Question

st
8. Show that the points (-4,-7),(-1,2), (8,5) and (5, 4) taken in order are the vertices of
a rhombus. And find its area.
1
(Hint: Area of rhombus = 1/2× product of its diagonals)
​

Answer 1

CORRECT QUESTION :-

Show that the points ( -4 , -7 ) , (-1 ,2 ) , (8 , 5 ) and (5, -4 ) are the vertices of rhombus and find its area

CONCEPT TO KNOW :-

In a Rhombus All sides are equal If we find distance between those points .If we get same distance the given vertices are the vertices of rhombus and we have to find diagonals because to find the area of Rhombus

Area of rhombus = 1/2 d1 d2

______________________

SOLUTION:-

By using distance formula = $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Distance between AB = $\sqrt{(-4+1)^2 +(-7-2)^2}$

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

AB = $\sqrt{9+81}$

AB = $\sqrt{90}$ = 3 $\sqrt{10}$

Distance between BC = $\sqrt{(-1-8)^2 + (2-5)^2}$

BC = $\sqrt{81 + 9}$

BC = $\sqrt{90}$ = 3 $\sqrt{10}$

Distance between CD = $\sqrt{(8-5)^2 +(5+4)^2}$

CD = $\sqrt{9+81}$

CD = $\sqrt{90}$ = 3 $\sqrt{10}$

Distance between AD = $\sqrt{(-4-5)^2 +(-7+4)^2}$

AD = $\sqrt{81+9}$

AD = $\sqrt{90}$ = 3 $\sqrt{10}$

Hence Distance between four sides are equal i.e

AB = BC = CD = AD = 3 $\sqrt{10}$

We can say that the above vertices are the vertices of rhombus

HENCE PROVED !

Now , we have to find the area of Rhombus

Area of rhombus = 1/2d1 d2

AC , BD are diagonals of a rhombus We can find distance between them and afterwards we shall find area of rhombus

AC = $\sqrt{(-4-8)^2 +(-7-5)^2}$

AC = $\sqrt{144+144}$

AC = $\sqrt{288}$

AC = 12$\sqrt{2}$

BD = $\sqrt{(-1-5)^2 + (2+4)^2}$

BD = $\sqrt{36+36}$

BD = $\sqrt{72}$

BD = 6$\sqrt{2}$

Area of rhombus = 1/2 d1 d2 = 1/2 AC BD

Area of rhombus = 1/2 12$\sqrt{2}$ 6$\sqrt{2}$

Area of rhombus = 1/2 72 × 2

Area of rhombus = 72 sq units

_________________