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Given a rectangle UVWX with UV = 10 cm and VX = 6 cm. Point Y is on UV such that VY = 4 cm. Point Z is on UW such that UZ = 2 cm. Calculate the lengths of c (hypotenuse of VYX), a (hypotenuse of UZY),

Mathematics

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Answers

3 parts
a

7.2 cm\text{7.2 cm}

b

6.3 cm\text{6.3 cm}

c

10.8 cm\text{10.8 cm}

Step-by-step solution

12 steps

Here are the calculations for the lengths:

Part a· 4 steps

Length of c

  1. Identify the right-angled triangle.

    Length cc is the hypotenuse of the right-angled triangle VYXVYX.

  2. Determine the lengths of the sides.

    The side VY=4VY = 4 cm.
    The side VX=6VX = 6 cm (given as the height of the rectangle).

  3. Apply the Pythagorean theorem.

    c2=VY2+VX2c^2 = VY^2 + VX^2 c2=42+62c^2 = 4^2 + 6^2 c2=16+36c^2 = 16 + 36 c2=52c^2 = 52

  4. Calculate cc and round to one decimal place.

    c=52≈7.211c = \sqrt{52} \approx 7.211 c≈7.2 cmc \approx \boxed{\text{7.2 cm}}

Part b

6.3 cm\text{6.3 cm}

Length of a

  1. Identify the right-angled triangle.

    Length aa is the hypotenuse of the right-angled triangle UZYUZY.

  2. Determine the lengths of the sides.

    The side UZ=2UZ = 2 cm.
    The side UY=UV−VY=10 cm−4 cm=6UY = UV - VY = 10 \text{ cm} - 4 \text{ cm} = 6 cm.

  3. Apply the Pythagorean theorem.

    a2=UZ2+UY2a^2 = UZ^2 + UY^2 a2=22+62a^2 = 2^2 + 6^2 a2=4+36a^2 = 4 + 36 a2=40a^2 = 40

  4. Calculate aa and round to one decimal place.

    a=40≈6.325a = \sqrt{40} \approx 6.325 a≈6.3 cma \approx \boxed{\text{6.3 cm}}

Part c

10.8 cm\text{10.8 cm}

Length of b

  1. Identify the right-angled triangle.

    Length bb is the hypotenuse of the right-angled triangle ZWXZWX.

  2. Determine the lengths of the sides.

    The side WX=10WX = 10 cm.
    The side ZW=UW−UZ=6 cm−2 cm=4ZW = UW - UZ = 6 \text{ cm} - 2 \text{ cm} = 4 cm.

  3. Apply the Pythagorean theorem.

    b2=ZW2+WX2b^2 = ZW^2 + WX^2 b2=42+102b^2 = 4^2 + 10^2 b2=16+100b^2 = 16 + 100 b2=116b^2 = 116

  4. Calculate bb and round to one decimal place.

    b=116≈10.770b = \sqrt{116} \approx 10.770 b≈10.8 cmb \approx \boxed{\text{10.8 cm}}
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