NCERT Exemplar for Class 9 Math Chapter 11 - Constructions (Book Solutions)

Sample questions

1. With the help of a ruler and a compass, it is possible to construct an angle of :

(A) 35° 

(B) 40° 

(C) 37.5° 

(D) 47.5°

Ans: Option (C) is the correct answer.

The angle 37.5° is the bisector of angle 75°. And we can draw 75° by first drawing 90° and 60° and then dividing the angle between these two angles. Thus, we will get 75o and then we will bisect 75° to get 37.5°.

2. The construction of a triangle ABC in which AB = 4 cm, A = 60° is not possible when difference of BC and AC is equal to:

(A) 3.5 cm 

(B) 4.5 cm 

(C) 3 cm 

(D) 2.5 cm

Ans: Option (B) is the correct answer.

The triangle inequality states that the difference of the lengths of any two sides of a triangle is always lesser than the length of the remaining (third) side. So, the length of the third side here is AB = 4 cm.

And the difference should be greater than 4 cm so the correct option is (B).

Exercise 11.1

1. With the help of a ruler and a compass it is not possible to construct an angle of:

(a) 37.5°

(b) 40°

(c) 22.5°

(d) 7.5°

Ans: Option (b) 40° is the correct answer.

It is not feasible to construct an angle of 40° with the aid of a ruler and a compass because it is only possible to construct an angle that is a multiple of 3 such as 30°, 60° and 90° and so on. angle can be constructed with the help of a ruler and a compass.

40° is not the multiple of 3, therefore it is not possible to construct. Hence, (b) is the correct answer.

2. The construction of a triangle ABC, given that BC = 6 cm, ∠B = 45° is not possible when difference of AB and AC is equal to:

(a) 6.9 cm

(b) 5.2 cm

(c) 5.0 cm

(d) 4.0 cm

Ans: Option (a) 6.9 cm is the correct answer.

The triangle inequality states that the difference of the lengths of any two sides of a triangle is always lesser than the length of the remaining (third) side.

Given BC = 6 cm and ∠B as the base angle, the difference between the other two sides AB and AC must be less than the length of side BC. But in this case the triangle is not possible when the difference is equal and greater than 6cm.

Hence, the correct answer is (a) 6.9 cm.

3. The construction of a triangle ABC, given that BC = 3 cm, ∠C = 60° is possible when difference of AB and AC is equal to:

(a) 3.2 cm

(b) 3.1 cm

(c) 3 cm

(d) 2.8 cm

Ans: Option (d) 2.8 cm is the correct answer.

The triangle inequality states that the difference of the lengths of any two sides of a triangle is always lesser than the length of the remaining (third) side.

Given BC = 3 cm and ∠C as the base angle, the difference between the other two sides AB and AC must not equal or exceed the length of side BC. 

Hence, the correct answer is (a) 2.8 cm.

Sample questions

Write True or False and give reasons for your answer.

1. An angle of 67.5° can be constructed.

Ans: True. 

The angle 67.5° is the bisector of angle 135°. And we can draw 135° by first drawing 90° and 180° and then dividing the angle between these two angles. Thus, we will get 135° and then we will bisect 135° to get 67.5°.

Exercise 11.2

Write True or False in each of the following. Give reasons for your answer:

1. An angle of 52.5° can be constructed.

Ans: The angle 52.5° is equal to $\dfrac{1}{4} \times 210^\circ $

And the angle 210° is equal to = 180° + 30° which we can construct easily so the angle 52.5° is a multiple of 3. 

Hence, the given statement is correct.

2. An angle of 42.5° can be constructed.

Ans: The angle 42.5° is equal to $\dfrac{1}{4} \times 170^\circ $

And the angle 170° cannot be constructed easily as the angle 42.5° is not a multiple of 3.

Hence, the given statement is incorrect.

3. A triangle ABC can be constructed in which AB = 5 cm, ∠A = 45° and BC + AC = 5 cm.

Ans: Here the side AB = BC + AC = 5 cm.

The triangle inequality states that the sum of the lengths of any two sides of a triangle is always greater than the length of the remaining (third) side.

So, we cannot construct a triangle in which AB = BC + AC.

Hence, the given statement is incorrect.

