Worksheets For Class 9 Mathematics Constructions

Constructions Class 9 Worksheets have been designed as per the latest pattern for CBSE, NCERT and KVS for Grade 9. Students are always suggested to solve printable worksheets for Mathematics Constructions Grade 9 as they can be really helpful to clear their concepts and improve problem solving skills. We at worksheetsbag.com have provided here free PDF worksheets for students in standard 9 so that you can easily take print of these test sheets and use them daily for practice. All worksheets are easy to download and have been designed by teachers of Class 9 for benefit of students and is available for free download.

Mathematics Constructions Worksheets for Class 9

We have provided chapter-wise worksheets for class 9  Mathematics Constructions which the students can download in Pdf format for free. This is the best collection of  Mathematics Constructions standard 9th worksheets with important questions and answers for each grade 9th  Mathematics Constructions chapter so that the students are able to properly practice and gain more marks in Class 9  Mathematics Constructions class tests and exams.

Basic Constructions
•
Some Constructions of Triangles
(1) Use only ruler and compass while drawing constructions.
(2) Protractor may be used for drawing non-standard angles.
(3) Constructions of a triangle given its base, a base angle and the difference of the other two sides.
(4) Constructions of a triangle given its perimeter and its two base angles.
(5) A triangle can be constructed if its perimeter and two base angles are given.
(6) Geometrical construction is the process of drawing a geometrical figure using only two instruments-an ungraduated ruler and a pair of compasses.
(7) Some specific angles like 15°,30°,45°,60°,75°,90° , etc. can be constructed without using protractor.
(8) A triangle can be constructed if its base, base angle and the sum of the two sides or the difference of the other two sides are given.

Question. 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) 67.5°

B

Question. 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

D

Question. 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

A

Question. A triangle ABC can be constructed in which ∠B = 60°, ∠C = 45° and AB + BC + AC = 12 cm.
Answer. The given statement is true, because ∠B + ∠C = 600 + 450 =1050 <1800

Question. A triangle ABC can be constructed in which AB = 5 cm, ∠A = 45° and BC + AC = 5 cm.
Answer. Since sum of two sides of a triangle is always greater than the third side, so we cannot construct a triangle in which AB = BC + AC.

Question. An angle of 52.5° can be constructed.
Answer. Since, 52.50 = 1/4 X 2100 and 210o = 180o + 30o which can be constructed.
Hence, the given statement is correct.

Question. A triangle ABC can be constructed in which BC = 6 cm, ∠C = 30° and AC – AB = 4 cm.
Answer. Because AC – AB (= 4 cm) < BC (= 6cm). i.e., AC < AB + BC or AB + BC > AC which is true.

Question. An angle of 42.5° can be constructed.
Answer. Since, 42.50 = 1/4 X 850 and 85o cannot be constructed by using ruler and compass.

Question. A triangle ABC can be constructed in which ∠B = 105°, ∠C = 90° and AB + BC + AC = 10 cm.
Answer. The given statement is false, because ∠B +∠C =1050 + 900 =1950 >1800

Question. Draw an angle of 110° with the help of a protractor and bisect it. Measure each angle.
Given: An angle ABC = 110o.
Required: To draw the bisector of ∠ABC

Steps of construction:
1. With B as centre and a convenient radius draw an arc to intersect the ray’s BA and BC at P and Q respectively.
2. With centre P and a radius greater than half of PQ, draw an arc.
3. With Centre Q and the same radius (as in step 2), draw another arc to cut the previous arc at R.
4. Draw ray BR. This ray BR is the required bisectors of ÐABC.

Question. Construct a triangle whose sides are 3.6 cm, 3.0 cm and 4.8 cm. Bisect the smallest angle and measure each part.
Step 1: Draw a line AB = 4.8 cm.
Step 2: Now, take raidus of 3 cm and centre ‘A’ draw an arc. And take radius of 3.6 cm and centre ‘B’ draw an arc that intersect our previous arc at ‘C’.
Step 3: Join CA and CB we get required triangle ABC.

