Question

2. If the area of square ABCD is 41/2 sq. units where A(-4, 2), B(-2, k), C(3, k), D(3, 1) then k=​

Answer 1

Answer:

The value of k is 6.

Step-by-step explanation:

The given data in the question is wrong(This will be explained in the last part of the explanation).

But solving with the concepts

Given,

Area of Square=$\frac{41}{2}$ sq. units

⇒s²=$\frac{41}{2}$, where s is the side of square          ∵Area of Square=side²

⇒s=$\sqrt{\frac{41}{2} }$------>eqn1

Now when two coordinates are given the distance between them is given by

$\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} )$  

Now let us consider two points such that one coordinate is known and the value of another coordinate has to be found

A(-4,2) &B(-2,k)

The distance between these two points is nothing but the side of the square

Thus from eqn1 and distance formula

$\sqrt{(-2-(-4))^{2}+(k-2)^{2} }$=$\sqrt{\frac{41}{2} }$

squaring on both sides

⇒(-2+4)²+(k-2)²=$\frac{41}{2}$

⇒2²+(k-2)²=$\frac{41}{2}$

⇒4+(k-2)²=$\frac{41}{2}$

⇒(k-2)²=$\frac{41}{2}$ - 4

⇒(k-2)²=$\frac{41-8}{2}$

⇒(k-2)²=$\frac{33}{2}$

⇒k-2=4.06

⇒k=6.06≅6

∴k=6

Now coming to the part where I said the data is wrong

Consider points A and D

The distance between them from the formula is found to be

AD=$\sqrt{(3-(-4))^{2}+(1-2)^{2} }$

⇒AD=$\sqrt{7^{2}+1^{2} }$

⇒AD=$\sqrt{50}$

As we can see the area now can be found to be 50 which is in contradiction with the fact  given in the question that area of the square is $\frac{41}{2}$ sq. units