Question

Maths Legends can Solve...!!​

Answer 1

Topic :-

3D - Coordinate Geometry

Given :-

The points (8, -5, 6), (11, 1, 8), (9, 4, 2) and (6, -2, 0) are vertices of a cyclic quadrilateral.

To Find :-

Area of the given quadrilateral.

Concept Used :-

Distance between two points

$\sf {Let\:points\:be\:A\:(x_1,y_1,z_1)\:and\:B\:(x_2,y_2,z_2).}$

$\sf {AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}}$

Solution :-

Let points be A (8, -5, 6), B (11, 1, 8), C (9, 4, 2) and D (6, -2, 0).

Calculating Distance between points,

$\sf {AB=\sqrt{(11-8)^2+(1-(-5))^2+(8-6)^2}}$

$\sf {AB=\sqrt{3^2+6^2+2^2}}$

$\sf {AB=\sqrt{9+36+4}}$

$\sf{AB=\sqrt{49}}$

$\sf{\bold{AB=7\:units}}$

$\sf {BC=\sqrt{(9-11)^2+(4-1)^2+(2-8)^2}}$

$\sf {BC=\sqrt{(-2)^2+(3)^2+(-6)^2}}$

$\sf {BC=\sqrt{4+9+36}}$

$\sf {BC=\sqrt{49}}$

$\sf{\bold{BC=7\:units}}$

$\sf {CD=\sqrt{(6-9)^2+(-2-4)^2+(0-2)^2}}$

$\sf {CD=\sqrt{(-3)^2+(-6)^2+(-2)^2}}$

$\sf {CD=\sqrt{9+36+4}}$

$\sf {CD=\sqrt{49}}$

$\sf{\bold{CD=7\:units}}$

$\sf {DA=\sqrt{(8-6)^2+(-5-(-2))^2+(6-0)^2}}$

$\sf {DA=\sqrt{(2)^2+(-3)^2+(6)^2}}$

$\sf {DA=\sqrt{4+9+36}}$

$\sf{\bold{DA=7\:units}}$

$\sf {AC=\sqrt{(9-8)^2+(4-(-5))^2+(2-6)^2}}$

$\sf {AC=\sqrt{(1)^2+(9)^2+(-4)^2}}$

$\sf {AC=\sqrt{1+81+16}}$

$\sf {AC=\sqrt{98}}$

$\sf{\bold{AC=7\sqrt{2}\:units}}$

$\sf {BD=\sqrt{(6-11)^2+(-2-1)^2+(0-8)^2}}$

$\sf {BD=\sqrt{(-5)^2+(-3)^2+(-8)^2}}$

$\sf {BD=\sqrt{25+9+64}}$

$\sf {BD=\sqrt{98}}$

$\sf{\bold{BD=7\sqrt{2}\:units}}$

We observe that all sides of given Quadrilateral are equal and diagonals are also equal to each other, which means given quadrilateral is a Square of side 7 units.

Area of Square = ( Side )²

Area of Square = ( 7 units)²

Area of Square = 49 sq. units

Answer :-

So, the area of given Quadrilateral which is a square is 49 sq. units.

Answer 2

Answer:

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