NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 10: Construction
Exercise 10.3

Constructions: Use ruler–compass methods. When a scale factor is given, construct a triangle similar to the given one using the ‘ray with equal parts and parallels’ technique.

Quick Links to Questions

Q. 1
Q. 2
Q. 3
Q. 4

Question. 1

1. Draw a line segment of length 7 cm. Find a point \(P\) on it which divides it in the ratio \(3:5\).

Answer

  1. Draw \(\overline{AB}\) with \(AB = 7\,\text{cm}.\)
  2. At \(A\), draw a ray \(AX\) making an acute angle with \(AB\).
  3. Mark eight equal steps \(A_1,A_2,\ldots,A_8\) on \(AX\) (since \(3+5=8\)).
  4. Join \(A_8\) to \(B\). Through \(A_3\) draw a line \(A_3P\,\parallel\,A_8B\) meeting \(AB\) at \(P\).
  5. Then \(P\) divides \(AB\) internally in the ratio \(3:5\).

Check (optional): \(AP = \dfrac{3}{8}\times 7 = 2.625\,\text{cm},\; BP = \dfrac{5}{8}\times 7 = 4.375\,\text{cm}.\)

Step by Step Solution

Step 1: We need to divide the line \(AB\) in the ratio \(3:5\). The total of the ratio is \(3+5 = 8\).

Step 2: By drawing 8 equal steps on the ray from \(A\), each step represents 1 part of the ratio.

Step 3: The point \(A_3\) corresponds to 3 parts from \(A\). When we join \(A_8\) to \(B\) and draw a parallel line from \(A_3\), the parallel line ensures similar triangles are formed.

Step 4: In similar triangles, the sides are in proportion. This makes the division on \(AB\) exactly the same as the ratio of steps (3 parts to 5 parts).

Step 5: Therefore, point \(P\) divides the line as required: \(AP : PB = 3 : 5\).

Why this works: The method of marking equal steps and drawing parallels is based on the concept of similar triangles. Parallel lines create equal ratios of sides, so the internal division comes out correctly.

Question. 2

2. Draw a right triangle \(ABC\) in which \(BC=12\,\text{cm},\; AB=5\,\text{cm}\) and \(\angle B=90^{\circ}\). Construct a triangle similar to it with scale factor \(\dfrac{2}{3}\). Is the new triangle also a right triangle?

Answer

  1. Draw \(\overline{AB}=5\,\text{cm}\) and \(\overline{BC}=12\,\text{cm}\) with \(AB\perp BC\). Join \(AC\) to get \(\triangle ABC\).
  2. From vertex \(B\), draw a ray \(BX\) making an acute angle with \(BA\).
  3. On \(BX\), mark three equal points \(B_1,B_2,B_3\).
  4. Join \(B_3\) to \(A\) and \(C\). Through \(B_2\) draw lines \(B_2A'\,\parallel\,B_3A\) and \(B_2C'\,\parallel\,B_3C\).
  5. \(\triangle AB C'\) or \(\triangle A'B C'\) (both constructed from \(B\)) is similar to \(\triangle ABC\) with scale factor \(\dfrac{2}{3}\).

Yes. Similarity preserves angles, hence the image of \(\angle B=90^{\circ}\) is also \(90^{\circ}\); the new triangle is right-angled at \(B\).

Step by Step Solution

Step 1: A scale factor of \(\tfrac{2}{3}\) means every side in the new triangle is reduced to \(\tfrac{2}{3}\) of the original length.

Step 2: We use the method of dividing a ray into equal parts. Here the denominator (3) tells us to mark 3 equal parts, and the numerator (2) tells us to stop at the 2nd division.

Step 3: By joining and drawing parallels, the sides shrink in the correct ratio. This ensures the triangles are similar.

Step 4: In similar triangles, all angles remain the same. Since \(\angle B\) was \(90^{\circ}\) in the original triangle, it remains \(90^{\circ}\) in the new triangle.

Therefore, the constructed triangle is also a right triangle.

Question. 3

3. Draw a triangle \(ABC\) in which \(BC=6\,\text{cm},\; CA=5\,\text{cm}\) and \(AB=4\,\text{cm}.\) Construct a triangle similar to it with scale factor \(\dfrac{5}{3}\).

Answer

  1. Construct \(\triangle ABC\) with the given side lengths.
  2. From vertex \(A\), draw a ray \(AX\) making an acute angle with \(AB\).
  3. On \(AX\), mark five equal points \(A_1,\ldots,A_5\).
  4. Join \(A_5\) to \(B\) and \(C\). Through \(A_3\) draw \(A_3B'\,\parallel\,A_5B\) and \(A_3C'\,\parallel\,A_5C\).
  5. Then \(\triangle AB'C'\sim\triangle ABC\) with scale factor \(\dfrac{5}{3}\) (enlargement).

Step by Step Solution

Step 1: We are asked to enlarge triangle \(ABC\) in the ratio \(5:3\). That means each side of the new triangle must be \(\dfrac{5}{3}\) times the side of the original triangle.

Step 2: To do this, we use the method of dividing a line into equal parts with the help of a ray and parallel lines.

Step 3: On the ray \(AX\), we marked 5 equal parts because the numerator of the ratio is 5.

Step 4: We join the 5th point (\(A_5\)) with \(B\) and \(C\). This gives the full size corresponding to '5' in the ratio.

Step 5: Since the ratio is \(\tfrac{5}{3}\), we take the 3rd division point (\(A_3\)) to represent the smaller triangle. By drawing parallels from \(A_3\), we ensure that the sides are reduced proportionally.

Step 6: The triangle \(AB'C'\) formed in this way is similar to the original triangle \(ABC\), and the sides are enlarged in the ratio \(\tfrac{5}{3}\).

Reason: This works because of the Basic Proportionality Theorem. Drawing parallels creates smaller triangles that are similar to the bigger triangle, keeping the ratio of corresponding sides exactly as required.

Question. 4

4. Construct a tangent to a circle of radius \(4\,\text{cm}\) from a point which is at a distance of \(6\,\text{cm}\) from its centre.

Answer

  1. Draw the circle with centre \(O\) and radius \(4\,\text{cm}\).
  2. Mark a point \(P\) such that \(OP=6\,\text{cm}\).
  3. Construct the midpoint \(M\) of \(OP\). With centre \(M\) and radius \(MO\), draw a circle; it meets the given circle at \(T\).
  4. Join \(PT\). Then \(PT\) is the required tangent. (Similarly, the other intersection gives the second tangent.)

Length (by right triangle \(\triangle OPT\)): \[ PT=\sqrt{OP^2-OT^2}=\sqrt{6^2-4^2}=\sqrt{20}=2\sqrt{5}\,\text{cm}. \]

Step by Step Solution

Step 1: We know a tangent touches the circle at exactly one point and makes a right angle (90°) with the radius at the point of contact.

Step 2: When we draw a circle with diameter \(OP\), by the property of a circle, the angle made at the circumference (here at point \(T\)) is a right angle, i.e., \(\angle OTP = 90^{\circ}\).

Step 3: Since \(\angle OTP = 90^{\circ}\), line \(PT\) is perpendicular to radius \(OT\). This means \(PT\) only touches the circle at \(T\) and does not cut through it.

Step 4: Therefore, \(PT\) is a tangent to the circle from the external point \(P\).

In simple words: We used the property of a circle drawn with diameter \(OP\) to locate the right angle at \(T\). This gave us the exact point where the tangent from \(P\) touches the circle.

Disclaimer:The questions are taken from NCERT Exemplar, which is the copyright of NCERT. The solutions provided here are prepared independently for educational purposes only. This material is not an official NCERT publication.