CBSE Class 10 Maths Chapter-11 Constructions – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 10 Maths Chapter 11 – Constructions prepared by expert Maths teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 10 Maths Chapter-11 Constructions Important Questions
CBSE Class 10 Maths Important Questions Chapter 11 – Constructions
3 Mark Questions
1. In each of the following, give the justification of the construction also:
Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.
Ans. Given: A line segment of length 7.6 cm.
To construct: To divide it in the ration 5 : 8 and to measure the two parts.
Steps of construction:
(a) From any ray AX, making an acute angle with AB.
(b) Locate 13 (=5 + 8) points A1, A2, A3, A4, A5, A6, A7, A8, A9, A10, A11, A12 and A13 on AX such that AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7 = A7A8 = A8A9 = A9A10 = A10A11 = A11A12 = A12A13.
(c) Join BA13.
(d) Through the point A5, draw a line parallel to A13B intersecting AB at the point C.
Then, AC : CB = 5 : 8
On measurement, AC = 3.1 cm, CB = 4.5 cm
Justification:
A5C
A13B [By construction]
[By Basic Proportionality Theorem]
But 
[By construction]
Therefore, 
AC : CB = 5 : 8
2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 
of the corresponding sides of the first triangle
Ans. To construct: To construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are of the corresponding sides of the first triangle.
Steps of construction:
(a) Draw a triangle ABC of sides 4 cm, 5 cm and 6 cm.
(b) From any ray BX, making an acute angle with BC on the side opposite to the vertex A.
(c) Locate 3 points B1, B2 and B3 on BX such that BB1 = B1B2 = B2B3.
(d) Join B3C and draw a line through the point B2, draw a line parallel to B3C intersecting BC at the point C’.
(e) Draw a line through C’ parallel to the line CA to intersect BA at A’.
Then, A’BC’ is the required triangle.
Justification:
B3CB2C’ [By construction]
[By Basic Proportionality Theorem]
But 
[By construction]
Therefore, 
………(i)
CAC’A’[By construction]
BC’A’ 
BCA[AA similarity]
[From eq. (i)]
3. Construct a triangle with sides 6 cm, 6 cm and 7 cm and then another triangle whose sides are
of the corresponding sides of the first triangle.
Ans. To construct: To construct a triangle of sides 5 cm, 6 cm and 7 cm and then a triangle similar to it whose sides are of the corresponding sides of the first triangle.
Steps of construction:
(a) Draw a triangle ABC of sides 5 cm, 6 cm and 7 cm.
(b) From any ray BX, making an acute angle with BC on the side opposite to the vertex A.
(c) Locate 7 points B1, B2, B3, B4, B5, B6and B7 on BX such that BB1 = B1B2 = B2B3= B3B4 = B4B5= B5B6 = B6B7.
(d) Join B5C and draw a line through the point B7, draw a line parallel to B5C intersecting BC at the point C’.
(e) Draw a line through C’ parallel to the line CA to intersect BA at A’.
Then, A’BC’ is the required triangle.
Justification:
C’A’CA [By construction]
ABCA’BC’ [AA similarity]
[By Basic Proportionality Theorem]
B7C’B5C [By construction]
BB7C’ BB5C [AA similarity]
But 
[By construction]
Therefore, 
4. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
Ans. To construct: To construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its lengths. Also to verify the measurements by actual calculation.
Steps of Construction:
(a) Join PO and bisect it. Let M be the mid-point of PO.
(b) Taking M as centre and MO as radius, draw a circle. Let it intersects the given circle at the point Q and R.
(c) Join PQ.
Then PQ is the required tangent.
By measurement, PQ = 4.5 cm
By actual calculation,
PQ = 
= 
= 
= 
= 4.47 cm
Justification: Join OQ. Then 
PQO is an angle in the semicircle and therefore,
PQO = 
PQ 
OQ
Since, OQ is a radius of the given circle, PQ has to be a tangent to the circle.
5. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of
Ans. To construct: A pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of
Steps of Construction:
(a) Draw a circle of radius 5 cm with centre O.
(b) Draw an angle AOB of 
(c) At A and B, draw angles which meet at C.
Then AC and BC are the required tangents which are inclined to each other at an angle of
Justification:
OAC = and OA is a radius. [By construction]
AC is a tangent to the circle.
OBC = and OB is a radius. [By construction]
BC is a tangent to the circle.
Now, in quadrilateral OACB,
AOB + OAC + OBC + ACB = 
[Angle sum property of a quadrilateral]
+ ACB =
+ ACB =
ACB = 
6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and B =
BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Ans. To construct: A right triangle ABC with AB = 6 cm, BC = 8 cm and B = BD is the perpendicular from B on AC and the tangents from A to this circle.
Steps of Construction:
(a) Draw a right triangle ABC with AB = 6 cm, BC = 8 cm and B = Also, draw perpendicular BD on AC.
(b) Join AO and bisect it at M (here O is the centre of circle through B, C, D).
(c) Taking M as centre and MA as radius, draw a circle. Let it intersects the given circle at the points B and E.
(d) Join AB and AE.
Then AB and AE are the required two tangents.
Justification: Join OE.
Then, AEO is an angle in the semicircle.
AEO =
AE OE
Since OE is a radius of the given circle, AE has to be a tangent to the circle. Similarly AB is also a tangent to the circle.
7. Prove that the tangents drawn at the ends of a chord of a circle make equal angles with chord.
Ans. Let NM be chord of circle with centre C.
Let tangents at M.N meet at the point O.
Since OM is a tangent
is a tangent
Again in 
Thus, tangents make equal angle with the chord
8. In the given figure, if AB = AC, prove that BE = EC.
Ans. Since tangents from an exterior point A to a circle are equal in length
Similarly, tangents from an exterior point B to a circle are equal in length
Similarly, for C
CE = CF ………….(3)
Now AB = AC
9. Find the locus of centre of circle with two intersecting lines.
Ans.
Let 
be two intersection lines.
Let a circle with centre P touch the two lines 
and 
at M and N respectively.
PM = PN [Radii of same circle]
P is equidistance from the lines and
Similarly, centre of any other circle which touch the two intersecting lines will be equidistant from and
P lies on 
a bisector of the angle between and
[The locus of points equidistant from two intersecting lines is the pair of bisectors of the angle between the lines]
Hence, locus of centre of circles which touch two intersecting lines is the pair of bisectors of the angles between the two lines.
10. In the given figure, a circle is inscribed in a quadrilateral ABCD in which 
. If AD = 23 cm, AB = 29 cm and DS = 5 cm, find the radius of the circle.
Ans. In the given figure,
and OQ^ BA
Also, OP = OQ =r
is a square
BP = BQ =r
But DR = DS = 5 cm …(i)
