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parallel to side MN below. O A B N We will show that ~. - PowerPoint PPT Presentation

D AY 81 P ROPORTIONAL TRIANGLES I NTRODUCTION In previous lessons, we have learned that a line parallel to one side of a triangle divides the other sides in equal proportions. In this lesson, we will use this idea to show how drawing such a


  1. D AY 81 – P ROPORTIONAL TRIANGLES

  2. I NTRODUCTION In previous lessons, we have learned that a line parallel to one side of a triangle divides the other sides in equal proportions. In this lesson, we will use this idea to show how drawing such a line results to a triangle which is similar to the original triangle.

  3. V OCABULARY Scalar This is any real number and not a vector or a matrix.

  4. Drawing a line parallel to one side of a triangle results to a triangle which is similar to the original triangle. Consider the βˆ†π‘π‘‚π‘ƒ with a line AB parallel to side MN below. O A B N 𝑁

  5. We will show that βˆ†π΅πΆπ·~βˆ†π‘π‘‚π‘ƒ. βˆ π‘π‘ƒπ‘‚ = βˆ π΅π‘ƒπΆ. Since βˆ π‘ƒ is shared. Now we show corresponding pairs of sides which include these angles are proportional. 𝑃𝑁 𝑃𝑂 That is, 𝑃𝐡 = 𝑃𝐢 Since 𝐡𝐢 βˆ₯ 𝑁𝑂 , then from triangle proportionality theorem, AB divides sides MO and NO proportionally. 𝐡𝑁 𝐢𝑂 Therefore, 𝑃𝐡 = 𝑃𝐢 … … (𝑗) Adding 1 to both sides of equation (𝑗) we get, 𝐡𝑁 𝐢𝑂 𝑃𝐡 + 1 = 𝑃𝐢 + 1 … … (𝑗𝑗) Equation (𝑗𝑗) can be written as, 𝐡𝑁 𝑃𝐡 𝐢𝑂 𝑃𝐢 𝑃𝐢 which can be written as below 𝑃𝐡 + 𝑃𝐡 = 𝑃𝐢 +

  6. 𝐡𝑁+𝑃𝐡 𝐢𝑂+𝑃𝐢 = … … 𝑗𝑗𝑗 𝑃𝐡 𝑃𝐢 Since 𝑃𝐡 + 𝐡𝑁 = 𝑃𝑁 and 𝑃𝐢 + 𝐢𝑂 = 𝑃𝑂 , we substitute these equations in equation (𝑗𝑗𝑗) to get, 𝑃𝑁 𝑃𝑂 𝑃𝐡 = 𝑃𝐢 Thus, the corresponding pairs of sides of the two triangles are proportional. Since the corresponding pairs of angles are equal and the corresponding pairs of sides are proportional then by S.A.S criterion βˆ†π΅πΆπ·~βˆ†π‘π‘‚π‘ƒ. This means that we multiply the length of each side of βˆ†π‘π‘‚π‘ƒ by the same scalar to get the length of the corresponding side of βˆ†π΅π‘ƒπΆ.

  7. Example Find the value of 𝑙 in the triangle below. 10 π‘—π‘œ 8 π‘—π‘œ 𝑙 10 π‘—π‘œ

  8. Solution 𝑙+10 10 10 = 8 8 𝑙 + 10 = 10 Γ— 10 8𝑙 + 80 = 100 8𝑙 = 20 𝑙 = 2.5 π‘—π‘œ

  9. HOMEWORK Find the value of π‘š in the diagram below 3 π‘—π‘œ π‘š 9 π‘—π‘œ 8 π‘—π‘œ

  10. A NSWERS TO HOMEWORK 2 π‘—π‘œ

  11. THE END

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