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uysha [10]
3 years ago
6

ABCD rhombus, AC:BD = 4:3 Perimeter ABCD = 20 cm Find: Area of ABCD

Mathematics
1 answer:
AlladinOne [14]3 years ago
6 0
Given that the perimeter of rhombus ABCD is 20 cm, the length of the sides will be:
length=20/4=5 cm
the ratio of the diagonals is 4:3, hence suppose the common factor on the diagonals is x such that:
AC=4x and BD=3x
using Pythagorean theorem, the length of one side of the rhombus will be:
c^2=a^2+b^2
substituting our values we get:
5²=(2x)²+(1.5x)²
25=4x²+2.25x²
25=6.25x²
x²=4
x=2
hence the length of the diagonals will be:
AC=4x=4×2=8 cm
BD=3x=3×2=6 cm
Hence the area of the rhombus wll be:
Area=1/2(AC×BD)
=1/2×8×6
=24 cm²
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Answer: y = 3.5 x+ 43.8

Step-by-step explanation:

Here x represents the number of years after 1990

Thus, we get the table that is used to find the equation will be,

x             0                  2                4               6                8

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\text{Where, }a=\frac{\sum y \sum x^2-\sum x \sum xy}{n(\sum x^2)-(\sum x)^2}

\text{And, }b = \frac{\sum xy - \sum x\sum y}{n\sum x^2 - (\sum x)^2}

By the above table,

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Read 2 more answers
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KonstantinChe [14]
Ans given below

-3r-12=-21
-3r=-21+12
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r=-9\-3
r=3
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