CBSETuts.com provides you Free PDF download of NCERT Exemplar of Class 10 Maths chapter 10 Constructions solved by expert teachers as per NCERT (CBSE) book guidelines. All the chapterwise questions with solutions to help you to revise complete CBSE syllabus and score more marks in Your board examinations.
You can also Download NCERT Solutions for class 10 Maths in Hindi to help you to revise complete Syllabus and score more marks in your examinations.
NCERT Exemplar Problems Class 10 Maths – Constructions
Exercise 10.1 Multiple Choice Questions (MCQs)
Question 1:
To divide a line segment AB in the ratio 5 : 7, first a ray AX is drawn, so that ∠BAX is an acute angle and then at equal distances points are marked on the ray AX such that the minimum number of these points is
(a) 8 (b) 10 (c) 11 (d) 12
Solution:
(d) We know that, to divide a line segment AB in the ratio m: n, first draw a ray AX which makes an acute angle ∠BAX, then marked m + n points at equal distance.
Here, m = 5, n = 7
So, minimum number of these points = m+n = 5 + 7 = 12.
Question 2:
To divide a line segment AB in the ratio 4 : 7, a ray AX is drawn first such that ∠BAX is an acute angle and then points A_{1} A_{2}, A_{3},… are located at equal distances on the ray AY and the point B is joined to
(a) A_{1}_{2} (b) A_{11} (c) A_{12} (d) A_{9
}Solution:
(b) Here, minimum 4+7 = 11 points are located at equal distances on the ray AX, and then B is joined to last point is A_{11}
Question 3:
To divide a line segment AB in the ratio 5 : 6, draw a ray AY such that ∠BAX is an acute angle, then draw a ray BY parallel to AY and the points A_{1}, A_{2}, A_{3},… and B_{1}, B_{2}, B_{3},… are located to equal distances on ray AY and BY, respectively. Then, the points joined are
(a) A_{5} and A_{6} (b) A_{6} and B_{5} (c) A_{4} and B_{5} (d) A_{5} and B_{4
}Solution:
(a) Given a line segment AB and we have to divide it in the ratio 5:6.
Steps of construction
 Draw a ray AX making an acute ∠BAX.
 Draw a ray BY parallel to AX by making ∠ABY equal to ∠BAX.
 Now, locate the points A_{1}, A _{2}, A_{3}, A _{4} and A_{5} (m= 5) on AX and B_{1}, B_{2}, B_{3}, B_{4}, B_{5} and B_{6} (n = 6) such that all the points are at equal distance from each other.
 Join B_{6}A_{5}. Let it intersect AB at a point C.
Then, AC:BC = 5:6
Question 4:
To construct a triangle similar to a given ΔABC with its sides of the corresponding sides of ΔABC, first draw a ray BX such that ∠CBX is an acute angle and X lies on the opposite side of A with respect to BC. Then, locate points B_{1}, B_{2}, B_{3},… on BX at equal distances and next step is to join
(a) B_{10} to C (b) B_{13 }to C (c) B_{7} to C (d)B_{4 }to C
Solution:
(c) Here, we locate points B_{1}, B_{2}, B_{3}, B_{4}, B_{5}, B_{6} and B_{7} on BX at equal distance and in next step join the last points is B_{7} to C.
Question 5:
To construct a triangle similar to a given ΔABC with its sides of the corresponding sides of ΔABC draw a ray BX such that ∠ CBX is an acute angle and X is on the opposite side of A with respect to BC. The minimum number of points to be located at equal distances on ray BX is
(a) 5 (b) 8 (c)13 (d) 3
Solution:
(b) To construct a triangle similar to a given triangle, with its sides of the corresponding sides of given triangle the minimum number of points to be located at equal distance is equal to the greater of m and n is
Hence, =
So, the minimum number of point to be located at equal distance on ray BX is 8.
Question 6:
To draw a pair of tangents to a circle which are inclined to each other at an angle of 60°, it is required to draw tangents at end points of those two radii of the circle, the angle between them should be
(a) 135° (b) 90° (c) 60° (d) 120°
Solution:
(d) The angle between them should be 120° because in that case the figure formed by the intersection point of pair of tangent, the two end points of thosetwo radii tangents are drawn) and the centre of the circle is a quadrilateral.
