Prove parallelogram side theorem for segments

Mathematics

1 answer:

spin [16.1K]3 years ago

6 0

The figure is attached.

To prove: In the ||gm ABCD, segment AB ≅ CD & segment BC ≅ AD.

Proof:

∵ ABCD is a parallelogram,
→ segment AB || segment DC
→ segment BC || segment AD
Now, the line AC is a transversal.
In Δ ABC and ΔADC∠1 = ∠4 (alternate interior angles)
similarly, ∠2 = ∠3 (alternate interior angles)
AC is common.
Hence, by ASA congruency property, we can say that,
Δ ABC ≅ ΔADC
⇒ segment AB ≅ CD & segment BC ≅ AD (congruent parts of congruent triangles)
Proved.

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The smallest number of tiles Quintin will need in order to tile his floor is 20

The given parameters;

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area of each square shape tiles, A = 2000 cm²

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width of the floor, W = 6 m = 600 cm

To find:

the smallest number of tiles Quintin will need in order to tile his floor

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The maximum number tiles needed (this will be possible if only one shape type is used)

$maximum \ number= \frac{Area \ of \ floor}{total \ area \ of \ one \ shape \ type} \\\\maximum \ number= \frac{600,000 \ cm^2}{10,000 \ cm^2} \\\\maximum \ number= 60$