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In this article, all the class 10 Mathematics board exam notes of the topic named “Constructions'' are provided. This will include the steps, basic procedure, and some sample questions for boards that are all you need to revise for construction before your board exams. We will discuss the steps on how to construct the division of the line segment, constructions of triangles using scale factor, and construction of tangents to a circle with 4 different cases. All these concepts are a part of your board’s syllabus, so you must revise them before your exams. So let’s get started.
Dividing a Line Segment
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- Bisecting (dividing into 2 halves):
Step 1: Taking more than half of a line segment, draw arches from both the ends of the line. The arches should be centered at the end of the line segment.
Step 2: Draw a dotted line to join the 2 points of intersections. This dotted line bisects the line segment.
- Divide a line segment in the ratio m:n where m and n are positive integers:
Let us assume m=3, n=1
Step 1: Draw a line segment AB and draw a ray named AC such that AC forms an acute angle with AB.
Step 2: You need to mark 4 (3 = m+n ) points that are at an equal distance from each other on AX. For that, take a length of less than ? rd of the length of AX, from A make an arch on Ax and from that arched point, make another arch on AX of the same distance. Keep doing this for (m+n) times. Name them A1, A2, A3, and A4 with the condition AA1 = A1A2 = A2A3= A3A4. Join A4 with B.
Step 3: Through the 3rd point i.e A3 draw a parallel line to A4B. You can do this by making an angle equal to
By following the similarity triangle rule, AC:CB = 3:1
Method 2 (alternative method)
Step 1: Draw a line PQ such that it makes an acute angle with PR.
Step 2: Draw another line SR parallel to PR so that
Step 3: Locate the points X1 as m=1 on PQ and Y1, Y2, Y3 as n = 3 on SR such that PX1 = RY1 = Y1Y2 = Y2Y3.
Step 4: Join X1Y3 such that it intersects PR at point A.
Thus PA: AR = 1:3
Construction of Similarity Triangles
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When two triangles are similar, the reduced ratio of any two corresponding sides is called the scale factor of similar triangles. For eg: Similar triangles whose scale factor is 2 : 1, then their ratios of corresponding sides are 6/3, 8/4, 10/5.
Construction of Similarity Triangles
Using Scale Factor m/n where m>n
Let’s assume that a triangle ABC is given and A’BC’ has to be made whose sides are m/n (m>n) times the corresponding sides of a given triangle.
Step 1: Draw a line segment AB and a ray AX such that AX makes an acute angle with AB. Draw AC in between AX and AB.
Step 2: Locate n consecutive points of equal length on AX. Let them be A1, A2, A3, A4...An
Step 3: Draw a line through Am parallel to AnC such that it intersects AC at C’.
Step 4: Through Am, draw a parallel line to BC so as to intersect AB at A’. Thus the required triangle is A′BC′, each of whose sides is m > n of the corresponding sides of ABC.
Construction of Triangles to a circle
Tangent is a line segment touching a point on the circle and forms 90 degrees with the radius that intersects the point.
To determine the number of tangents of a circle:
- When the Point is within the circle -- No tangent
- When the Point lies on the circumference of circle -- 1 tangent
- When the Point outside the circle -- 2 tangents intersecting
Drawing tangents when the point lies outside the circle
- When the Center is known
Step 1: Draw a circle with center O and let S be the exterior point from where the tangent has to be drawn.
Step 2: Join OS and bisect it. Let P be the midpoint of the OS.
Step 3: Consider this P as the center of the circle. So OP = PS = radius. From P, draw the circle with OP as the radius. Let it intersect the given circle at two different points A and B.
Step 4: Join AS and BS. These 2 are the tangents to the circle.
- When the Centre is unknown
Step 1: Find the center by drawing 2 non parallel chords of the circle. The point where the perpendicular bisectors of the chords intersect forms the center of the circle. Then follow the steps as mentioned above.
Tangents when the point lies outside the circle
Drawing a tangent when the point lies on the circumference of the circle
- When the Center is known
Step 1: Draw a circle with O as the center of any radius. Let P be any point on the circle. Extend the radius in the direction such that it cuts tP.
Step 2: Draw a perpendicular through the radius of the circle. This will be the only tangent to the circle.
- When the Centre in unknown
Step 1: Find the center by drawing 2 non-parallel chords of the circle. The point where the perpendicular bisectors of the chords intersect forms the center of the circle. Then follow the steps mentioned above.
Sample Questions
Ques: Mention the steps required to draw a circle of radius 3 cm, taking a point 7 cm from the center, and draw 2 tangents to the circle. Measure the length of both the tangents. (2 marks)
Ans. Let O be the center of the circle, P be the point 7 cm away. Let S be a point on the circle such that SOP forms a right angled triangle at
So we get, SP = PT= 6.3 cm
Steps of construction:
- Take O as the centre and draw a circle of radius 3 cm.
- From O, locate point P at a distance of 7 cm. Join OP.
- Bisect OP.
- Let M be the midpoint of OP.
- Considering M at the midpoint and MP, OM as the radius, draw a circle of radius OM.
- Draw this circle such that it should intersect with the precious circle at S and T
- Join SP, TP which form the required tangents.
- So SP = 6.3 cm
Previous Year Questions
Ques: Find the coordinates of a point A, where AB is diameter of a circle whose centre is (2, -3) and B is the point (1, 4). (2019)
Ans. Let’s take the points of A being (x,y) and O as (2,-3) the center point, therefor by midpoint formula,
(3, -10) are the coordinates of A.
Ques: Find the ratio in which the segment joining the points (1, -3) and (4, 5) is divided by x-axis? Also, find the coordinates of this point on the x-axis. (2019)
Ans. Let’s take the points given as A (1, -3) and B (4, -5) and the line segment that joins that is connected by these points is divided by the x-axis, so the point coordinates that belongs to the point of intersection will be as P (a, b).
So, by section formula, we get the y coordinate of the point P as,
Ques: Write the coordinates of a point P on the x-axis which is equidistant from point A(-2, 0) and B(6, 0). ( Outside Delhi 2019)
Ans. Let’s take the coordinates of P on the x-axis be (x, 0)
We know A (-2, 0) and B (6, 0)
Here we have PA = PB
When we square both the side we acquire,
(x+2)2 + (0)2= (x-6)² + (0)2
(x+2)² = (x-6)²
x² + 4 + 4x = x² +36 + 12x
4 + 4x = 36 - 12x
16x = 32
x = 2
Therefore the coordinates of P = (2, 0)
Ques: Find the ratio in which P(4, m) divides the line segment joining the points A(2, 3) and B(6, -3). Hence find m. (2018)
Ans. Let’s take P as the point that divides the line segment AB in the ratio form of k:1
Therefore,
Ques: If the distance between the points (4, k) and (1, 0) is 5, then what can be the possible values of k? (2017)
Ans. The distance given between two points (4, k) and (1, 0) = 5
Ques: Prove that tangents drawn at the ends of a diameter of a circle are parallel to each other. (2017)
Ans. PQ is given as the diameter of the circle that has its center as O
The two lines AB and CD are both tangents at P and Q respectively
To prove AB || CD
Proof: At point P, AB can be referred to as a tangent to the circle and OP forms the radius through the point contact
∠OPD = 90º
In the same way CD forms a tangent to the circle at point Q and OQ forms the radius through the point of contact
∠OQD = 90º
∠OPD = 90º= ∠OQD = 90º
But both of them becomes a pair of the alternate angles
AB || CD
Hence Proved.
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