Area And Diagonals

We know that ${ D }_{ 1 }+{ D }_{ 2 } = 20$

Using the AM-GM Inequality we get that:

$\frac { { D }_{ 1 }+{ D }_{ 2 } }{ 2 } \geqq \sqrt { { D }_{ 1 }*{ D }_{ 2 } }$

$10 \geqq \sqrt { { D }_{ 1 }*{ D }_{ 2 } }$

${ D }_{ 1 }*{ D }_{ 2 } \leqq 100$

The area of a quadrilateral is $\frac { 1 }{ 2 } { D }_{ 1 }*{ D }_{ 2 }$ , which is always less than or equal to 50 (as proven by AM-GM)

Thus the max is 50.

solution:

let d1 & d2 be the diagonals

1. d1 + d2 = 20

2. A = d1d2sin(a) / 2 ; area of quadrilateral using diagonals

3. let (a) the angle be 90 deg so that sin(a) is maximum or 1

4. A = d1d2/2 ; substitute d2 = 20 - d1

5. 2A = 20d1 - d1^2

6. find the derivative with respect to d1 so 2dA/dd1 = 20 - 2d1

7. for maxima set 2dA/dd1 = 0 ; so d1 = 10 and d2 = 10

8. A = 10*10/2 = 50 ....check

As sum of diagonals is 20 and area will be max iff product of diagonals is max.. i.e we should have two numbers having sum equal to 20 and max product and they are 10 and 10,.as area is defined as half of the product of diagonals i.e. 50

this is my simple equation...

Given:

d1+ d2 = 20

diagonals of square are equal, so:

( d1+ d2 ) / 2 = 10..the measure of its diagonal is 10

equation:

in a 45-45-90 triangle ... to get the shorter leg:

shorter leg = hypotenuse/ sqr. root of 2

so the measure of the shorter leg is just the same as the side of a square

s= hypotenuse / sqr. root of 2

s = 10 / sqr. root of 2

s = 5 sqr. root of 2

so we get the measure of the side...to solve for the area:

A= { s }^{ 2 }

A={ 5 sqr. root of 2 }^{ 2 }

A = 50

i can say that the maximum value of area of the square is 50...thank you

What I did was, I cosidered the quadrilateral a rhombus therefore Area of the rhombus: =1/2(d1×d2) =1/2(10×10) [as the diagonals are equal in a rhombus] =50 Btw...Don't know if I used the correct method or not..

Assuming the diagonals intersect at right angles the area is (1/2) * d1 * d2 .Now the diagonals intersect each other perpendicularly therefore d1=d2=10 and so the area is 50.

also approached the same way

With a given perimeter, square will have the greatest area among the quadrilaterals.

Solve the problem considering the quadrilateral as square.

okay i wouldn't call this a solution, but just a smart way to approach the question

consider the quadrilateral is a square and find the area.

area = 50

now look at the options...50 is the max possible choice hence answer is 50

Square is the biggest quadrilateral, We know that the side of square is = diagonal/root2 so if diagonal of a square is 10 it means the side should be 5 root2 so the area is (5 root2)^2 = 50

the answer is 50.

SOME OF THE DIAGONALS IS 20. QUADRILATERAL WILL HAVE MAXIMUM AREA IF TWO DIAGONALS ARE EQUAL. IF TWO DIAGONALS ARE EQUAL THEN IT IS A SQUARE. EACH DIAGONAL EQUAL TO 10( SUM IS 20) HENCE AREA=d1 -d2/2=(10 10)/2=100/2=50

let us consider, the shape of the quadrilateral a square shape ,so the diagonal of the two diagonal d=10 & d'=10. we know that area of a square =(1/2) * d * d'=50 which represents maximum area.

Maximum area with same perimeter or same diagonal lenght is always a SQUARE. Lenght of one diagonal is 20/2=10. If "a" is side of square then area is a^2. By Pythagoras Theorem , a^2+a^2=10^2 2(a^2)=100 a^2=50 hence Area is 50. Square can be proved with maximum area for given perimeter or diagonal by using Derivatives.

For a given length of the diagonals, the shape covering the maximum area is a square... hence ,a+b=20 => a=b=10 (Diagonals of a square being equal) =>area =(Diagonal^2)/2 =100/2 =50 Ans.

through differential calculus,

let x and y be the diagonals.
x + y = 20
we know that the area of the quadrilateral is xy/2 = A
we need to maximize the area, thus we need to find its derivative and equate to zero, but we need to transform all variables into one unique variable, x = 20 - y
so, [(20 - y)y]/2 = A = 10y - y^2/2
A' = 10 - y
0 = 10 - y
y = 10; x = 10
(10)(10)/2 = A = 50 sq. units

The quadrilateral with the maximum area must be a SQUARE. Since the sum of the diagonals is 20, each diagonal must be 10.
It follows that the side of the square is 5sqrt2. This gives us an area of 50.

it s square of side root50 n so area s 50...