NCERT Solutions Class 10 Maths Chapter 11 Constructions
NCERT Solutions Class 10 Maths Chapter 11 Constructions: National Council of Educational Research and Training Class 10 Maths Chapter 11 Solutions – Constructions. NCERT Solutions Class 10 Maths Chapter 11 PDF Download.
NCERT Solutions Class 10 Maths Chapter 11: Overview
Board |
NCERT |
Class |
10 |
Subject |
Maths |
Chapter |
11 |
Chapter Name |
Constructions |
Topic |
Exercise Solutions |
NCERT Solutions Class 10 Maths
Chapter 11 – Constructions
Exercise 11.1
In each of the following, give the justification of the construction also:
(1) Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.
Solution:
∴ Draw line segment AB o length 7.6 cm
Then draw any Ax, making an acute angle with AB.
Mark 13 = (5 + 8) points A1, A2, A3 …… As on Ax such that AA1 = A1 A2 = A2 A3 …… by drawing equal —–.
Then join B A13
Since we want the ratio 5:8
Through point A5 (—- = 5), we draw a line parallel to A13 B by making ∠AA13 B from point A5
Thus AC: CB = 5:8
On measuring AC and BC by scale
AC = 2.9 cm and BC = 4.7 cm.
Since, A13 B is parallel to A5,
AA5/A5A13 = AC/CB
∴ AA5/A5A13 = 5/8
∴ AC/CB = 5/8
Thus C divides AB at the ratio 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 2/3 of the corresponding sides of the first triangle.
Solution:
It this figure AB = 6cm
AC = 5cm
BC = 4cm
Let’s first construct ∆ABC with sides 4cm, 5cm, 6cm.
(1) Draw base AB or side 6cm
(2) With A as center, and 5cm as radius, draw an arc with B a centre, and 4cm as radius draw an arc
(3) Let C be the point where the two arc — join AC and BC
Thus ∆ABC is the required triangle
Now, Let’s make a similar triangle with scale factor = 2/3
Draw any ray AD making an acute angle with AB on the side opposite to the vertex C.
Mark 3 points (The greater of 2 and 3 in 1/3): A1, A2, A3 on A× so AA1 = A1 A2 = A2 A3
Join A3 B and draw a line trough A2 (the 2nd point, 2 being smaller of 2 and 3 in 1/3) parallel to A, B.
To intersect AB at B’
Draw a line through B’ parallel to the line BC to intersect AC at C’
Thus ∆AB’C’ is the required triangle.
Since scale factor is 2/3,
We need to prove, AB’/AB = AC’/AC = B’C’/BC = 2/3
By construction
AB’/AB = AA2/AA3 = 2/3 —– (i)
Also, B’C is parallel to BC
So, we well make the same angle with line AB
∴ ∠AB’C’ = ∠ABC (Corresponding angles) —– (ii)
Now, In ∆AB’C and ∆ABC
∠A = ∠A = (common)
∠AB’C’ = ∠ABC [From (ii)]
∆AB’C’ ~ ∆ABC (A similarity )
Since corresponding sides of similar triangles are in the same ratio.
AB’/AB = AC’/AC = B’C’/BC
So, AB’/AB = AC’/A = B’C’/BC = 2/3 [From (i)]
Thus our construction is justified.
(3) Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle.
Solution:
Let first construct ∆ABC with sides 5cm, 6cm, 7cm.
(1) Draw base AB of sides 5cm
(2) With A as center and 7cm as radius draw an arc. With B as center and 6cm as radius draw an arc
(3) Let C be the point where the two areas intersects join AC and BC.
Thus, ∆ABC required triangle
Now, Let’s Make a similar triangle with scale factor = 7/5
Draw any ray A× making an acute angle with AB on the side opposite to the vertex c.
Mark 7 (The greater of 7 and 5 in 7/5) points.
A1, A2, A3, A4, A5, A6, A7, on A× so that AA1 = A1A2 = ……. = A6A7
Join A5 B (5th point as 5 is smaller in 7/5) and draw a line through A7 Parallel to A5 B:
To intersect AB extended at B’
Draw a line thorough B’ Parallel to the line BC to intersect AC extended at C’
Thus ∆AB’ C’ is the required triangle
Since need to prove
AB’/AB = AC’/AC = B’C’/BC = 7/5
By construction, AB’/AB = AA7/AA5 = 7/5 —– (i)
Also, B’C’ is parallel to BC
So, the will make the same angle with line AB.
