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mars1129 [50]
3 years ago
12

A triangle has an area of 900m2 . If a parallelogram has the same height and base as the triangle, what is the area of the paral

lelogram?
Mathematics
1 answer:
timama [110]3 years ago
4 0

Answer:

1800 m^{2} is the area of parallelogram.

Step-by-step explanation:

Given that:

Area of a triangle = 900 m^{2}

To find:

Area of a parallelogram which has same height and base as that of the given triangle.

Solution:

First of all, let us have a look at the formula for Area of a parallelogram:

Area_{Par} = Base \times Height ...... (1)

So as to find the area of a parallelogram, we need to have the product of Base and Height of Parallelogram.

Now, let us have a look at the formula for area of a triangle:

Area_{Tri} = \dfrac{1}{2} \times Base \times Height

Given that height and base of triangle and parallelogram are equal to each other.

So,the product of base and height will also be equal to each other.

900 = \dfrac{1}{2} \times Base \times Height\\\Rightarrow Base \times Height = 2 \times 900\\\Rightarrow Base \times Height = 1800\ m^2

By equation (1):

<em>Area of parallelogram = 1800 </em>m^{2}<em> </em>

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Lady_Fox [76]

Answer:

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Step-by-step explanation:

The problem is a classic example of a telescoping series of products, a series in which each term is represented in a certain form such that the multiplication of most of the terms results in a massive cancelation of subsequent terms within the numerators and denominators of the series.

The simplest form of a telescoping producta_{k} \ = \ \displaystyle\frac{t_{k}}{t_{k+1}}, in which the products of <em>n</em> terms is

a_{1} \ \times \ a_{2} \ \times \ a_{3} \ \times \ \cdots \times \ a_{n-1} \ \times \ a_{n} \ = \ \displaystyle\frac{t_{1}}{t_{2}} \ \times \ \displaystyle\frac{t_{2}}{t_{3}} \ \times \ \displaystyle\frac{t_{3}}{t_{4}} \ \times \ \cdots \ \times \ \displaystyle\frac{t_{n-1}}{t_{n}} \ \times \ \displaystyle\frac{t_{n}}{t_{n+1}} \\ \\ \-\hspace{5.55cm} = \ \displaystyle\frac{t_{1}}{t_{n+1}}..

In this particular case, t_{1} \ = \ 2 , t_{2} \ = \ 3, t_{3} \ = \ 4, ..... , in which each term follows a recursive formula of t_{n+1} \ = \ t_{n} \ + \ 1. Therefore,

\displaystyle\frac{t_{2}}{t_{1}} \times \displaystyle\frac{t_{3}}{t_{2}} \times \displaystyle\frac{t_{4}}{t_{3}} \times \cdots \times \displaystyle\frac{t_{n}}{t_{n-1}} \times \displaystyle\frac{t_{n+1}}{t_{n}} \ = \ \displaystyle\frac{3}{2} \times \displaystyle\frac{4}{3} \times \displaystyle\frac{5}{4} \times \cdots \times \displaystyle\frac{2005}{2004} \times \displaystyle\frac{2006}{2005} \\ \\ \-\hspace{5.95cm} = \ \displaystyle\frac{2006}{2} \\ \\ \-\hspace{5.95cm} = 1003

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Answer:

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Step-by-step explanation:

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Finished Problems = 8

Remaining = 175 - 8 = 167

8 problems in 30 minutes, means:

30/8 = 15/4 = 3.75 minutes per problem

If he works for 45 minutes, at that rate, he can solve:

45/3.75 = 12 problems per day

To finish 167 problems, he will need:

167/12 = 13.91

Rounded, that is <u>14 days more to finish</u>

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Answer:

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The dimensions that would result to maximum area will be found as follows:
let the length be x, the width will be 32-x
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Answer:

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