Math, asked by tayyabakhalid1412, 2 months ago


ABCD is a square.
A is the point (-2, 0) and C is the point (6,4).
AC and BD are diagonals of the square, which intersect at M.
a Find the coordinates of M, B and D.
b Find the area of ABCD.​

Answers

Answered by tennetiraj86
14

Answer :

Hope it helps you

ABCD is a square.

A is the point (-2, 0) and C is the point (6,4).AC and BD are diagonals of the square, which intersect at M.

To find :-

a)Find the coordinates of M, B and D.

b )Find the area of ABCD.

Solution :-

Given that

ABCD is a square

The coordinates of A = (-2,0)

The coordinates of C = ( 6,4)

Given that

AC and BD are diagonals of the square, which intersect at M.

So M is the mid point of AC and BD

Mid Point of AC :-

Let (x1, y1)=(-2,0)=> x1=-2 and y1=0

Let (x2, y2)=(6,4)=> x2=6 and y2=4

We know that

Mid Point formula =( {x1+x2}/2 , {y1+y2}/2)

Mid point of AC = ({-2+6}/2,{0+4}/2)

=>(4/2,4/2)

=> (2,2)

Mid point of AC = Mid point of BD

Coordinates of M = (2,2)

Let B =(x1,y1)

Let D=(x2, y2)

Mid point of BD

= ({x1+x2}/2 , {y1+y2}/2) = (2,2)

=>( x1+x2)/2 = 2 and (y1+y2)/2 = 2

=> x1+x2 = 4 and y1+y2 = 4

=> x2 = 4-x1 ----(1)and y2 = 4-y1--------(2)

We know that

All sides are equal in a square

=> AB = BC = CD = DA

We know that

Distance formula=√[(x2-x1)²+(y2-y1)²] units

AB = √[(x1+2)²+(y1-0)²]

=> AB = √[(x1+2)²+y1²]

DA = √[(x2+2)²+(y2-0)²]

=> DA =√[(x2+2)²+y2²]

From (1)&(2)

AB = AD

=> √[(x1+2)²+y1²] = [(x2+2)²+y2²]

On squaring both sides then

=> (x1+2)²+y1² = (x2+2)²+y2²

=> x1²+4x1+4+y1²=(4-x1+2)²+(4-y1)²

=> x1²+4x1+4+y1²=(6-x1)²+(4-y1)²

=> x1²+4x1+y1²+4=36+x1²-12x1+16+y1²-8y1

=> 16x1+8y1=48

=> 8(2x1+y1)=6(8)

=> 2x1+y1 = 6

=> y1 = 6-2x1------(4)

The diagonal = √2 × side

=> AC = √2 AD

=> √[(6+2)²+(4-0)²] = √2√[(x2+2)²+(y2-0)²

=> √(64+16)=√2(x2+2)²+y2²

=>√80=√2(x2+2)²+y2²

On squaring both sides

=> 80=2((x2+2)²+y2²)

=> 80/2 = (x2+2)²+y2²

=> 40 = (4-x1+2)²+(4-y1)²

=> 40 = (6-x1)²+(4-y1)²

=> 40 = 36+x1²-12x1+16-8y1+y1²

=> 40 = 52-12x1-8y1+x1²+y1²

from (4)

=> 40 = 52-12x1-8(6-2x1)+x1²+(6-2x1)²

=> 40-52=-12x1-48+16x1+x1²+36-24x1+4x1²

=> -12 = 5x1²-20x1-12

=> -12+12 = 5x1²-20x1

=> 0 = 5x12-20x1

=> 5x1² = 20x1

=> x1= 20/5

=> x1 = 4

from (4)

y1 = 6-2(4)

=> y1 = 6-8

=> y1=-2

The coordinates of B=(x1, y1)=(4,-2)

and

From (1)&(2)

x2 = 4-x1

=> x2 = 4-4

=> x2 = 0

and

y2 = 4-y1

=> y2 = 4-(-2)

=> y2 = 4+2

=> y2 = 6

The coordinates of D = (x2, y2)=(0,6)

Area of a square = side × side

=> AB²= BC²=CD²=DA²

By distance formula

AB = √[(4+2)²+(-2-0)²]

=> AB =√[6²+(-2)²]

=> AB = √(36+4)

=> AB =√40

=>AB²=(√40)²

=> AB = 40 sq.units

Answer:-

The coordinates of M = (2,2)

The other vertices are (4,-2) and (0,6)

Area of the square = 40 sq.units

Used formulae:-

In a square

All sides are equal.

Diagonals are equal

Area of a square = side×side

Mid Point formula = ( {x1+x2}/2 , {y1+y2}/2)

Distance formula

=√[(x2-x1)²+(y2-y1)²] units

Answered by farzena
0

Answer:

a) (5,6)

b) k = 8

Step-by-step explanation:

this is the proper answer.

from book

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