4. A triangle ABC can be constructed in which BC = 6 cm, ∠C = 30° and AC – AB = 4 cm.

Ans: The triangle inequality states that the difference of the lengths of any two sides of a triangle is always lesser than the length of the remaining (third) side.

Here the difference is, AC – AB (= 4 cm) < BC (= 6cm).

Hence, the given statement is correct.

5. A triangle ABC can be constructed in which ∠B = 105°, ∠C = 90° and AB + BC + AC = 10 cm.

Ans: The sum of angles $\angle B + \angle C = {105^ \circ } + {90^ \circ } = {195^ \circ } > {180^ \circ }$ 

And we know that the sum of all the angles of a triangle is 180° 

Hence, the given statement is incorrect.

6. A triangle ABC can be constructed in which ∠B = 60°, ∠C = 45° and AB + BC + AC = 12 cm.

Ans: The sum of the angles $\angle B + \angle C = {60^ \circ } + {45^ \circ } = {105^ \circ } < {180^ \circ }$

So, a triangle is possible with the given values.

Hence, the given statement is incorrect.

Sample Questions

1. Construct a triangle ABC in which BC = 7.5 cm, B = 45° and AB – AC = 4 cm.

Ans:

Steps of Construction:

1. Draw the base of the triangle BC = 7.5 cm

2. At point B make an angle \[\angle XBC = 45^\circ \].

3. Cut the line segment BD equal to AB – AC = 4cm from ray we drew at point B (BX).

4. Join DC and draw the perpendicular bisector of the line segment DC. Let’s name is PQ

5. Let PQ intersect BX at a point A. 

6. Join AC.

Then ABC is the required triangle

Exercise 11.3

1. Draw an angle of 110° with the help of a protractor and bisect it. Measure each angle.

Ans: Given: An angle \[\angle ABC = {110^ \circ }\]

We have to construct: The bisector of \[\angle ABC\]

Steps of construction:

1. Draw an arc with B as the centre and a reasonable radius to intersect the rays BA and BC at P and Q, respectively.

2. Draw an arc with a radius greater than half of PQ and a centre P.

3. Draw another arc with Centre Q and the same radius (as in step 2) to cut the previous arc at R.

3. Draw a BR ray. The needed bisector of \[\angle ABC\] is the ray BR.

The new angles formed will be half of the angle \[\angle ABC\]. The measure of these angles is $\dfrac{1}{2} \times {110^ \circ } = {55^ \circ }$

2. Draw a line segment AB of 4 cm in length. Draw a line perpendicular to AB through A and B, respectively. Are these lines parallel?

Ans: Given: A line segment AB of length 4 cm.

We have to construct: Perpendiculars to AB through A and B, respectively. 

Steps of construction:

1. Draw AB = 4 cm.

2. Draw an arc with A as the centre and any appropriate radius, cutting AB at P.

3. Draw an arc cutting the arc drawn in step 2 at Q with P as the centre and the same radius.

4. Draw an arc with Q as the centre and the same radius as steps 2 and 3, cutting the arc formed in step 3 at R.

5. Draw an arc with Q as the centre and the same radius.

6. Draw an arc with R as the centre and the same radius, cutting the arc made in step 5 at X.

7. Make a drawing of OX and produce it to C and D. This is the perpendicular to AB at A that is required.

8. Now draw the line EF perpendicular through B by repeating steps 2–7. This is the perpendicular to B that is necessary.

Because CD and EF are perpendicular to each other on AB. As a result, \[\angle CAB = \angle EBA = 90^\circ .\]

Yes, these lines are parallel since the sum of the interior angles on the same side of the triangle is the same.

The transverse angle is 180°.

3. Draw an angle of 80° with the help of a protractor. Then construct angles of (i) 40° (ii)160° and (iii)120°.

Ans: Steps of construction:

1. OA, draw a ray.

2. We will complete the construction with the aid of a protractor.

3. Draw an arc to connect rays OA and OB at positions P and Q, using O as the centre and any reasonable radius.

4. Bisect the angle $\angle BOA$. Let the ray OC be the bisector of angle $\angle BOA$ then $\angle COA = \dfrac{1}{2}\angle BOA = \dfrac{1}{2} \times {80^ \circ } = {40^ \circ }$

5. Draw an arc with Q as the centre and a radius equal to PQ to cut the extended arc PQ at R. To form ray OD, join OR and produce it, then $\angle DOA = 2\angle BOA = 2 \times {80^ \circ } = {160^ \circ }$

