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PIT_PIT [208]
3 years ago
7

The area of the kite is 48cm squared what are the lengths of the diagonals

Mathematics
1 answer:
Anika [276]3 years ago
5 0
We see from the attached, that kite area = product of the diagonals / 2
The diagonals could be 12 by 8 or
6 by 16 or
3 by 32, etc
It cannot be narrowed down any further.

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Katyanochek1 [597]

Answer:

Option b is correct (8,13).

Step-by-step explanation:

7x - 4y = 4

10x - 6y =2

it can be represented in matrix form as\left[\begin{array}{cc}7&-4\\10&-6\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}4\\2\end{array}\right]

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i.e, AX=B

or X= A⁻¹ B

A⁻¹ = 1/|A| * Adj A

determinant of A = |A|= (7*-6) - (-4*10)

                                    = (-42)-(-40)

                                    = (-42) + 40 = -2

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Adj A=  \left[\begin{array}{cc}-6&4\\-10&7\end{array}\right]

A⁻¹ =  \left[\begin{array}{cc}-6&4\\-10&7\end{array}\right]/ -2

A⁻¹ =  \left[\begin{array}{cc}3&-2\\5&-7/2\end{array}\right]

X= A⁻¹ B

X=  \left[\begin{array}{cc}3&-2\\5&-7/2\end{array}\right] *\left[\begin{array}{c}4\\2\end{array}\right]

X= \left[\begin{array}{c}(3*4) + (-2*2)\\(5*4) + (-7/2*2)\end{array}\right]

X= \left[\begin{array}{c}12-4\\20-7\end{array}\right]

X= \left[\begin{array}{c}8\\13\end{array}\right]

x= 8, y= 13

solution set= (8,13).

Option b is correct.

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