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hammer [34]
3 years ago
13

In a parallelogram a diagonal of the length 20 cm is perpendicular to one of the sides. Find the longer side of parallelogram if

its perimeter is 80 cm.
Mathematics
1 answer:
vovangra [49]3 years ago
5 0
If diagonals of a parallelogram are perpendicular to each other, this is a rhombus. A rhombus has all sides equal.
In the current scenario, the shorter rhombus is 20 cm long and has a perimeter of 80 cm

Side of rhombus = 80/4 = 20 cm
A quarter of rhombus is right angled triangle with hypotenuse being side of rhombus.
Half length of long diagonal = sqrt (20^2-10^2) = 17.32 cm
Length of longer rhombus = 17.32*2 = 34.64 cm
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<h2>Good morning,</h2>

<em><u>__________________</u></em>

<em><u>Answer</u></em>:

a+a+a = 3a.

4b-b = 3b.

3x²+x² = 4x².

<u><em>explanation</em></u>

a+a+a = 3a.

4b-b = b(4 - 1) = b × 3 = 3b.

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4 years ago
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andreev551 [17]

Step-by-step explanation:

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3 years ago
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mrs_skeptik [129]

Answer:

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Step-by-step explanation:

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6 0
3 years ago
Read 2 more answers
A square with sides of 3 squareroot 2 is inscribed in a circle. what is the area of one of the sectors formed by the radii to th
mrs_skeptik [129]
Drawing this square and then drawing in the four radii from the center of the cirble to each of the vertices of the square results in the construction of four triangular areas whose hypotenuse is 3 sqrt(2).  Draw this to verify this statement.  Note that the height of each such triangular area is (3 sqrt(2))/2.

So now we have the base and height of one of the triangular sections.

The area of a triangle is A = (1/2) (base) (height).  Subst. the values discussed above,   A = (1/2) (3 sqrt(2) ) (3/2) sqrt(2).  Show that this boils down to A = 9/2.

You could also use the fact that the area of a square is (length of one side)^2, and then take (1/4) of this area to obtain the area of ONE triangular section.   Doing the problem this way, we get (1/4) (3 sqrt(2) )^2.  Thus, 
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4 0
3 years ago
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Vladimir [108]

\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}

\frac{4}{5}  +  \frac{3}{15} -  \frac{2}{3}   \\

<u>Taking </u><u>LCM</u>

\frac{12 + 3 - 10}{15}  \\  \\ \longrightarrow \:  \cancel\frac{5}{15}  \\  \\ \longrightarrow \:  \frac{1}{3}

hope helpful ~

8 0
2 years ago
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