∴ ∠AB’C’ = ∠ABC (Corresponding angle) —– (ii)
Now, ∆AB’C’ and ∆ABC
∠A = ∠A (Common)
∠AB’C’ = ∠ABC [From (i)]
∆AB’C’ ~ ∆ABC (AA similarity)
Since corresponding sides of similar triangles are in the same ratio
AB’/AB = AC’/AC = B’C’/BC
So, AB’/AB = AC’/AC = B’C’/BC = 7/5 (From (i)]
Thus our construction is justified.
(4) Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1 ½ time the corresponding sides of the isosceles triangle.
Solution:
∆ABC is required triangle
Now, we need to make a triangle which is 1 1/2 time its size
∴ Scale factor = 1 1/2 = 2/3 > 1
(1) Draw any ray B × making an acute angle with BC on the side to the vertex A.
(2) Mark 3 (The greater of 3 and 2 in 3/2) points B1, B2, B3 on B × so that BB1 = B1B2 = B2B3
(3) Join B2C (2nd point as 2 is smaller in 3/2) and draw a line through B3 parallel to B2C, to intersect BC extended at C’
(4) Draw a line through C’ parallel to the line AC to intersect AB extended at A’.
Thus, ∆A’BC’ is the required triangle.
Since scale factor is 3/2.
We need to prove,
A’B/AB = A’C’/AC = BC’/BC = 3/2
By Construction,
BC’/BC = BB3/BB2 = 3/2 —– (i)
Also, A’C’ is parallel to AC
So, they will make the same angle with line BC
∴ ∠A’C’B = ∠ACB (Corresponding angles) —– (ii)
Now in ∆A’BC’ and ∆ABC
∠B = ∠B (Common)
∠A’C’B = ∠ACB (From (ii)
∆A’BC’ ~ ∆ABC (AA similarity)
Since corresponding sides or similar triangles are in the same ratio.
A’B/AB = A’C’/AC = BC’/BC
So, A’B/AB = A’C’/AC = BC’BC = 3/2 [From (i)]
Thus our construction is justified
(5) Draw a triangle ABC with side BC = 6cm, AB = 5cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.
Solution:
(1) Draw base BC of side 6cm
(2) Draw ∠B = 60°
(3) Taking B as center, 5 cm as radius, we draw an arc.
Let, the point where arc intersects the ray be point A.
(4) Join AC.
∆ABC is required triangle.
Now, we need to make a triangle which is 3/4 times its size.
∴ Scale factor = 3/4 < 1
Draw any ray B × making an acute angle with BC on the side opposite to the vertex A.
Mark 4 (The greater of 3 and 4 in 3/4) points.
B1, B2, B3, B4 on B× So that BB1 = B1B2 = B2B3 = B3B4 Point, 3 being smaller of 3 and 4 in 3/4) parallel to B4C.
To intersect BC ay C’
Draw a line through C’ parallel to the line AC to intersect BA at A’.
Thus ∆A’BC’ is the required triangle
Justification/Since scale factor is 3/4.
We need prove A’B/AB = A’C’/AC = BC’/BC = 3/4
By construction,
BC’/BC = BB3/BB4 = 3/4 —- (i)
Also, A’C’ is parallel to AC
So, they will make the same angle with line BC.
∴ ∠A’C’B = ∠ABC
∠B = ∠B (Common
∠A’C’B = ∠ACB [From (ii)]
∆A’BC’ ~ ∆ABC (AA similarity)
Since corresponding sides of similar triangles are in the same ratio.
A’B/AB = A’C’/AC = BC’/BC
So, A’B/AB = A’C’/AC = BC’/BC = 3/4 [From (i)]
Thus our construction is justified.
(6) Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of ∆ABC.
Solution:
Let’s first draw a rough diagram,
To construct ∆ABC, we first need to find ∠C.
Finding ∠C
In ∆ABC, ∠A + ∠B + ∠C = 180°
105° + 45° + ∠C = 180°
Or, ∠C = 180° – 150° = 30
(1) Draw base BC of side 7cm
(2) Draw ∠B = 45°
(3) Draw ∠C = 30°
(4) Let point A be the point where the two rays intersect.
∴ ∆ABC is the required triangle.
∴ Now, we need to make a triangle which is 4/3 times its size.
(1) Draw any ray B × making an acute angle with BC on the side opposite to the vertex A.
