NCERT SOLUTIONS FOR CLASS 10 MATHS CONSTRUCTIONS CHAPTER 11 EX 11.1
Question 1.
Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.
Solution:
Ex 11.1 Class 10 Maths Question 2.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 23 of the corresponding sides of the first triangle.
Solution:
Ex 11.1 Class 10 Maths Question 3.
Construct a triangle with sides 5 cm, 6 cm, and 7 cm and then another triangle whose sides are 75 of the corresponding sides of the first triangle.
Solution:
Ex 11.1 Class 10 Maths Question 4.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 112 times the corresponding sides of the isosceles triangle.
Solution:
Ex 11.1 Class 10 Maths Question 5.
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 34 of the corresponding sides of the triangle ABC.
Solution:
Ex 11.1 Class 10 Maths Question 6.
Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 43 times the corresponding sides of ∆ABC.
Solution:
Ex 11.1 Class 10 Maths Question 7.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 53F times the corresponding sides of the given triangle.
Solution:
Steps of Construction:
1. Construct a ∆ABC, such that BC = 4 cm, CA = 3 cm and ∠BCA = 90°
2. Draw a ray BX making an acute angle with BC.
3. Mark five points B1, B2, B3, B4 and B5 on BX, such that BB1 = B1B2 = B2B3 = B3B4 = B4B5.
4. Join B3C.
5. Through B5, draw B5C’ parallel to B3C intersecting BC produced at C’.
6. Through C’, draw C’A’ parallel to CA intersecting AB produced at A’.
Thus, ∆A’BC’ is the required right triangle.
Justification:
Class 10 Maths Constructions Mind Maps
Construction
Construction implies drawing geometrical figures accurately such that triangles, quadrilateral and circles with the help of ruler and compass.
Division of a Line Segment
A line segment can be divided in a given ratio (both internally and externally)
Example:
Divide a line segment of length 12 cm internally in the ratio 3:2.
Solution :
Steps of construction :
(i) Draw a line segment AB = 12 cm. by using a ruler.
(ii) Draw a ray making a suitable acute angle ∠BAX with AB.
(iii) Along AX, draw 5 ( = 3 + 2) arcs intersecting the ray AX at A1? A2, A3, A4 and A5 such that
AA1 = A1A2 = A2A3 = A3A4 = A4A5
(iv) Join BA5.
(v) Through A3 draw a line A3P parallel to A5B making ∠AA3P = ∠AA5B, intersecting AB at point P.
The point P so obtained is the required point, which divides AB internally in the ratio 3 : 2.
Similar Triangles
(i) This Construction involves two different situation.
(a) Construction of a similar triangle smaller than the given triangle.
(b) Construction of a similar triangle greater than the given triangle.
(ii) The ratio of sides of the triangle to be constructed with the corresponding sides of the given triangle is called scale factor.
Example:
Draw a triangle ABC with side BC = 7 cm. ∠B = 45°, ∠A = 105°. Construct a triangle whose sides are (4/3) times the corresponding side of ∆ABC.
Solution :
Steps of construction :
(i) Draw BC = 7 cm.
(ii) Draw a ray BX and CY such that ∠CBX= 45° and
∠BCY = 180° – (45° + 105°) = 30°
Suppose BX and CY intersect each other at A.
∆ABC so obtained is the given triangle.
(iii) Draw a ray BZ making a suitable acute angle with BC on opposite side of vertex A with respect to BC.
(iv) Draw four (greater of 4 and 3 in 4/3) arcs intersecting the ray BZ at B1, B2, B3, B4 such that BB1 = B1B2 = B2B3 = B3B4.
(v) Join B3 to C and draw a line through B4 parallel to B3C, intersecting the extended line segment BC at C’.
(vi) Draw a line through C’ parallel to CA intersecting the extended line segment BA at A’. Triangle A’BC’ so obtained is the required triangle.
Tangents to a Circle
Two tangents can be drawn to a given circle from a point outside it.
Example:
Draw a circle of radius 4 cm. Take a point P outside the circle. Without using the centre of the circle, draw two tangents to the circle from point P.
Solution :
Steps of construction :
(i) Draw a circle of radius 4 cm.
(ii) Take a point P outside the circle and draw a secant PAB, intersecting the circle at A and B.
(iii) Produce AP to C such that AP = CP.
(iv) Draw a semi-circle with CB as diameter.
(v) Draw PD ⊥ CB, intersecting the semi-circle at D.
(vi) With P as centre and PD as radius draw arcs to intersect the given circle at T and T’
(vii) Join PT and PT’. Then, PT and PT’ are the required tangents.
Note:
If centre of a circle is not given, then it can be located by finding point of intersection of perpendicular bisector, of any two nonparallel chords of a circle.