Now, we measure all internal angles and we get ∠ABC is smallest angle,
So, we bisect ∠ABC
Step 4: Take any radius (less than half of AB) and centre ‘B’ draw an arc that intersect our line AB at P and line BC at Q.
Step 5: With same radius and centre ‘P’ and ‘Q’ draw arc which intersect at ‘R’.
Step 6: Join BR and extended BR that line intersects AC at ‘D’.
Now, we can easily measure each other angle with the help of protractor.

Question. Construct a triangle ABC in which BC = 5 cm, ∠B = 60° and AC + AB = 7.5 cm.
Given: In ΔABC , BC = 5 cm, AC + AB = 7.5 cm and ∠B = 600
Required: To construct ΔABC

Steps of construction:
1. Draw a ray BX and cut off a line segment BC = 5 cm from it.
2. At B, construct 0 ÐXBY = 60 .
3. With B as centre and radius = 7.5 cm, draw an arc at meet BY at D.
4. Join CD.
5. Draw the perpendicular bisector of CD, intersecting BD at A.
6. Join AC. Then, ABC is the required triangle.

Question. 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?
Answer. Given: A line segment AB of length 4cm.
Required: To draw perpendicular to AB through A and B, respectively.
Steps of construction:
1. Draw AB = 4 cm.
2. With A as centre and any convenient radius, draw an arc, cutting Ab at P.
3. With P as centre and the same radius, draw an arc cutting the arc drawn in step 2 at Q.

4. With Q as centre and the same radius as in steps 2 and 3, draw an arc, cutting the arc drawn in step 3 at R.
5. With Q as centre and the same radius, draw an arc.
6. With R as centre and the same radius, draw an arc, cutting the arc drawn in step 5 at X.
7. Draw OX and produced it to C and D.
8. Now, repeat the steps from 2 to 7 to draw the line EF perpendicular through B.
Yes, these lines are parallel because sum of the interior angles on the same side of the transversal is 180o.

Question. Draw an angle of 80° with the help of a protractor. Then construct angles of (i) 40° (ii) 160° and (iii) 120°.
1. Draw a ray OA.
2. With the help of a protractor, construction ÐBOA = 800
3. Taking O as centre and any suitable radius, draw an arc to intersect rays OA and OB at points P and Q respectively.

4. Bisect ∠BOA as done in Q1. Let ray OC be the bisector of ∠BOA, then ∠ROA = 1/2 ∠BOA = 1/2 × 80 = 400
5. With Q as centre and radius equal to PQ, draw an arc to cut the extended arc PQ at R. Join OR and produce it to form ray OD, then ∠DOA = 2∠BOA = 2 × 800 = 400
6. Bisect ∠DOB as in Q1. Let OE be the bisector of ∠DOB is then ∠EOA = ∠EOB + ∠BOA = 1/2 ∠DOB + ∠BOA = 1/2 (80)0 +800 + 400 + 800 = 1200

Question. Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm.
1. Take AB = 5cm.
2. Draw AY ⊥ AB.
3. With A as centre and radius = 3.5cm, describe an arc cutting AY at D.

4. With D as centre and radius 5cm, describe an arc with B as centre and radius 3.5cm, describe another arc intersecting the first arc at C.
5. Join BC and DC. ABCD is the required rectangle.

Question. Construct a rhombus whose side is of length 3.4 cm and one of its angles is 45°.
1. Take AB = 3.4 cm.
2. At A and B, construct 0 ∠BAM = 450 and ∠TBP = 450 respectively.
3. From AM cut off AD = 3.4 cm and from BP cut Off BC = 3.4 cm.
4. Join

4. Join AD, DC and BC. ABCD is the required rhombus.

Question. Construct a square of side 3 cm.
1. Take AB = 3 cm.
2. At A, draw AY ⊥ AB.
3. With A as centre and radius = 3cm, describe an arc cutting AY at D.