6. Bisect the angle $\angle DOB$. Let OE be the bisector of $\angle DOB$ is then $\angle EOA = \angle EOB + \angle BOA = \dfrac{1}{2} \times {80^ \circ } + {80^ \circ } = {120^ \circ }$

4. Construct a triangle whose sides are 3.6 cm, 3.0 cm and 4.8 cm. Bisect the smallest angle and measure each part.

Ans: Steps of construction: 

1. Draw a line AB = 4.8 cm.

2. Draw an arc with a radius of 3 cm and a centre of 'A'. Draw an arc with a radius of 3.6 cm and a centre of 'B' that intersects our previous arc at 'C.'

3. Join CA and CB we get the required triangle ABC.

We now measure all internal angles and find that $\angle ABC$ is the least.

So, we will bisect the angle $\angle ABC$.

4. Using any radius (less than half of AB) and the centre 'B,' draw an arc that intersects our AB line at P and our BC line at Q.

5. Draw an arc with the same radius and centre as 'P' and 'Q' that intersects at 'R'. 

6. Join BR and extend BR so that the line crosses AC at 'D.'

With the use of a protractor, we can now readily measure each other's angles.

5. Construct a triangle ABC in which BC = 5 cm, ∠B = 60° and AC + AB = 7.5 cm.

Ans: Given: In$\Delta ABC$ , BC = 5 cm, AC + AB = 7.5 cm and angle $\angle B = {60^ \circ }$

To construct $\Delta ABC$

The steps of construction are:

1. Drawing a ray BX and cutting out a line segment BC = 5 cm from it

2. Construct $\angle XBY = {60^ \circ }$ at B.

3. Draw an arc at meet BY at D with B as the centre and a radius of 7.5 cm.

4. Join CD.

5. Draw a perpendicular bisector from CD to A, intersecting BD.

6. Join AC.

7. The needed triangle is then ABC which we just constructed.

6. Construct a square of side 3 cm.

Ans: Steps of construction:

1. Take a side AB = 3 cm.

2. At A, draw a perpendicular $AY \bot AB$

3. Draw an arc cutting AY at D with A as the centre and a radius of 3cm.

4. Draw an arc intersecting at C with the centres B and D and radii of 3 cm.5. Join BC and DC. 

5. ABCD is our required square.

7. Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm.

Ans: Steps of construction:

1. Take a side AB = 5 cm.

2. Draw a perpendicular $AY \bot AB$ at A.

3. Draw an arc cutting AY at D with A as the centre and a radius of 3.5 cm.

4. Draw an arc with D as the centre and a radius of 5 cm. Draw another arc with B as the centre and a radius of 3.5 cm, intersecting the first arc at C.

5. Join BC and DC.

ABCD is our required rectangle.

8. Construct a rhombus whose side is of length 3.4 cm and one of its angles is 45°.

Ans:  Steps of construction:

1. Take a side AB = 3.4 cm.

2. At the points A and B, construct $\angle BAM = {45^ \circ }$ and $\angle TBP = {45^ \circ }$ respectively.

3. We will now cut off AD = 3.4 cm from AM and BC = 3.4 cm from BP.

4. Join AD, DC and BC.

ABCD is our required rhombus.

Sample Question: 

1. Construct an equilateral triangle if its altitude is 6 cm. Give justification for your construction.

Ans:  Steps of Construction:

1. Draw a line XY of a proper large point.

2. Mark any point D on the line XY we drew on step 1.

3. At point D, draw a perpendicular $\overline {DP}  \bot XY$ with the help of a ruler and a compass and then cut a segment DA = 6 cm on $\overline {DP} $.

4. At point A, construct the sides AB and AC which meet the line XY at points B and C respectively such that the angle $\angle DAB = \angle DAC = {30^ \circ }$.

