NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 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.

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

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Question. 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 \(B\).

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Question. 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).

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

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