Constructions Class 10 Important Questions | NCERT Maths Chapter-11

Constructions Class 10 Important Questions

The start of class 10 marks the beginning of the foundation for class 11 and class 12. It is very important study the basics in order to understand each and every chapter properly. In this page, we have provided all the important question for cbse class 9 that could be asked in the examination. Students also need to study the ncert solutions for class 10 in order to gain more knowledge and understanding the lessons. Questions and Answers are way to learn the new things in a proper way. NCERT textbooks downloads for class 9 in pdf are also available for the students if they need more help. By downloading these books, they can study from it. Our experts also prepared revision notes for class 9 so that students should see the details of each and every chapters. Class 9 important questions are the best to revise all the chapters in the best way.

Question 1.
Draw two concentric circles of radii 3 cm and 5 cm. Construct a tangent to smaller circle from a point on the larger circle. Also measure its length
Solution:

Now after measuring, PA and PB comes out to be 4 cm.
Steps of construction of tangents:

1. Take point O. Draw 2 concentric circles of radii 3 cm and 5 cm respectively.
2. Locate point P on the circumference of larger circle.
3. Join OP and bisect it. Let M be mid-point of OP.
4. Taking M as centre and MP as radius, draw an arc intersecting smaller circle at A and B.
5. Join PA and PB. Thus, PA, PB are required tangents

Question 2.
Construct a triangle ABC in which BC = 6 cm, AB = 5 cm and∠ABC = 60°. Then construct another triangle whose sides are 3/4 times the corresponding sides of ΔABC.
Solution:

Steps of construction:

1. Draw ΔABC with side BC = 6 cm, AB = 5 cm, ∠ABC = 60°.
2. Draw ray BX making an acute angle with BC on opposite side of vertex A.
3. Locate 4 points P1 P2, P3, P4 on line segment BY.
4. Join P4C and draw a line through P3, parallel to P4C intersecting BC at C’.
5. Draw a line through C’ parallel to AC intersecting AB at A’. ΔA’BC’ is the required triangle

Question 3.
Draw a triangle ABC with BC = 7 cm, ∠B = 45° and ∠A = 105°. Then construct a triangle whose sides are 4/5 times the corresponding sides of ΔABC.
Solution:

Given, ∠B = 45°,∠A = 105°
Sum of all interior angles in Δ = 180°
∠A +∠B + ∠C = 180°
∠C = 30°
Steps of construction:

1. Draw ΔABC with side BC = 7 cm,∠B = 45°, ∠C = 30°.
2. Draw a ray BX making an acute angle with BC on opposite side of vertex A.
3. Locate 5 points P1, P2, P3, P4, P5 on BZ.
4. Join P5C. Draw line through P4 parallel to P5C intersecting BC at C’.
5. Through C’, draw line parallel to AC intersecting AB at A’. ΔA’BC’ is the required triangle

Question 4.
Draw a circle of radius 4 cm. Draw two tangents to the circle inclined at an angle of 60° to each other
Solution:

Steps of construction:
1. Draw a circle of radius 4 cm with centre O.
2. Take point A on circle. Join OA.
3- Draw line AP perpendicular to radius OA.
4. Draw ∠AOB = 120° at O.
5. Join A and B at P, to get 2 tangents. Here ∠APB = 60°.

Question 5.
Draw an isosceles ΔABC in which BC = 5.5 cm and altitude AL = 3 cm. Then construct another triangle whose sides are  3/4 of the corresponding sides of ΔABC
Solution:

Steps of construction:

1. Draw BC = 5.5 cm.
2. Construct AP the perpendicular bisector of BC meeting BC at L.
3. Along LP cut off LA = 3 cm.
4. Join BA and CA. Then ΔABC so obtained is the required ΔABC.
5. Draw an acute angle CBY and cut 4 equal lengths as BA1 = A1A2 = A2A3 = A3A4 and join CA4.
6. Now draw a line through A3 parallel to CA4 intersecting BC at C’.
7. Draw a line through C’ and parallel to AC intersecting AB at A’. BA’C’ is the required triangle.

Question 6.
Draw a triangle with sides 5 cm, 6 cm and 7 cm. Then draw another triangle whose 4/5 sides are y of the corresponding sides of first triangle
Solution:

1. Draw a line segment AB of length 7 cm.
   Then using A as centre and distance 5 cm draw an arc C.
   Also draw an arc using B as centre and with distance 6 cm, which intersect earlier drawn arc at C. Join AC and BC.
2. Draw an acute angle BAZ and cut AZ as AA1 = A1A2 = A2A3 = A3A4 = A4A5 and join BA5.
3. Through A4 draw a line parallel BA5 intersecting AB at B’.
4. Through B’ draw a line parallel to BC intersecting AC at C’. AAB’C’ is the required triangle.

