm. Then construct another triangle whose sides are 2/3 times the corresponding sides of first triangle.
Steps of Construction:
- Draw ∆ABC with AC = 6 cm, AB = 5 cm, BC = 4 cm.
- Draw ray AX making an acute angle with AÇ.
- Locate 3 equal points A1, A2, A3 on AX.
- Join CA3.
- Join A2C’ || CA3.
- From point C’ draw B’C’ || BC.
∴ ∆AB’C’ is the required triangle.
