Question 15 - End-of-Chapter Exercises - Chapter 6 Class 9 - Measuring Space: Perimeter and Area (Ganita Manjar

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Question 15 (i) Three problems about fitting congruent shapes together: (i) Rectangle ABCD has sides 𝑎,𝑏, and rectangle PQRS has sides 2𝑎,2𝑏. Show that PQRS has 4 times the area of ABCD . Does this mean that 4 copies of rectangle ABCD will fit into rectangle PQRS? Check and see Now, Area of ABCD =𝑎 × 𝑏=𝒂𝒃 Area of PQRS =2𝑎 × 2𝑏 =𝟒𝒂𝒃 Thus, Area of PQRS = 4 × Area of ABCD Can they fit? If we draw a vertical line down the middle and a horizontal line across the middle of PQRS, you create a 2 × 2 grid. Exactly 4 copies of 𝑨𝑩𝑪𝑫 will fit perfectly inside. Let’s look at it in detail Step 1 of 5 The Rectangles Area of ABCD=a×b=ab. Let's double the sides to create PQRS on the right. Step 2 of 5 The Scaled Rectangle Area of PQRS . Yes, 4ab is exactly 4 times larger. Step 3 of 5 Can they fit? Yes. If you draw a vertical line down the middle and a horizontal line across the middle of PQRS...Step 4 of 5 The Grid ...you create a grid! Let's generate 4 identical copies of on the left to test this grid. Step 5 of 5 Check and see! Exactly 4 copies of ABCD will fit perfectly inside!Question 15 (ii) Three problems about fitting congruent shapes together: (ii) △ABC has sides 𝑎,𝑏,𝑐, and △PQR has sides 2𝑎,2𝑏,2𝑐. Show that △PQR has 4 times the area of △ABC. Does this mean that 4 copies of △ABC will fit into △PQR ? Check and see! Finding Area using Heron's formula Area ∆ ABC = √(𝒔(𝒔−𝒂)(𝒔−𝒃)(𝒔−𝒄)) For ∆ PQR Since each side is doubled, the semi-perimeter is also doubled (𝟐𝒔) Let’s put values in Heron's formula: Area ∆ PQR = √(𝟐𝒔(𝟐𝒔−𝟐𝒂)(𝟐𝒔−𝟐𝒃)(𝟐𝒔−𝟐𝒄)) = √(2𝑠 × 2(𝑠−𝑎) × 2(𝑠−𝑏) × 2(𝑠−𝑐)) = √(2 × 2 × 2 × 2 × 𝑠(𝑠−𝑎)(𝑠−𝑏)(𝑠−𝑐) ) = √(4 × 4 × 𝑠(𝑠−𝑎)(𝑠−𝑏)(𝑠−𝑐) ) = √(4^2 × 𝑠(𝑠−𝑎)(𝑠−𝑏)(𝑠−𝑐) ) = √(4^2 × 𝑠(𝑠−𝑎)(𝑠−𝑏)(𝑠−𝑐) ) You can factor a 2 out of every single parenthesis, which gives you 16 inside the square root ( √16=4 ). The area is exactly 𝟒 times larger. Can they fit? Yes. If you find the midpoints of all three sides of △𝑃𝑄𝑅 and draw lines connecting them, you will break the large triangle into a perfect tessellation of exactly 𝟒 small △𝐴𝐵𝐶 copies! Step 1 of 5 Doubling the Triangle (Sides 2a, 2b, 2c) If we use Heron's formula on the small triangle (semi-perimeter s ), the area is: √((s(s-a)(s-b)(s-c))). Step 2 of 5 The Doubled Semiperimeter For the large triangle, every side is doubled, so the semi-perimeter is also doubled (2s). If you plug this into Heron's formula: √((2s(2s-2a)(2s-2b)(2s-2c)) Step 3 of 5 Factoring the Equation You can factor a 2 out of every single parenthesis, which gives you 16 inside the square root . The area is exactly 4 times larger. Step 4 of 5 Can they fit? Yes. If you find the midpoints of all three sides of and draw lines connecting them...Step 5 of 5 Check and see! ...you will break the large triangle into a perfect tessellation of exactly 4 small copies!