4. With B and D as centres and radii equal to 3 cm, draw arc intersecting at C.
5. Join BC and DC. ABCD is the required square.

Question. A triangle if its perimeter is 10.4 cm and two angles are 45° and 120°.
Steps of construction:

1. Draw XY = 10.4 cm.
2. Draw ∠LXY = 450 and ∠MYX =1200
3. Draw angle bisector of ∠LXY.
4. Draw angles bisector of ÐMYX such that it meets the angle bisector of ∠LXY at point A.
5. Draw the perpendicular bisector of AX such that it meets XY at B.
6. Draw the perpendicular bisector of AY such that it meets XY at C.
7. Join AB and AC.
Thus, ABC is the required triangle.

Question. A right triangle when one side is 3.5 cm and sum of other sides and the hypotenuse is 5.5 cm.
Answer. In ΔABC, base BC = 3.5 cm, the sum of other side and hypotenuse i.e., AB + AC = 5.5 and ∠ABC = 900
Required: Construct the ΔABC.
Steps of construction:
1. Draw a ray BX and cut off a line segment BC = 3.5 cm from it.
2. Construct ∠XBY = 900
3. From BY cut off a line segment BD = 5.5 cm.

4. Join CD.
5. Draw the perpendicular bisector of CD intersecting BD at A.
6. Join AC. Then, ABC is the required triangle.

Question. An equilateral triangle if its altitude is 3.2 cm.
1. Draw a line l.
2. Mark any point D on the line l.

3. At point D, draw DX ⊥ l and cut DA = 3.2 cm from DX
4. At the point A, construct AB and AC which meets the l at points B and C respectively such that ∠DAB = 300 and ∠DAC = 300 Then ΔABC is the required equilateral triangle because ∠ABC =1800 − (900+ 300 ) = 600
∠ACB =180 − (900 + 300) = 600
And ∠BAC = 300 + 300 = 600 .

Question. A triangle PQR given that QR = 3cm, ∠PQR = 45° and QP – PR = 2 cm.
Answer. 1. Draw a ray OX and cut off a line segment QR = 3 cm.
2. AT Q, construction ∠YQR = 45o.
3. From QY, cut off QS = 2 cm.

4. Join RS.
5. Draw perpendicular bisector of RS to Meet QY at P.
6. Join PR. Then PQR is the required triangle.

Question. A rhombus whose diagonals are 4 cm and 6 cm in lengths.
1. Take AC = 6 cm.
2. Draw BD the right bisectors of AC.

3. Cut off MB = MD = 2 cm.
4. Join AB, BC, CD and DA.
Hence, ABCD is the required rhombus.

Mathematics Constructions Worksheets for Class 9 as per CBSE NCERT pattern

Parents and students are welcome to download as many worksheets as they want as we have provided all free. As you can see we have covered all topics which are there in your Class 9 Mathematics Constructions book designed as per CBSE, NCERT and KVS syllabus and examination pattern. These test papers have been used in various schools and have helped students to practice and improve their grades in school and have also helped them to appear in other school level exams. You can take printout of these chapter wise test sheets having questions relating to each topic and practice them daily so that you can thoroughly understand each concept and get better marks. As Mathematics Constructions for Class 9 is a very scoring subject, if you download and do these questions and answers on daily basis, this will help you to become master in this subject.

Benefits of Free Constructions Class 9 Worksheets

1. You can improve understanding of your concepts if you solve NCERT Class 9 Mathematics Constructions Worksheet,
2. These CBSE Class 9 Mathematics Constructions worksheets can help you to understand the pattern of questions expected in Mathematics Constructions exams.
3. All worksheets for Mathematics Constructions Class 9 for NCERT have been organized in a manner to allow easy download in PDF format
4. Parents will be easily able to understand the worksheets and give them to kids to solve
5. Will help you to quickly revise all chapters of Class 9 Mathematics Constructions textbook
6. CBSE Class 9 Mathematics Constructions Workbook will surely help to improve knowledge of this subject

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