Triangle and Rectangle

Question 1

On the rectangle below, the dimensions are 8 cm x 6 cm. BN and DM are perpendicular to diagonal AC. Calculate the length of MN.

Question 2

Look at the diagram below. The length of AB is 20 cm and the length of CE is 5 cm. Calculate the shaded area.

6 thoughts on “Triangle and Rectangle”

  1. 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

    1. what i mean with the height of larger triangle is DE,
      “remained height” is CD.

      sorry, not paying attention to the question clearly.

      1. 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

  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)

    1. 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

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