(2) Mark 4 (The greater of 4 and 3 in 3/4) points B1, B2, B4 on B × so that BB1 = B1 B2 = B2 B3 = B3 B4
(3) Join B3C (3rd point as 3 is smaller in 4/3) and draw a line through B4 parallel t B3C to intersect BC extended at C’
(4) Draw a line through C’ Parallel to the line AC to intersect AB extended at A’
Thus ∆A’BC’ is required triangle
Justification:
Since scale factor is 4/3,
We need to prove A’B/AB = A’C’/AC = BC’/BC = 4/3
By construction,
BC’/BC = BB4/BB3 = 4/3 —— (i)
Also, A’C’ is parallel to AC
So, they will make the same angle with line BC
∴ ∠A’C’B = ∠ACB (Corresponding angles) —- (ii)
Now, In ∆A’BC’ and ∆ABC
∠B = ∠B (Common)
∠A’C’B = ∠ACB [From (ii)]
∆A’BC’ ~ ∆ABC (AA similarity)
Since, corresponding sides of similar triangles are in the same, ratio.
A’B/AB = A’C’/AC = BC’/BC
So, A’B/AB = A’C’/AC = BC’/BC = 4/3 [From (i)]
Thus our construction is justified.
(7) Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle sides are 5/3 time the corresponding sides of the given triangle.
Solution:
(1) Draw base BC of side 3cm
(2) Draw ∠B = 90°
(3) Taking B as center, 4cm as radius, we draw an arc
Let the point where arc intersects the ray be point A.
(4) Join AC
∴ ∆ABC is required triangle.
Now, we need to make a triangle which is 5/3 times its size.
∴ Scale factor = 5/3 >1
(1) Draw any ray B × making an acute angle with BC on the side opposite to the vertex A.
(2) Mark 5 (The greater of 5 and 3 in 5/3) points.
B1, B2, B3, B4, B5, on B × so that BB1 = B1 B2 = B2 B3 = B3 B4 = B4 B5
(3) Join B3 C (3rd point as 3 is smaller in 5/5) and draw a line through B5 parallel to B3 C, to intersect BC extended at C’
(4) Draw a line through C’ parallel to the line AC to intersect AB extended at A’. Thus, ∆A’BC’ is required triangle
Justification:
Since scale factor is 5/3, we need to prove A’B/AB = A’C’/AC = BC’/BC = 5/3 by construction, BC’/BC = BB5/BB3 = 5/3 —– (i)
Also, A’C’ is parallel to AC. So, they will make the same angle with line BC.
∠A’C’B = ∠ACB (Corresponding angles) —– (ii)
Now, in ∆A’BC and ∆ABC
∠B = ∠B (Common)
∠A’C’B = ∠ACB [From (ii)]
∆A’BC’ ~ ∆ABC (AA similarity)
Since corresponding sides of similar triangles are in the same
A’B/AB = A’C’/AC = BC’/BC So, A’B/AB = A’C’/AC = BC’/BC = 5/3 [From (i)]
∴ Justified
Exercise 11.2
In each of the following, give also the justification of the construction:
(1) Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Solution:
(1) Draw a circle of radius 6cm
(2) Draw point P, 10cm away from center
(3) Join Po.
Let M be the midpoint of PO.
(4) Taking M as centre and MO as radius, draw a circle
(5) Let it intersect the given circle at point Q and R.
(6) Join PQ and PR
∴ PQ and PR are the required two tangents. After measuring lengths of tangents PQ and QR are 8cm each.
Justification:
We need to prove that PQ and PR are the tangents to the circle. Join OQ and OR.
Now, ∠PQO is an angle in the semi circle of the circle and we know that. Angle in a semi circle is a right angle
∴ ∠PQO = 90°
= OQ⊥PQ
Since, OQ is the radius of the circle.
PQ has to be a tangent of the circle (Since tangent is perpendicular to radius)
Similarly PR is a tangent of the circle.
(2) 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.
Solution:-
(1) Draw a circle of radius 4cm with center O.
(2) Draw concentric circle of radius 6cm with center O.
(3) Mark point P on the larger circle.
Now, Join OQ and OR
Since tangent is perpendicular to radius ∠PQO = 90° and ∠PRO = 90°
Thus ∆PQO is right angled triangle, and PO = radius of bigger circle = 6cm and OQ = radius of smaller circle = 4cm
By Pythagoras theorem
PO2 = PQ2 + OQ2
62 = PQ2 + 42
Or, 36 = PQ2 + 16
Or, PQ2 = 36 – 16 = 20
Or, PQ2 = √20 = 2√5 = 2×2.236 = 4.47 cm
Similarity PR = 4.47 cm
Justification
We need to prove that PQ and PR are the tangents to the circle.
