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iogann1982 [59]
3 years ago
6

The scale of a drawing is 2 cm : 1 mm. Is the scale drawing larger or smaller than the actual object? Explain.

Mathematics
2 answers:
PSYCHO15rus [73]3 years ago
7 0
The scale drawing is smaller than the actual object because a scale is an estimation on where an object is. We can't draw 1 mile that would be impossible to draw a whole drawing based on one location.
barxatty [35]3 years ago
3 0
Its 2cm for every 1mm... it would be a larger drawing.
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For the parallelogram ABCD the extensions of the angle bisectors AG and BH intersect at point P. Find the area of the parallelog
natita [175]

Answer:

The area of a parallelogram is 360 in.²

Step-by-step explanation:

Where DG = GH

GP = 12 in.

AB = 39 in.

∠DAB + ∠ABC = 180° (Adjacent angles of a parallelogram)

Whereby ∠DAB is bisected by AG and ∠ABC is bisected by BH

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We note that ∠AGD = ∠GAB (Alternate angles of parallel lines)

∴ ∠AGD = ∠AGD since ∠AGD = ∠GAB (Bisected angle)

Hence AD = DG (Side length of isosceles triangle)

The bisector of ∠ADG is parallel to BH and will bisect AG at point Q

Hence ΔDAQ  ≅ ΔDGQ ≅ ΔGPH and AQ = QG = GP

Hence, AP = 3 × GP = 3 × 12 = 36

cos(\angle GAB) = \dfrac{AP}{AB} = \dfrac{36}{39}

\angle GAB = cos^{-1} \left (\dfrac{36}{39}  \right )

∠GAB = 22.62°

cos(\angle GAB) =  \dfrac{36}{39} = \dfrac{12}{GH}

GH =  \dfrac{39}{36} \times {12}

GH = 13 in.

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GH = √(GP² + HP²)

∠DAB = 2 × 22.62° = 45.24°

The height of the parallelogram = AD × sin(∠DAB) =  13 × sin(45.24°)

The height of the parallelogram = 120/13 =  9.23 in.

The area of a parallelogram = Base × Height = (120/13) × 39 = 360 in.²

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