Question 7.
Draw a ΔABC in which AB = 4 cm, BC = 5 cm and AC = 6 cm. Then construct another triangle whose sides are 3/5 of the corresponding sides of ΔABC
Solution:

Steps of construction:

1. Draw a line segment AC = 6 cm.
2. Draw an arc with A as centre and radius equal to 5 cm.
3. Draw an arc with C as centre and radius equal to 4 cm intersecting the previous drawn arc at B.
4. Join AB and CB, then ΔABC is required triangle.
5. Below AC make an acute angle CAX.
6. Along AX mark of 5 points A1, A2, A3, A4, A5 such that AA1= A1A2 = A2A3= A3A4 = A4A5.
7. Join A5C.
8. From A3 draw A3D | | A3C meeting AC at D.
9. From D, draw ED | | BC meeting AB at E. Then we have ΔEDA which is the required triangle.

Question 8.
Draw a triangle with sides 4 cm, 5 cm and 6 cm. Then construct another triangle whose sides are 2/5 of the corresponding sides of given (first) triangle
Solution:

Steps of construction:

1. Draw a line segment BC = 6 cm.
2. Draw an arc with B as centre and radius equal to 5 cm.
3. Draw an arc with C as centre and radius equal to 4 cm intersecting the previous one at A.
4. Join AB and AC, then ΔABC is the required triangle.
5. Below BC, make an acute angle CBX.
6. Along BX, mark off 5 points B1, B2, B3, B4, B5 such that BB1 = B1B2 = B2B3 = B3B4 = B4B5.
7. Join B5C.
8. From B2, draw B2D ¦ ¦B5C, meeting BC at D.
9. From D, draw ED ¦ ¦ AC, meeting BA at E. Then we have ΔEDB which is the required triangle.

Question 9.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 8 cm and 6 cm. Then construct another triangle whose sides are 3/4 times the sides of first triangle
Solution:

Steps of construction:

1. Draw line BC = 8 cm then at B draw a line making angle of 90°.
2. Cut a length of 6 cm and name it A. Join AC. ΔABC is the right triangle.
3. Below BC make an acute angle ∠CBX.
4. Along BX mark off 4 points B1, B2, B3, B4 such that BB1=B1B2 = B2B3 = B3B4.
5. Join B4C.
6. From B3 draw B3D ¦ ¦ B4C meeting BC at D.
7. From D draw ED 11 AC meeting BA at E. Now we have ΔEBD which is the required triangle whose sides are 3/4 of the corresponding sides of ΔABC

2015

Long Answer Type Questions [4 Marks]

Question 10.
Construct a triangle ABC with BC = 7 cm, ∠B = 60° and AB = 6 cm. Construct another triangle whose sides are 3/4 times the corresponding sides of ΔABC
Solution:

Steps of construction:

1. Draw a line segment BC = 7 cm.
2. Draw ∠B = 60° at point B. Thus∠XBC = 60°
3. Take an arc of 6 cm, with B as centre mark an arc on BX to get point A.
4. Join AC.
5. ΔABC is constructed triangle.
6. Draw an acute angle CBY below BC.
7. Take points P1, P2, P3, P4, at BY such that BP1 = P1P2=P2P3=P3P4
8. Join P4C with line.
9. Draw a line parallel to P4C through the point P3 which intersects BC at C’.
10. Join P3C’ with line
11. Draw a line parallel to AC through the point C’ which intersects AB at point A’. ΔA’BC’ is the required triangle whose sides are 3/4 times the corresponding sides of ΔABC.

Question 11.
Construct an isosceles triangle whose base is 6 cm and altitude 4 cm. Then construct another triangle whose sides are 3/4 times the corresponding sides of the isosceles triangle.
Solution:

Steps of construction:

1. Draw a line segment BC = 6 cm.
2. Draw perpendicular bisector of BC which intersects BC at point D.
3. Take an arc of 4 cm, with D as centre mark on arc ⊥ bisector as point A.
4. Join AB and AC. ΔABC is constructed isosceles Δ.
5. Draw an acute angle CBY below BC.
6. Take points P1,P2,P3, P4 at BY such that BP1 = P1P2 = P2P3=P3P4
7. Join P4C with dotted line.
8. Draw a line parallel to P4C through the point P3 which intersects BC at C’.
9. Join P3C’ with dotted line.
10. Draw a line parallel to AC through the point C’ which intersects AB at A’.
    ΔABC is the required triangle whose sides are 3/4 times the corresponding sides of ΔABC.