Join OQ and OR
Now, ∠PQO is an angle in the semi circle of the O circle and we know that, Angle in a semi circle is a right angle.
∴ ∠PQO = 90°
= OQ⊥PQ
Since OQ is the radius of the circle, PQ has to be tangent of the circle (Since tangent is perpendicular to radius)
Similarity, PR is a tangent of the circle
(3) Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
Solution:
(1) Draw a circle of radius 3cm
(2) Draw diameter of circle, and extend it and mark points P and Q, 7 cm from the center.
(3) Taking M as centre the given circle at points A and B
(4) Let, it intersect the given circle at points A and B
(5) Join PA and PB.
Now we draw tangent from point Q
(6) Similar we draw tangent from point Q.
∴ QC and QD are the tangents from point Q
Justification
We need to prove that PA, PB, QC, QD are the tangents to the circle
Join OA, OB, OC and OD
Now, ∠PAO is an angle in the semi-circle of the circle and we know that,
Angle in a semi-circle is a right angle
∴ ∠PAO = 90°
= OA⊥PA
Since OA is the radius of the circle;
PA has to be a tangent of the circle. (Since tangent is perpendicular to radius)
Similarity we can prove PB, QC, QD are tangents of the circle.
(4) Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.
Solution:
(1) Draw a circle of radius 5cm
(2) Draw horizontal radius OQ
(3) Draw angle 120° from point O
Let the ray of angle intersects the circle at point R.
(5) Draw 90° from point R.
(6) Where the two areas intersect, mart as point P.
∴ PQ is the tangent to the circle.
Similarity PR is the tangent to the circle
(5) Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Solution:
(1) Draw line segment AB or length 8cm
(2) Taking A as center, draw a circle of radius 4cm
(3) Taking B as center, draw a circle of radius 3cm
Now, we need to draw tangent from point A to right circle, and from point B to the left circle, we need to draw perpendicular bisector of line AB.
(4) Make perpendicular bisector of AB
Let M be the midpoint of AB
(5) Taking M as centre and MA as radius draw a circle
(6) Let circle intersect left circle at P, Q.
(7) Let circle intersect right circle at R, S.
(8) Join BP, BQ, AR and AS
∴ AR, AS and BP, BQ are the required tangents.
Justification:
We need to prove that BP, BQ, AR AS are the tangents to the circle.
Join OP, OR and OS
Now, ∠APB is an angle in the semi-circle of the circle and we know that,
Angle in a semi-circle is a right angle.
∴ ∠ABP = 90°
= AP⊥BP
(6) Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Solution:
AB = 6cm, BC = 8cm and ∠B = 90°
First we draw ∆ABC
Now, Let’s draw a circle and construct tangents
(1) Draw perpendicular bisector of line BC.
Let the line intersect BC at point E.
(2) Now, E is the mid-point of BC
Taking E as center, and BE as radius, draw a circle
(3) Taking M as center and MA as radius draw a circle.
Justification:
We need to prove that AB and AQ are the tangents to the circle. Join EQ
Now, ∠AQE is an angle in the semi-circle of the circle and we know that.
Angle in a semi-circle is a right angle.
∴ ∠AQE = 90°
= EQ⊥AQ
(7) Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
Solution:
As perpendicular to the chord passes through the center
So, to find center
(1) We construct two non-parallel chords.
(2) Then, finding the pain of intersection of their perpendicular bisectors.
Steps to draw center of circle:
(1) Draw a circle with the help of a bangle
(2) Draw two non parallel chords AB and CD
(3) Draw perpendicular bisector of AB.
(4) Draw perpendicular bisector of CD
(5) Where the two perpendicular bisectors, intersects mark it as center of circle o.
Now, we need to draw tangents to this circle
Construction
(1) Draw point P outside the circle
(2) Join PO make perpendicular bisector of PO let M be midpoint of PO
(3) Taking M as centre and MO as radius, draw a circle
(4) Let it intersect the given circle at points Q and R
(5) Join PQ and PR
∴ PQ and PR are the required two tangents.
Justification:-
We need to prove that PQ and PR are the tangents to the circle
Join OQ and OR
Now, ∠PQO is an angle in the semi-circle of the circle and the we know that,
Angle in a semi-circle is a right angle.
∠PQO = 90
= OQ⊥PQ
Since OQ is the radius of the circle,
PQ has to be a tangent of the circle.
Similarly, PR is a tangent of the circle.