Question

Find the area of quadrilateral ABCD whose vertices are A (1, 0), B (5, 3), C (2, 7), D ( -2, 4)

Answer 1

ATQ,

A → (1, 0)

B → (5, 3)

C → (2, 7)

D → (-2, 4)

Since we are asked to find the area of the quadrilateral, we'll draw a line joining any two opposite vertices to form two triangles.

We'll find the area of both these triangles separately, then add them up to get the area of the quadrilateral.

Area of ΔACD:

A (x₁, y₁) → (1, 0)

D (x₂, y₂) → (-2, 4)

C (x₃, y₃) → (2, 7)

• ar(ΔABD) = ¹/₂ (x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂))

• ar(ΔABD) = ¹/₂ (1 (4 - 7) + -2 (7 - 0) + 2 (0 - 4))

• ar(ΔABD) = ¹/₂ (1 (-3) + -2 (7) + 2 (-4))

• ar(ΔABD) = ¹/₂ (-3 - 14 -8)

• ar(ΔABD) = ¹/₂ (-25)

• ar(ΔABD) = -12.5

• (Area can't be negative, therefore ↓)

• ar(ΔABD) = 12.5 sq.units.

Area of ΔABC:

A (x₁, y₁) → (1, 0)

B (x₂, y₂) → (5, 3)

C (x₃, y₃) → (2, 7)

• ar(ΔABD) = ¹/₂ (x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂))

• ar(ΔABD) = ¹/₂ (1 (3 - 7) + 5 (7 - 0) + 2 (0 - 3))

• ar(ΔABD) = ¹/₂ (1 (-4) + 5 (7) + 2 (-3))

• ar(ΔABD) = ¹/₂ (-4 + 35 - 6)

• ar(ΔABD) = ¹/₂ (35 - 10)

• ar(ΔABD) = ¹/₂ (25)

• ar(ΔABD) = 12.5 sq.units.

Area of ABCD = Area of ABC + Area of ACD

Area of ABCD = 12.5 + 12.5

Area of ABCD = 25 sq.units

Hope you understood! (◕‿◕)

Answer 2

The veritces of quadrilateral ABCD are A (1,0) ,B (5,3) , C (2,7) and D (-2,4).

AB

$=\sqrt{(5-1)^{2} +(3-0)^{2} }$

$=\sqrt{4^2+3^2}$

$=\sqrt{16+9}$

$=\sqrt{25}$

$=5 \:units$

AD

$=\sqrt{(-2-1)^{2}+(4-0)^{2}$

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

$=\sqrt{9+16}$

$=\sqrt{25}$

$=5 \: units$

BC

$=\sqrt{(2-5)^{2}+(7-3)^{2} }$

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

$=\sqrt{16+9}$

$=\sqrt{25}$

$= 5 \: units$

CD

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

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

$=\sqrt{16+9}$

$=\sqrt{25}$

$= 5 \: units$

AB = AD = BC = CD

So quatrilateral ABCD is a square .

Area of square = Side × Side

                       = 5²

                       = 25 sq.units

HOPE IT HELPS U .........