NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 10: Construction - Exercise 10.3
Question 2

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 \(B\).

Detailed Answer with Explanation:

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.

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 10: Construction – Exercise 10.3 | Detailed Answers