From figure it is quadrilateral,
∠POQ + ∠PRQ = 180° [∴ sum of opposite angles are 180°]
60°+ θ = 180°
θ=120
Hence, the required angle between them is 120°.
Exercise 10.2 Very Short Answer Type Questions
Question 1:
By geometrical construction, it is possible to divide a line segment in the ratio √3 :.
Solution:
True
Hence, the geometrical constrution is possible to divide a line segment in the ratio 3 : 1
Question 2:
To construct a triangle similar to a given ΔABC with its sides of the corresponding sides of ΔABC, draw a ray BX making acute angle with BC and X lies on the opposite side of A with respect of BC. The points B_{1}, B_{2}, …, B_{7} are located at equal distances on BX, B_{3}is joined to C and then a line segment B_{6}C’ is drawn parallel to B_{3}C, where C’ lines on BC produced. Finally line segment A’C’ is drawn parallel to AC.
Solution:
False
Steps of construction
 Draw a line segment BC with suitable length.
 Taking B and C as centres draw two arcs of suitable radii intersecting each other at A
 Join BA and CA ΔABC is the required triangle.
 From B draw any ray BX downwards making an acute angle CBX.
 Locate seven points B_{1}, B_{2}, …, B_{7} on SX, such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4} = B_{4}B_{5 }= B_{5}B_{6} = B_{6}B_{7}.
 Join B_{3}C and from B_{7} draw a line B_{7}C’ B_{3}C intersecting the extended line segment BC at C’.
 From point C’ draw C’A’ CA intersecting the extended line segment BA at A’.Then, ΔA’BC’ is the required triangle whose sides are of the corresponding sides of ΔABC.
Given that, segment B_{6}C’ is drawn parallel to B_{3}C. But from our construction is never possible that segment B_{6}C’ is parallel to B_{3}C because the similar triangle A’BC’ has its sides of the corresponding sides of triangle ABC. So, B_{7}C’ is parallel to B_{3}C.
Question 3:
A pair of tangents can be constructed from a point P to a circle of radius 3.5 cm situated at a distance of 3 cm from the centre.
Solution:
False
Since, the radius of the circle is 3.5 cm i.e., r = 3.5 cm and a point P situated at a distance of 3 cm from the centre i.e.,d= 3 cm
We see that, r > d
i.e., a point P lies inside the circle. So, no tangent can be drawn to a circle from a point lying inside it. ‘
Question 4:
A pair of tangents can be constructed to a circle inclined at an angle of 170°.
Solution:
Exercise 10.3 Short Answer Type Questions
Question 1:
Draw a line segment of length 7 cm. Find a point P on it which^{1} divides it in the ratio 3:5.
Solution:
Steps of construction
 Draw a line segment AB = 7 cm.
 Draw a ray AX, making an acute ∠BAX
 Along AX, mark 3+ 5= 8 points
A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}, A_{8} such that
AA_{1 }= A_{1}A_{2 }= A_{2}A_{3} = A _{3}A_{4} = A _{4}A_{5} = A _{5}A_{6} = A _{6}A _{7 }= A _{7}A_{8}  Join A_{8}B_{ }
 From A_{3}, draw A_{3}C  A_{8}B meeting AB at C.
[by making an angle equal to ∠BA _{8}A at A _{3}]
Then, C is the point on AB which divides it in the ratio 3 : 5,
Question 2:
Draw a right ΔABC in which BC = 12 cm, AB = 5 cm and ∠B = 90°.Construct a triangle similar to it and of scale factor Is the new triangle also a right triangle?
Solution:
Steps of construction

 Draw a line segment BC = 12 cm,
 From 6 draw a line AB = 5 cm which makes right angle at B.
 Join AC, ΔABC is the given right triangle.
 From B draw an acute ∠CBY downwards.
 On ray BY, mark three points B_{1}, B_{2}and B_{3}, such that BB_{1} = B_{1}B_{2} = B_{2}B_{3}.
 Join B_{3} C.
 From point B_{2} draw B_{2}N  B_{3}C intersect BC at N.
 From point N draw NM  CA intersect BA at M. ΔMBN is the required triangle. ΔMBN is also a right angled triangle at B.
