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• okay. let the "remained height" as x.
  So, the height of larger triangle = 5+x

  Thus,
  Area of the bigger triangle = 0.5*20*(5+x) = 50+10x
  Area of smaller triangle (unshaded) = 0.5*20*x = 10x
  Subtract the bigger triangle with the smaller triangle, which gives us:
  50+10x-10x = 50 cm^2

  • what i mean with the height of larger triangle is DE,
    "remained height" is CD.

    sorry, not paying attention to the question clearly.

    • Revised:
      okay. let the CD as x.
      So, DE = 5+x

      Thus,
      Area of the triangle AEB = 0.5*20*(5+x) = 50+10x
      Area of unshaded triangle (triangle ACB)= 0.5*20*x = 10x
      Subtract the triangle AEB with triangle ACB, which gives us:
      50+10x-10x = 50 cm^2

      • hehehe...
        Okay Josh.......the award goes to youuuu :) for question number 2

• Question 1:

  Area ABCD = Area ADC + Area ABC
  8 x 6 = 0.5 x MD x 10 + 0.5 x BN x 10
  48 = 5MD + 5BN
  (MD = BN, since ADC congruent to ABC by SSS, they both has the same base, and equally add up to the whole rectangle)
  48 = 10 MD
  MD = 4.8 = BN

  MC =square root (DC squared - DM squared) = sqrt (64 - 23.04) = sqrt 40.96 = 6.4
  NC = square root (BC squared - BN squared) = sqrt (36 - 23.04) = sqrt 12.96 = 3.6

  MN = MC - NC = 6.4 - 3.6 = 2.8

  Question 2:

  (Sedang dikerjakan)

  • Excellent!! and also you can use the formulas obtained from similar triangles that says BC^2 = CN x AC and AB^2 = AN x AC.
    And the award goes to youuuu :) for question number 1