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Ket [755]
3 years ago
11

A postcard in the shape of a parallelogram has an area of 12in^2. What are two possible lengths of bases and heights for the pos

tcard
Mathematics
1 answer:
Tpy6a [65]3 years ago
8 0

A postcard is in the shape of a parallelogram. A parallelogram  is a quadrilateral with two pair of parallel sides, opposite sides and opposite angles are equal.

Since, the postcard has an area of 12 square inches.

Since, area of parallelogram = base \times height

As area of parallelogram is 12, it means that the product of base and height is 12 square inches.

So, the possible dimensions of postcard are 3 inches and 4 inches and 2 inches and 6 inches.

So, base = 3 inches , height = 4 inches or base = 4 inches , height = 3 inches.

So, base = 2 inches, height = 6 inches or base = 6 inches , height = 2 inches.

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3 years ago
Product of 3.24<br><br> 0.45 x 0.72 <br><br> 0.45 x 72<br><br> 4.5 x 0.7<br><br> 4.5 x 7.2
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Answer:

4.5 x 0.72

Step-by-step explanation:

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2 years ago
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In triangle $ABC$, let angle bisectors $BD$ and $CE$ intersect at $I$. The line through $I$ parallel to $BC$ intersects $AB$ and
Umnica [9.8K]

Answer:

41

Step-by-step explanation:

If you work through a series of obscure calculations involving area and the radius of the incircle, they boil down to a simple fact:

... For MN║BC, perimeter ΔAMN = perimeter ΔABC - BC = AB+AC

.. = 17+24 = 41

_____

Wow! Thank you for an interesting question with a not-so-obvious answer.

_____

<em>A little more detail</em>

The point I that you have defined is the incenter—the center of an inscribed circle in the triangle. Its radius is the distance from I to any side, such as BC, for example.

If we use "Δ" to represent the area of the triangle and "s" to represent the semi-perimeter, (AB+BC+AC)/2, then the incircle has radius Δ/s. The area Δ can be computed from Heron's formula by ...

... Δ = √(s(s-a)(s-b)(s-c)) . . . . where a, b, c are the side lengths

For this triangle, the area is Δ = √38480 ≈ 196.1632 units². That turns out to be irrelevant.

The altitude to BC will be 2Δ/(BC), so the altitude of ΔAMN = (2Δ/(BC) -Δ/s). Dividing this by the altitude to BC gives the ratio of the perimeter of ΔAMN to the perimeter of ΔABC, which is 2s.

Putting these ratios and perimeters together, we get ...

... perimeter ΔAMN = (2Δ/(BC) -Δ/s)/(2Δ/(BC)) × 2s

... = (2/(BC) -1/s) × BC × s = 2s -BC

... perimeter ΔAMN = AB +AC

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3 years ago
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Evgesh-ka [11]
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D. If you were to say the given ratio aloud, it would go something like this: 4 to 3, 4,3 or 4 sports video games to 3 driving ones. So, the answer is 4 to 3, or answer D.

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