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2 years ago

Find the area of the figure

Mathematics

1 answer:

fiasKO [112]2 years ago

7 0

Step-by-step explanation:

there is one central rectangle :

from (-4, 2) to (2, -2).

that means it dimensions are 6×4 (for x from -4 to +3, for y from +2 to -2).

and then there are 3 additional triangles

(2, -4) (2, 2) (6, 2)

(-4, -2) (2, -2) (2, -4)

(-6, 2) (-4, 2) (-4, -2)

with the leg lengths

(2, -4) to (2, 2) = 6, and (2, 2) to (6, 2) = 4

(-4, -2) to (2, -2) = 6, and (2, -2) to 2, -4) = 2

(-6, 2) to (-4, 2) = 2, and (-4, 2) to (-4, -2) = 4

the area of the rectangle is

6×4 = 24 square units

the area of the first triangle is

6×4/2 = 3×4 = 12 square units

the area of the second triangle is

6×2/2 = 6 square units

the area of the third triangle is

2×4 = 8 square units.

so, the total area is

24 + 12 + 6 + 8 = 50 square units.

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Please answer question three and four if you can :)<br> Show full working out ty;)

nikklg [1K]

Step-by-step explanation:

Let D be the mid point of side BC, [B(2, - 1), C(5, 2)].

Therefore, by mid-point formula:

$D = ( \frac{2 + 5}{2}, \: \: \frac{ - 1 + 2}{2} ) = ( \frac{7}{2}, \: \: \frac{ 1}{2} ) \\ \therefore D= (3.5, \: \: 0.5) \\ \& \: A=(-1,\:\:4)...(given) \\\\ now \: by \: distance \: formula \\ \\ Length \: of \: segment \: AD \\ = \sqrt{( - 1 - 3.5)^{2} + {(4 - 0.5)}^{2} } \\ = \sqrt{(4.5)^{2} + {(3.5)}^{2} } \\ = \sqrt{20.25 + 12.25 } \\ = \sqrt{32.5} \\ \red{ \boxed{\therefore Length \: of \: segment \: AD = 5.7 \: units}}$

4 (a)

Equation of line AB[A(2, 1), B(-2, - 11)] in two point form is given as:

$\frac{y-y_1}{y_1-y_2} =\frac{x-x_1}{x_1 - x_2} \\\\\therefore \frac{y-1}{1-(-11)} =\frac{x-2}{2 - (-2) } \\\\\therefore \frac{y-1}{1+11} =\frac{x-2}{2 +2} \\\\\therefore \frac{y-1}{12} =\frac{x-2}{4} \\\\\therefore \frac{y-1}{3} =\frac{x-2}{1} \\\\\therefore y-1= 3(x - 2)\\\\\therefore y= 3x - 6+1\\\\\therefore y= 3x - 5\\\\ \huge \purple {\boxed {\therefore 3x - y-5=0}} \\$

is the equation of line AB.

Now we have to check whether C(4, 7) lie on line AB or not.

Let us substitute x = 4 & y = 7 on the Left hand side of equation of line AB and if it gives us 0, then C lies on the line.

$LHS = 3x - y-5\\=3\times 4-7-5\\= 12-12\\=0\\= RHS$

Hence, point C (4, 7) lie on the straight line AB.

4(b)

Like we did in 4(a), first find the equation of line AB and then substitute the coordinates of point C in equation and if they satisfy the equation, then all the three points lie on the straight line.

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