Question 3:
Draw a ΔABC in which BC = 6 cm, CA = 5 cm and AB = 4 cm. Construct a triangle similar to it and of scale factor
Solution:
Steps of construction
 Draw a line segment BC = 6 cm.
 Taking Sand C as centres, draw two arcs of radii 4 cm and 5 cm intersecting each other at A.
 Join BA and CA. ΔABC is the required triangle.
 From B, draw any ray BX downwards making at acute angle.
 Mark five points B_{1}, B_{2},B_{3}, B_{4} and B_{5} on BX, such that
BB, = B,B_{2} = B_{2}B_{3} = B_{3}B_{4} = B_{4}B_{5}.  Join B_{3}C and from B_{5} draw B_{5}M  B_{3}C intersecting the extended line segment BC at
 From point M draw MN  CA intersecting the extended line segment BA at N.
Then, ΔNBM is the required triangle whose sides is equal to of the corresponding
sides of the ΔABC.
Hence, ΔNBM is the required triangle.
Question 4:
Construct a tangent to a circle of radius 4 cm from a point which is at a distance of 6 cm from its centre.
Solution:
Given, a point M’ is at a distance of 6 cm from the centre of a circle of radius 4 cm.
Exercise 10.4 Long Answer Type Questions
Question 1:
Two line segments AB and AC include an angle of 60°, where AB = 5 cm and AC = 7 cm. Locate points P and Q on AB and AC, respectively such that AP = AB and AQ = AC. Join P and Q and measure the length PQ.
Solution:
Given that, AB = 5 cm and AC = 7 cm
Steps of construction
 Draw a line segment AB = 5 cm.
 Now draw a ray AZ making an acute ∠BAZ = 60°.
 With A as centre and radius equal to 7 cm draw an arc cutting the line AZ at C.
 Draw a ray AX, making an acute ∠BAX
 Along AX, mark 1+3 = 4 points A_{1}, A_{2}, A_{3}, and A_{4 }Such that AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4}
 Join A_{4}B
 From A_{3} draw A_{3}P  A_{4}B meeting AB at P [by making an angle equal to ∠AA_{4}B]
Then, Pis the point on AB which divides it in the ratio 3:1.
So, AP: PB = 3:1  Draw a ray AY, making an acute ∠CAY
 Along AY, mark 3+1 = 4 points B_{1}, B_{2}, B_{3} and B_{4}.
Such that AB_{1} = B_{1}B_{2} = B_{2}B_{3} =B_{3}B_{4}  Join B_{4}C_{ }
 From B_{1} draw B_{1}Q  B_{4}C meeting AC atQ. [by making an angle equal to ∠AB_{4}C]
Then, Q is the point on AC which divides it in the ratio 1 : 3.
So AQ:OC = 1:3  Finally, join PQ and its measurment is 3.25 cm.
Question 2:
Draw a parallelogram ABCD in which BC = 5 cm, AB = 3 cm and ∠ABC = 60°, divide it into triangles BCD and ABD by the diagonal BD. Construct the triangles BD’C’ similar to ΔBDC with Scale factor . Draw the line segment D’A’ parallel to DA, where A’ lies on extended side BA. Is A’BC’D’ a parallelogram?
Solution:
Steps of construction
 Draw a line segment AB = 3 cm.
 Now, draw a ray BY making an acute ∠ABY = 60°.
 With B as centre and radius equal to 5 cm draw an arc cut the point C on
 Again draw a ray AZ making an acute ∠ZAX’ = 60°. [∴ BY  AZ, ∴ ∠YBX’ = TAX’ = 60°]
 With A as centre and radius equal to 5 cm draw an arc cut the point D on AZ.
 Now, join CD and finally make a parallelogram ABCD
 Join BD, which is a diagonal of parallelogram ABCD
 From B draw any ray BX downwards making an acute ∠CBX.
 Locate 4 points B_{1}, B_{2}, B_{3}, B_{4} on BX, such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4},
 Join B_{4}C and from B_{3}C draw a line B_{4}C’  B_{3}C intersecting the extended line segment BC at C’.
