CBSE Class 10 Constructions Important Formulas and concepts for exams

CBSE Class 10 Constructions Important Formulas and concepts for exams. There are many more useful educational material which the students can download in pdf format and use them for studies. Study material like concept maps, important and sure shot question banks, quick to learn flash cards, flow charts, mind maps, teacher notes, important formulas, past examinations question bank, important concepts taught by teachers. Students can download these useful educational material free and use them to get better marks in examinations.  Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

This Class 10 Maths notes page on Chapter 11, Constructions, covers the important construction steps for dividing a line in a given ratio and constructing a triangle similar to a given triangle with a scale factor of 3/4 or 5/3, constructing tangents to a circle from an external point, and constructing a tangent at a given point on the circumference.

CLASS X : CHAPTER - 11 CONSTRUCTIONS

IMPORTANT CONCEPTS

To construct a triangle similar to a given triangle as per given scale factor.

Example 1: Construct a triangle similar to a given triangle ABC with its sides equal to \( \frac{3}{4} \) of the corresponding sides of the triangle ABC (i.e., of scale factor \( \frac{3}{4} \)).

Steps of Construction:
1. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
2. Locate 4 (the greater of 3 and 4 in \( \frac{3}{4} \)) points B₁, B₂, B₃ and B₄ on BX so that BB₁ = B₁B₂ = B₂B₃ = B₃B₄.
3. Join B₄C and draw a line through B₃ (the 3rd point, 3 being the smaller of 3 and 4 in \( \frac{3}{4} \)) parallel to B₄C to intersect BC at C′.
4. Draw a line through C′ parallel to the line CA to intersect BA at A′.
5. Then △A′BC′ is the required triangle.

[Figure: Triangle ABC with B at the bottom left and C on the right. Ray BX runs along the base towards the right, with points B₁, B₂, B₃ and B₄ marked on it at equal spacing. The line B₄C is joined, a parallel line is drawn through B₃ meeting BC at C′, and a line through C′ parallel to CA meets BA at A′. The required triangle is A′BC′.]

Example 2: Construct a triangle similar to a given triangle ABC with its sides equal to \( \frac{5}{3} \) of the corresponding sides of the triangle ABC (i.e., of scale factor \( \frac{5}{3} \)).

Steps of Construction:
1. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
2. Locate 5 points (the greater of 5 and 3 in \( \frac{5}{3} \)) B₁, B₂, B₃, B₄ and B₅ on BX so that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅.
3. Join B₃ (the 3rd point, 3 being the smaller of 3 and 5 in \( \frac{5}{3} \)) to C, and draw a line through B₅ parallel to B₃C, intersecting the extended line segment BC at C′.
4. Draw a line through C′ parallel to CA, intersecting the extended line segment BA at A′.
5. Then △A′BC′ is the required triangle.

[Figure: Triangle ABC with the ray BX drawn from B, points B₁ to B₅ marked on BX, and the line B₃C joined. A line through B₅ parallel to B₃C meets the extended BC at C′, and a line through C′ parallel to CA meets the extended BA at A′. The required triangle is A′BC′, which is larger than triangle ABC.]

To construct the tangents to a circle from a point outside it.

Given: We are given a circle with centre O and a point P outside it. We have to construct two tangents from P to the circle.

Steps of Construction:
1. Join PO and draw the perpendicular bisector of it. Let M be the midpoint of PO.
2. Taking M as centre and PM or MO as radius, draw a circle. Let it intersect the given circle at the points A and B.
3. Join PA and PB.
4. Then PA and PB are the required two tangents.

[Figure: Left, a point P outside a circle with centre O, the segment PO with its midpoint M and the perpendicular bisector through M. Right, a second circle with centre M and radius MO cuts the given circle at A and B, and the lines PA and PB are the two tangents from P.]

To construct a tangent to a circle at a given point when the centre of the circle is known.

Given: We have a circle with centre O and a point P anywhere on its circumference. We have to construct a tangent through P.

Steps of Construction:
1. Draw a circle with centre O and mark a point P anywhere on it. Join OP.
2. Draw a perpendicular line through the point P and name it XY.
3. XY is the required tangent to the given circle passing through P.

[Figure: Left, a circle with centre O and point P on its circumference, joined by the radius OP. Right, the same circle with the line XY drawn through P perpendicular to OP, with X above and Y below P. XY is the tangent at P.]