Justification

Then the triangle $ABC$ is the required equilateral triangle because:

$\angle ABC = {180^ \circ } - \left( {{{90}^ \circ } + {{30}^ \circ }} \right) = {60^ \circ }$

$\angle ACB = {180^ \circ } - \left( {{{90}^ \circ } + {{30}^ \circ }} \right) = {60^ \circ }$

And the final angle 

\[\angle BAC = {30^ \circ } + {30^ \circ } = {60^ \circ }\]

Exercise 11.4

Construct each of the following and give justification:

1. A triangle if its perimeter is 10.4 cm and two angles are 45° and 120°.

Ans: Steps of construction:

1. Draw a line segment XY = 10.4 cm.

2. Draw the angle $\angle LXY = {45^ \circ }$ and $\angle MYX = {120^ \circ }$ with the help of a protractor.

3. Draw angle bisector of angle $\angle LXY = {45^ \circ }$.

4. Draw angles bisector of the angle $\angle MYX = {120^ \circ }$ such that it meets the angle bisector of $\angle LXY$ at point A.

5. Draw the perpendicular bisector of the line segment AX such that it meets XY at B.

6. Draw the perpendicular bisector of the line segment AY such that it meets XY at C.

Join AB and AC.

7. Thus, ABC is our required triangle.

2. A triangle PQR given that QR = 3 cm, ∠PQR = 45° and QP – PR = 2 cm.

Ans: Steps of construction:

1. Draw a ray OX and cut off a line segment QR = 3 cm from it.

2. At Q, construct an angle $\angle YQR = {45^ \circ }$ with the help of a protractor.

3. On QY, cut off a line segment of QS = 2 cm.

4. Join RS.

5. Draw the perpendicular bisector of RS to Meet QY at the point P.

6. Join PR. 

Then PQR is our required triangle.

3. A right triangle when one side is 3.5 cm and the sum of the other side and the hypotenuse is 5.5 cm.

Ans: In $\Delta ABC$ base BC = 3.5 cm, the sum of other side and hypotenuse i.e., AB + AC = 5.5 cm and the angle $\angle ABC = {90^ \circ }$

To construct: A triangle $ABC$

Steps of construction:

1. Draw a ray BX and from it cut off a line segment BC = 3.5 cm from it.

2. Construct $\angle XBY = {90^ \circ }$ with the help of a ruler and compass, or by the standard method of drawing a right-angle 

3. From BY we will cut off a line segment BD of length equal to 5.5 cm.

4. Join CD.

5. Draw the perpendicular bisector of the segment CD which will be intersecting BD at point A.

6. Join AC.

Then, ABC is our required triangle.

4. An equilateral triangle if its altitude is 3.2 cm.

Ans: Steps of Construction:

1. Draw a line $l$ of a proper large point.

2. Mark any point D on the line $l$ we drew on step 1.

3. At point D, draw a perpendicular $\overline {DX}  \bot l$ with the help of a ruler and a compass and then cut a segment DA = 3.2 cm on $\overline {DX} $.

4. At point A, construct the sides AB and AC which meet the line $l$ at points B and C respectively such that the angle $\angle DAB = \angle DAC = {30^ \circ }$.

Then the triangle $ABC$ is the required equilateral triangle because:

$\angle ABC = {180^ \circ } - \left( {{{90}^ \circ } + {{30}^ \circ }} \right) = {60^ \circ }$

$\angle ACB = {180^ \circ } - \left( {{{90}^ \circ } + {{30}^ \circ }} \right) = {60^ \circ }$

And the final angle 

\[\angle BAC = {30^ \circ } + {30^ \circ } = {60^ \circ }\]

5. A rhombus whose diagonals are 4 cm and 6 cm in lengths.

Ans:  To construct: A rhombus ABCD with diagonals 4 cm and 6 cm. 

Steps of construction:

1. First draw the diagonal AC = 6 cm.

2. We will draw BD the right-angle bisectors of the diagonal AC.

3. Now we will cut off MB = MD = 2 cm.

4. Then join AB, BC, CD and DA.

Hence, ABCD is our required rhombus.

Introduction to NCERT Exemplar

NCERT Exemplar for Class 9 Math Chapter 11- Constructions begins with the introduction of what Geometrical Constructions are. In this, the drawings are done only with the help of a non graduated ruler and a compass. The construction of the different triangles has been explained.

Benefits Of Solving NCERT Solutions For Class 9 Math

• Regular practice will make you stand out from the rest of the crowd as you will get familiar with the different kinds of problems

• Most of these problems come for examinations and so solving them will actually help you secure good marks

• These solutions will act as a swift mode of revision before assessments

• Time management gets worked upon and so solving them will ensure that you complete your assessments on time.