Question 1 - Page 72 - Sierpinski Gasket - Chapter 4 Class 8 - Exploring Some Geometric Themes (Ganita Prakash II

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Question 1 - Page 72 Show that by joining the midpoints of an equilateral triangle, we divide it into 4 identical equilateral triangles. [Hint: Note that the corner triangles are isosceles.] STEP 1: Triangle BED has 2 equal sides (BD = BE, midpoints). It is isosceles. STEP 2: Angle B is . In an isosceles triangle, angles opposite equal sides are equal. STEP 3: Therefore, angles BED and BDE are also . Triangle BED is equilateral. Conclusion: All 4 small triangles are identical equilateral triangles. Okay, let’s do this Start with a large equilateral triangle. All its angles are 〖𝟔𝟎〗^∘. Mark the midpoints of each of the three sides. Look at one of the corner triangles formed. Because you used midpoints, the two sides forming the corner point are equal in length. This makes the corner triangle an isosceles triangle. An isosceles triangle with a 60^∘ angle between its equal sides must have its other two angles equal to (180^∘−60^∘)/2=60^∘. Since all three angles are 60^∘, the corner triangle is an equilateral triangle with a side length exactly half of the original triangle. This applies to all three corner triangles. The shape left in the very center is bounded by the bases of these three identical corner triangles, meaning its sides are also exactly the same length. Therefore, you have 4 identical equilateral triangles!