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zmey [24]
3 years ago
10

Find the area of the image below: Figure ABCDE is shown. A is at 6, 0. B is at negative 4, 5. C is at 2, 5. D is at 2, 9. E is a

t 6, 9. 41 square units 44 square units 52 square units 56 square units

Mathematics
2 answers:
salantis [7]3 years ago
6 0
<h2>Answer:</h2>

The area of image is: 41 square units

<h2>Step-by-step explanation:</h2>

The area of the image is equal to:

Area of square CDEF+Area of right angled triangle ΔBFA

In square CDEF we have:

The lengths of the side of the square= 4 units

Hence, the area of square is: side²

i.e. Area of square= 4²=16 square units

Also, in ΔBFA we have:

The base of the triangle is: BF=CB+CF=6+4= 10 units

Also, height of the triangle is: FA= 5 units

Hence, we have area of ΔBFA as:

Area= 1/2×BF×FA

i.e.

Area= 1/2×10×5

i.e.

Area= 25 square units

Hence, the area of image is: 16+25

i.e.

Area of image= 41 square units

KatRina [158]3 years ago
3 0
Where is the figure??
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Answer:

97^0

Step-by-step explanation:

Given that the angles A, B, C in a Triangle form an arithmetic sequence where A=23 degrees.

The nth term of an Arithmetic sequence is given by the formula: T_n=a+(n-1)d

Where:

a=first term

n=number of term

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In this case,

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Angle B, T_2=23+(2-1)d=(23+d)^0\\Angle C, T_3=23+(3-1)d=(23+2d)^0

The sum of angles in a triangle is 180 degrees. Therefore:

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