Question 12.
Draw a line segment AB of length 7 cm. Taking A as centre, draw a circle of radius 3 cm and taking B as centre, draw another circle of radius 2 cm. Construct tangents to each circle from the centre of the other circle
Solution:

Required tangents are

1. BP and BQ
2. AR and AS.
   Steps of construction:

• Draw AB = 7 cm. Taking A and B as centres, draw two circles of 3 cm and 2 cm radius.
• Bisect line AB. Let mid-point of AB be C.
• Taking C as centre, draw circle of AC radius which will intersect circles at P, Q, R, S. Join BP, BQ, AR, AS

Question 13.
Construct a ΔABC in which AB = 6 cm, ∠A = 30° and ∠B = 60°. Construct another ΔAB’C’ similar to ΔABC with base AB’ = 8 cm
Solution:

Steps of construction:

1. Draw AB = 6 cm and make an angle of 60° at point B and 30° at point A.
2. Make any acute angle BAX at point A.
3. Cut four arcs A1,A2, A3, A4 on line AX such that AA1=A1A2=A2A3=A3A4
4. Join B to A3.
5. Draw line from A4 parallel to A3B cutting AB extended to B’.
6. Draw line from B’ | | BC cuts AC at C’.

Question 14.
Construct a right triangle ABC with AB = 6 cm, BC = 8 cm and ∠B = 90°. Draw BD, the perpendicular from B on AC. Draw the circle through B, C and D and construct the tangents from A to this circle
Solution:

Thus, AP and AB are the required tangents
Steps of construction:

1. Draw BC = 8 cm, ∠B = 90°.
2. Take an arc of 6 cm, with B as centre, mark an arc on point A. Join AB.
3. Draw BD ⊥ AC. Bisect line BC at E as mid-point of BC.
4. Taking E as centre and EC as its radius, draw circle which will intersect AC at D. Join BD.
5. Mark point P on circle. Join A to P.

Question 15.
Construct a triangle ABC in which AB = 5 cm, BC = 6 cm and ∠ABC = 60°. Now construct another triangle whose sides are 5/7  times the corresponding sides of ΔABC.
Solution:
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Steps of construction:

1. Draw BC = 6 cm.
2. Take any radius (less than half of BC) and centre B, draw an arc intersecting BC at P. With same radius and centre P, draw another arc intersecting previous arc at Q.
3. Join BQ, extend it to D.
4. Take radius = 5 cm and centre B, we draw arc intersecting BD at A. Join AC, get ΔABC.
5. Draw line BX, as ∠CBX is any acute angle. Draw 7 equal radius arcs on line BX intersecting at B1, B2, B3, B4, B5, B6, B7 as BB1=B1B2=B2B3=B3B4=B4B5=B5B6=B6B7
6. Join B7 to C. Draw line from B5 as parallel to B7C intersecting BC at C’.
7. Draw line from C’as parallel to AC intersecting AB at A’.

Question 16.
Draw a circle of radius 3 cm. From a point P, 7 cm away from its centre draw two tangents to the circle. Measure the length of each tangent.
Solution:

We know, radius perpendicular to tangent with OA = 3 cm, OP = 7 cm
In right ΔOAP, (OP)² = (OA)² + (PA)²
PB = PA = 2√10 cm
Steps of construction:

1. Take point O as centre, draw circle of radius 3 cm. Locate point P, 7 cm away from its centre O. Join OP.
2. Bisect OP. Let Q be mid-point of PO.
3. Taking Q as centre and QO as radius, draw a circle.
4. Let this circle intersect previous circle at A and B.
5. Join AP, BP which are required tangents.
   .’. AP = 6.3 cm (approx.)

Question 17.
To a circle of radius 4 cm, draw two tangents which are inclined to each other at an angle of 60°.
Solution:
Refer to Ans 4.

Question 18.
Draw a circle of radius 3.5 cm. Draw two tangents to the circle which are perpendicular to each other
Solution:

Steps of construction:

1. Draw a circle of radius 3.5 cm with centre O.
2. Take point A on circle. Join OA.
3. Draw perpendicular to OA at A.
4. Draw radius OB, making an angle of 90° with OA.
5. Draw perpendicular to OB at point B. Let these perpendiculars intersect at C.
   Hence, CA and CB are required tangents inclined at angle of 90°.

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