Question

(A) What is the position of C
(B) whats is the distance of the pole B from the corner 0 of the park
(C)
(D)
(E)
Answer all 5
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Answer 1

i. Position of C is(5, 4)

ii. The distance of pole B from corner O of the park is $6\sqrt{2}$  units

iii. The position of D is (1, 5)  

iv. The distance between poles A and C is $3\sqrt{2}$ units  

v. The distance between poles B and D is √26.

Given:

Three electric poles are named A, B, and C  

To find:

(i) What is the position of pole C?

(ii) What is the distance of pole B from corner O of the park?

(iii) Find the position of the fourth pole D so that four points A, B C, and D form a parallelogram

(iv) What is the distance between poles A and C?

(v) What is the distance between poles B and D?

Solution:

From given figure

The coordinates of A, B, and C can be taken as follows

A(2, 7) B(6, 6), and C(5, 4)

(i) What is the position of pole C?  

Therefore, Position of C is(5, 4)

(ii) What is the distance of pole B from corner O of the park?

Here the corner of the park O(0, 0) and pole B(6, 6)

Distance of pole B to O, OB = $\sqrt{(0 -6)^{2} +(0 -6)^{2} }$

= $\sqrt{36 + 36 }$ = $\sqrt{2(36)}$ = $6\sqrt{2}$  units

The distance of pole B from corner O of the park is $6\sqrt{2}$  units

(iii) Find the position of the fourth pole D so that four points A, B C, and D form a parallelogram

Let (a, b) be the D position such that ABCD is a parallelogram

AD = BC  [ Since ABCD is a parallelogram ]

=> $\sqrt{(a -2)^{2} +(b -7)^{2} } = \sqrt{(5-6)^{2} +(4 -6)^{2} }$    

Do squaring on both sides

=> $(a -2)^{2} +(b -7)^{2} = (5-6)^{2} +(4 -6)^{2}$  

=> $(a -2)^{2} +(b -7)^{2} = 1 + 4$  

=> $(a -2)^{2} +(b -7)^{2} = 5$        

From the given options, the point that will satisfy the expression is (1, 5)

Therefore, The position of D is (1, 5)  

(iv) What is the distance between poles A and C?

AC =  $\sqrt{(5 -2)^{2} +(4 -7)^{2} }$

= $\sqrt{(3)^{2} +(3)^{2} }$ = $\sqrt{2(9) }$ = $3\sqrt{2}$ units  

The distance between poles A and C is $3\sqrt{2}$ units  

(v) What is the distance between poles B and D?

BD = $\sqrt{(1 -6)^{2} +(5 -6)^{2} }$  = $\sqrt{25 +1 } = \sqrt{26}$

The distance between poles B and D is √26.

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Answer 2

Answer:

(A)  The coordinate is C(5,4)

(B) The distance between the origin to point B is 6√2 units

(C) The coordinate is D(1,5)

(D) The distance between points A and C is 3√2 units

(E) The distance between points B and D is √26 units

Step-by-step explanation:

Given,

The coordinates of quadrilateral ABCD are A(2,7), B(6,6), and C(5,4).

(A) From the given diagram we can determine the coordinate of point C is (5,4).

(B) Coordinates of B(6,6)

So, the distance from the origin is

$OB = \sqrt{(6-0)^{2} + (6-0)^{2} }\\\\ OB = \sqrt{36+36}\\\\ OB = 6\sqrt{2}\ units$

(C) If the quadrilateral ABCD is a parallelogram then, the diagonal of the parallelogram will bisect each other. So,

The diagonals of a parallelogram are AC and BD, and the midpoint of AC is

$midpoint\ of\ AC = (\frac{2+5}{2},\frac{7+4}{2} )\\\\midpoint\ of\ AC = (\frac{7}{2}, \frac{11}{2} )$

Now, the midpoint of BD will also be equal as both diagonals bisect each other i.e., the midpoint of BD = (7/2, 11/2)

Let's assume the coordinate of D is (x,y). So,

$\frac{6+x}{2} = \frac{7}{2} \ \ \ \ and \ \ \ \frac{6+y}{2}=\frac{11}{2} \\\\ \implies 6+x = 7\ \ \ \ and\ \ \ \ 6+y=11\\\\\implies x=1\ \ \ and\ \ \ y = 5$

The coordinate is D(1,5).

(D) The distance between A(2,7) and C(5,4) is

$AC = \sqrt{(5-2)^{2} + (4-7)^{2} }\\\\AC = \sqrt{3^{2}+3^{2} } \\\\AC= 3\sqrt{2}\ units$

(E) The distance between B(6,6) and D(1,5) is

$BD = \sqrt{(1-6)^{2}+ (5-6)^{2} }\\\\BD = \sqrt{5^{2}+1^2 }\\\\BD = \sqrt{26}\ units$

To learn more about quadrilateral, click on the link below:

https://brainly.com/question/23935806

To learn more about distance, click on the link below:

https://brainly.in/question/5971539

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