NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 10: Construction - Exercise 10.4
Question 2

Question. 2

Draw a parallelogram \(ABCD\) in which \(BC=5\,\text{cm}\), \(AB=3\,\text{cm}\) and \(\angle ABC=60^\circ\). Divide it into triangles \(BCD\) and \(ABD\) by diagonal \(BD\). Construct \(\triangle BD'C'\) similar to \(\triangle BDC\) with scale factor \(\dfrac{4}{3}\). Draw \(D'A'\parallel DA\) with \(A'\) on the extension of \(BA\). Decide whether \(A'BC'D'\) is a parallelogram.

Answer:

Final answer: Yes, \(A'BC'D'\) is a parallelogram.

Detailed Answer with Explanation:

Step-by-step Explanation (Beginner level)

  1. Start with given measurements: Draw side \(AB = 3\,\text{cm}\). At point \(B\), make an angle of \(60^\circ\). On this new line, mark point \(C\) such that \(BC = 5\,\text{cm}\).
  2. Complete parallelogram \(ABCD\): From \(A\), draw a line parallel to \(BC\). From \(C\), draw a line parallel to \(AB\). Let these two lines meet at point \(D\). Now \(ABCD\) is a parallelogram (because opposite sides are parallel).
  3. Draw diagonal: Join points \(B\) and \(D\) to draw diagonal \(BD\). This divides the parallelogram into two triangles: \(\triangle ABD\) and \(\triangle BCD\).
  4. Enlarge triangle \(BDC\): We need to construct another triangle \(BD'C'\) that is similar to \(BDC\) with scale factor \(\tfrac{4}{3}\). - Extend the ray \(BD\) beyond \(D\). - On this extension, locate point \(D'\) such that \(BD':BD = 4:3\). (You can do this using the basic proportionality method: draw an auxiliary ray from \(B\), mark 3 equal parts, then mark 4 equal parts, and connect back to \(D'\).)
  5. Find point \(C': Through \(D'\), draw a line parallel to \(DC\). Through \(B\), draw a line parallel to \(BC\). These two lines meet at \(C'\). Now \(\triangle BD'C'\) is similar to \(\triangle BDC\) with scale factor \(\tfrac{4}{3}\) (because they have the same angles).
  6. Locate point \(A': Through \(D'\), draw a line parallel to \(DA\). Extend line \(BA\) forward beyond \(A\). Let this parallel from \(D'\) meet the extended \(BA\) at point \(A'\).

Check if \(A'BC'D'\) is a parallelogram

  • In the original parallelogram, we know \(DC \parallel AB\) and \(DA \parallel BC\).
  • In the new figure, we constructed: - \(D'C' \parallel DC\), so \(D'C' \parallel AB\). Hence, \(A'B \parallel D'C'\). - \(D'A' \parallel DA\), so \(D'A' \parallel BC\).
  • Therefore, in quadrilateral \(A'BC'D'\), both pairs of opposite sides are parallel.

Conclusion: By definition, a quadrilateral with both pairs of opposite sides parallel is a parallelogram. So, \(A'BC'D'\) is indeed a parallelogram.

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NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 10: Construction – Exercise 10.4 | Detailed Answers