NCERT Solutions Class 10 Maths Chapter 11 – Constructions Exercise 11.1

NCERT Solutions for Class 10 Maths Chapter 11 – Constructions Exercise 11.1 are designed to help students develop a strong understanding of basic construction techniques in geometry. The exercise focuses on constructing triangles, bisecting angles, and drawing tangents to circles, which are foundational concepts in geometrical constructions.

Exercise 11.1 involves:

NCERT Solutions Class 10 Maths Chapter 11 – Constructions Exercise 11.1

  • Constructing a triangle when three sides are given (SSS criterion).

  • Bisecting an angle of a given angle in a triangle.

  • Constructing tangents to a circle from an external point.

The solutions are explained in a step-by-step manner, making it easier for students to understand the geometrical constructions involved and apply them effectively in Class 10 exams.

Q1. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.

Difficulty level: Easy
What is known/given?

  • Length of the line segment and the ratio to be divided.

What is unknown?

  • Construction

Reasoning:

  • Draw the line segment of the given length.

  • Then draw another line which makes an acute angle with the given line.

  • Divide the line into m + n parts, where m and n are the ratio given.

  • By the Basic Proportionality Theorem, "If a straight line is drawn parallel to a side of a triangle, then it divides the other two sides proportionally."

Solution:

  1. Draw the line segment AB = 7.6 cm.

  2. Draw ray AX, making an acute angle with AB.

  3. Mark 13 points on AX such that AA1 = A1A2 = A2A3 = ... = A12A13.

  4. Join BA13.

  5. Through A5 (since we need the ratio 5:8), draw line AC5 parallel to BA13, where C lies on AB.

  6. Thus, AC:CB = 5:8. We find AC = 2.9 cm and CB = 4.7 cm.

Q2. 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.

Difficulty level: Medium
What is known/given?

  • Sides of the triangle and the ratio of corresponding sides of two triangles.

What is unknown?

  • Construction

Reasoning:

  • Draw the line segment of the largest length (6 cm).

  • Measure 5 cm and 4 cm separately and cut arcs from the two ends of the line segment such that they cross each other at one point.

  • Then draw another line making an acute angle with the given line (6 cm).

  • Divide the line into (m + n) parts, where m and n are the ratio given.

  • By the Basic Proportionality Theorem, "If a straight line is drawn parallel to a side of a triangle, then it divides the other two sides proportionally."

Solution:

  1. Draw BC = 6 cm.

  2. With B and C as centers and radii 5 cm and 4 cm respectively, draw arcs to intersect at A.

  3. Draw ray BX making an acute angle with BC.

  4. Mark 3 points (B1, B2, B3) on BX such that BB1 = B1B2 = B2B3.

  5. Join BC3 and draw the line through B2 parallel to BC3 meeting BC at C’.

  6. Draw a line through C’ parallel to CA to meet BA at A’.

  7. Thus, triangle A'BC' is the required triangle similar to triangle ABC.

Q3. 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.

Difficulty level: Medium
What is known/given?

  • Sides of the triangle and the ratio of corresponding sides of two triangles.

What is unknown?

  • Construction

Reasoning:

  • Draw the line segment of the largest length (7 cm).

  • Measure 5 cm and 6 cm separately and cut arcs from the two ends of the line segment such that they cross each other at one point.

  • Then draw another line which makes an acute angle with the given line (7 cm).

  • Divide the line into m + n parts, where m and n are the ratio given.

  • By the Basic Proportionality Theorem, "If a straight line is drawn parallel to a side of a triangle, then it divides the other two sides proportionally."

Solution:

  1. Draw BC = 7 cm.

  2. With B and C as centers and radii 5 cm and 6 cm respectively, draw arcs to intersect at A.

  3. Draw ray BX making an acute angle with BC.

  4. Mark 7 points on BX such that BB1 = B1B2 = ... = B7.

  5. Join B5 to C and draw BC' parallel to BC5 intersecting the extension of BC at C’.

  6. Draw C’A’ parallel to CA to intersect BA at A’.

  7. Thus, triangle A'B'C' is the required triangle similar to triangle ABC.

Q4. Construct an isosceles triangle whose base is 8 cm and altitude is 4 cm, and then another triangle whose sides are 1.5 times the corresponding sides of the isosceles triangle.

Difficulty level: Medium
What is known/given?

  • Base and altitude of an isosceles triangle and the ratio of corresponding sides of two triangles.

What is unknown?

  • Construction

Reasoning:

  • Draw the line segment of the base (8 cm).

  • Draw the perpendicular bisector of the line and mark a point on the bisector 4 cm from the base.

  • Connect this point from both ends.