 From point C’ draw C’D’ CD intersecting the extended line segment BD at D’. Then, AD’BC’ is the required triangle whose sides are of the corresponding sides of ΔDBC
 Now draw a line segment D’A’ parallel to DA, where A’ lies on extended side BA i.e ray BX’.
 Finally, we observe that A’BCD’ is a parallelogram in which A’D’ = 6.5,cm A’B = 4 cm and ∠A’BD’ = 60° divide it into triangles BCD’ and A’BD’ by the diagonal BD.
Question 3:
Draw two concentric circles of radii 3 cm and 5 cm. Taking a point on outer circle construct the pair of tangents to the other. Measure the length of a tangent and verify it by actual calculation.
Solution:
Given, two concentric circles of radii 3 cm and 5 cm with centre 0. We have to draw pair of
tangents from point P on outer circle to the other.
Steps of construction
 Draw two concentric circles with centre 0 and radii 3 cm and 5 cm.
 Taking any point P on outer circle. Join OP.
 Bisect OP, let M’ be the midpoint of .
Taking M’ as centre and OM’ as radius draw a circle dotted which cuts the inner circle at M and P’.  Join P M and PP’. Thus, PM and PP’ are the required tangents.
 On measuring PM and PP’, we find that PM = PP’ = 4 cm.
Question 4:
Draw an isosceles triangle ABC in which AB = AC = 6 cm and BC = 5 cm. Construct a triangle PQR similar to AABC in which PQ = 8 cm. Also justify the construction.
Solution:
Let ΔPQR and ΔABC are similar triangles, then its scale factor between the corresponding sides is
Steps of construction

 Draw a line segment BC = 5 cm.
 Construct OQ the perpendicular bisector of line segment BC meeting BC at P’.
 Taking B and C as centres draw two arcs of equal radius 6 cm intersecting each other at A
 Join BA and CA. So, ΔABC is the required isosceles triangle.
 From B, draw any ray BX making an acute ∠CBX
 Locate four points B_{1}, B_{2}, B_{3} and B_{4} on BX such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}
 Join B_{3}C and from B_{4} draw a line B_{4}R  B_{3}C intersecting the extended line segment BC at R.
 From point R, draw RPCA meeting BA produced at P
Then, ΔPBR is the required triangle.
Question 5:
Draw a ΔABC in which AB = 5 cm, BC = 6 cm and ∠ABC = 60°. Construct a triangle similar to ABC with scale factor Justify the construction.
Solution:
Question 6:
Draw a circle of radius 4 cm. Construct a pair of tangents to it, the angle between which is 60°. Also justify the construction. Measure the distance between the centre of the circle and the point of intersection of tangents.
Solution:
In order to draw the pair of tangents, we follow the following steps
Steps of construction
 Take a point 0 on the plane of the paper and draw a circle of radius OA = 4 cm.
 Produce OA to B such that OA = AB = 4 cm.
 Taking A as the centre draw a circle of radius AO = AB = 4 cm.
Suppose it cuts the circle drawn in step 1 at P and Q.  Join BP and BQ to get desired tangents.
Question 7:
Draw a ΔABC in which AB = 4 cm, SC = 6 cm and AC = 9 cm. Construct a triangle similar to ΔABC with scale factor Justify the construction. Are the two triangles congruent? Note that, all the three angls and two sides of the two triangles are equal.
Solution:
Steps of construction

 Draw a line segment BC = 6 cm.
 Taking B and C as centres, draw two arcs of radii 4 cm and 9 cm intersecting each other at A.
 Join BA and CA, ΔABC is the required triangle.
 From B, draw any ray BX downwards making an acute angle.
 Mark three points B_{1}, B_{2}, B_{3} on BX, such that BB_{1} = B_{1}B_{2} = B_{2}B_{3}.
 Join B_{2}C and from B_{3} draw B_{3}M  B_{2}C intersecting the extended line segment BC at
 From point M, draw MNCA intersecting the extended line segment BA to N.
Then,ΔNBM is the required triangle whose sides are equals to of the corresponding sides of the ΔABC
The two triangles are not congruent because, if two triangles are congruent, then they have same shape and same size. Here, all the three angles are same but three sides are not same one side is different.
NCERT Exemplar ProblemsNCERT Exemplar MathsNCERT Exemplar Science
Tanmai says
Super