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Fantom [35]
3 years ago
6

Find the area of the kite

Mathematics
1 answer:
dexar [7]3 years ago
4 0

Given:

A figure of a kite.

To find:

The area of the given kite.

Solution:

The area of a kite is the half of the product of its diagonals.

A=\dfrac{d_1d_2}{2}             ...(i)

Where, d_1,d_2 are two diagonals of the kite.

From the given figure it is clear that the length of one diagonal is the sum of 12 and 8.

d_1=12+8

d_1=20

Let the second diagonal be 2x. The first diagonals bisect the second diagonal. So, the length of one parts of the diagonal is x.

Diagonals of a kite are perpendicular to each other. Using Pythagoras theorem, we get

Hypotenuse^2=Perpendicular^2+Base^2

13^2=x^2+12^2

169-144=x^2

25=x^2

5=x

The length of second diagonal is:

d_2=2x

d_2=2(5)

d_2=10

Substituting d_1=20,d_2=10 in (i), we get

A=\dfrac{20\times 10}{2}

A=\dfrac{200}{2}

A=100

Therefore, the area of the kite is 100 square units.

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Both circle A and circle B have a central angle measuring 50°. The area of circle A's sector is 36π cm2, and the area of circle
slega [8]

Answer:

A) 3/4

Step-by-step explanation:

Given: Both circle A and circle B have a central angle measuring 50°.

           The area of circle A's sector is 36π cm2.

            The area of circle B's sector is 64π cm2.

We know, area of the circle= \pi r^{2}

lets assume the radius of circle A be "r_1" and radius of circle B be "r_2"

As given, Area of circle A and B´s sector is 36π and 64π repectively.

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Cancelling out the common factor

⇒ \frac{r_1^{2}  }{r_2^{2} }  = \frac{36 }{64}

⇒ (\frac{r_1  }{r_2} )^{2} = \frac{36 }{64}

Taking square on both side.

Remember; √a²= a

⇒ (\frac{r_1  }{r_2} ) =\sqrt{ \frac{36 }{64}}

⇒ (\frac{r_1  }{r_2} ) = \frac{6}{8}

⇒\frac{r_1  }{r_2}  = \frac{3}{4}

Hence, ratio of the radius of circle A to the radius of circle B is 3:4 or 3/4.

5 0
3 years ago
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