  • Then draw another line making an acute angle with the given line. Divide the line into m + n parts, where m and n are the ratio given.

  • By the Basic Proportionality Theorem, "If a straight line is drawn parallel to a side of a triangle, then it divides the other two sides proportionally."

Solution:

  1. Draw BC = 8 cm.

  2. Through D (mid-point of BC), draw the perpendicular bisector and cut an arc at D such that DA = 4 cm.

  3. Join BA and CA to form triangle ABC.

  4. Draw ray BX so that ∠CBX is acute.

  5. Mark 3 points on BX such that BB1 = B1B2 = B2B3.

  6. Join B2 to C and draw BC’ parallel to BC2.

  7. Draw C’A’ parallel to CA to meet BA at A’.

  8. Thus, triangle A'BC' is the required triangle similar to triangle ABC.

Q5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm, and ∠ABC = 60°. Then construct a triangle whose sides are 2/3 of the corresponding sides of triangle ABC.

Difficulty level: Medium
What is known/given?

  • Two sides and the angle between them and the ratio of corresponding sides of two triangles.

What is unknown?

  • Construction

Reasoning:

  • Draw the triangle with the given conditions.

  • Then draw another line making an acute angle with the base line. Divide the line into m + n parts, where m and n are the ratio given.

  • By the Basic Proportionality Theorem, "If a straight line is drawn parallel to a side of a triangle, then it divides the other two sides proportionally."

Solution:

  1. Draw BC = 6 cm.

  2. At B, make ∠CBY = 60° and mark A on BY such that BA = 5 cm.

  3. Join AC to form triangle ABC.

  4. Draw ray BX making an acute angle with BC.

  5. Mark 4 points (B1, B2, B3, B4) on BX such that BB1 = B1B2 = B2B3 = B3B4.

  6. Join B4 to C and draw BC' parallel to BC4.

  7. Draw C'A' parallel to CA to meet BA at A’.

  8. Thus, triangle A'BC' is the required triangle similar to triangle ABC.

Q6. Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, and ∠A = 105°. Then construct a triangle whose sides are 4/3 times the corresponding sides of triangle ABC.

Difficulty level: Medium
What is known/given?

  • One side and two angles of a triangle and the ratio of corresponding sides of two triangles.

What is unknown?

  • Construction

Reasoning:

  • Draw the triangle with the given conditions.

  • Then draw another line making an acute angle with the base line. Divide the line into m + n parts, where m and n are the ratio given.

  • By the Basic Proportionality Theorem, "If a straight line is drawn parallel to a side of a triangle, then it divides the other two sides proportionally."

Solution:

  1. Draw BC = 7 cm.

  2. At B, make ∠CBY = 45° and at C, make ∠BCZ = 30° to intersect at A.

  3. Thus, triangle ABC is constructed.

  4. Draw ray BX such that ∠CBX is acute.

  5. Mark 4 points (B1, B2, B3, B4) on BX such that BB1 = B1B2 = B2B3 = B3B4.

  6. Join B3 to C and draw BC' parallel to BC3.

  7. Draw C'A' parallel to CA to meet BA at A’.

  8. Thus, triangle A'BC' is the required triangle similar to triangle ABC.

FAQs: Class 10 Maths Chapter 11 – Constructions Exercise 11.1

Q1. What is the focus of Exercise 11.1?
Answer:
Exercise 11.1 focuses on geometrical constructions, such as constructing triangles using the SSS criterion, bisecting angles, and drawing tangents to a circle.

Q2. What does the SSS criterion for triangle construction mean?
Answer:
The SSS criterion states that a triangle can be constructed if the lengths of all three sides are given. Using the side lengths, you can construct the triangle accurately using a compass and ruler.

Q3. How do I bisect an angle in a triangle?
Answer:
To bisect an angle in a triangle:

  1. Place the compass at the vertex of the angle.

  2. Draw arcs to intersect both sides of the angle.

  3. Without changing the radius, draw arcs from the intersection points of the previous arcs.

  4. The point where these arcs intersect gives the angle bisector.

Q4. How do I construct a tangent to a circle from an external point?
Answer:
To construct a tangent from an external point

PP

to a circle with center

OO

:

  1. Draw a line from the center of the circle

    OO to the external point

    PP.

  2. Find the midpoint

    MM of

    OPOP and draw a perpendicular bisector from

    MM to

    OPOP.

  3. The line through this bisector will be the tangent to the circle at the point of contact.

Q5. How do NCERT Solutions help with exam preparation?
Answer:
NCERT Solutions provide clear, step-by-step construction methods that help students grasp the process of geometrical constructions efficiently. Regular practice of these solutions strengthens the understanding of construction techniques, ensuring students can solve construction problems in Class 10 exams.