Question

Show that the points (– 4, – 7), (– 1, 2), (8, 5) and (5, – 4) taken in

order are the vertices of a rhombus. Find its area​

Answer 1

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GIVEN :

• Points (-4,-7) , (-1,2) , (8,5) and (5,-4) are the vertices of a quadrilateral taken in order.

TO SHOW :

• It is a rhombus

TO FIND :

• Area of a rhombus

FIGURE :

$\setlength{\unitlength}{1 cm}\begin{picture}(20,15)\thicklines\qbezier(1,1)(1,1)(6,1)\qbezier(1,1)(1,1)(1.6,4)\qbezier(1.6,4)(1.6,4)(6.6,4)\qbezier(6,1)(6,1)(6.6,4)\qbezier(6.6,4)(6.6,4)(1,1)\qbezier(1.6,4)(1.6,4)(6,1)\put(0.7,0.5){\sf A(-4,-7)}\put(6,0.5){\sf B(-1,2)}\put(1.4,4.3){\sf D(5,-4)}\put(6.6,4.3){\sf C(8,5)}\end{picture}$

SOLUTION :

• Finding length of it's diagonals and sides :

Side AB :

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

$:\implies\:\sf \sqrt{[-1\:-\:(-4)]^2\:+\:[2\:-\:(-7)]^2}$

$:\implies\:\sf \sqrt{(-1\:+\:4)^2\:+\:(2\:+\:7)^2 }$

$:\implies\:\sf \sqrt{(3)^2\:+\:(9)^2 }$

$:\implies\:\sf \sqrt{9\:+\:81}$

$:\implies\:\bf \sqrt{90}\:units$

Side BC :

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

$:\implies\:\sf \sqrt{[8\:-\:(-1)]^2\:+\:[5\:-\:2]^2}$

$:\implies\:\sf \sqrt{(8\:+\:1)^2\:+\:(3)^2 }$

$:\implies\:\sf \sqrt{(9)^2\:+\:(3)^2 }$

$:\implies\:\sf \sqrt{81\:+\:9}$

$:\implies\:\bf \sqrt{90}\:units$

Side CD :

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

$:\implies\:\sf \sqrt{(5\:-\:8)^2\:+\:[-4\:-\:5]^2}$

$:\implies\:\sf \sqrt{(-3)^2\:+\:(-9)^2 }$

$:\implies\:\sf \sqrt{9\:+\:81}$

$:\implies\:\bf \sqrt{90}\:units$

Side DA :

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

$:\implies\:\sf \sqrt{[5\:-\:(-4)]^2\:+\:[-4\:-\:(-7)]^2}$

$:\implies\:\sf \sqrt{(5\:+\:4)^2\:+\:(-4\:+\:7)^2 }$

$:\implies\:\sf \sqrt{(9)^2\:+\:(3)^2 }$

$:\implies\:\sf \sqrt{81\:+\:9}$

$:\implies\:\bf \sqrt{90}\:units$

Diagonal BD :

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

$:\implies\:\sf \sqrt{[5\:-\:(-1)]^2\:+\:[-4\:-\:2]^2}$

$:\implies\:\sf \sqrt{(5\:+\:1)^2\:+\:(-6)^2 }$

$:\implies\:\sf \sqrt{(6)^2\:+\:(-6)^2 }$

$:\implies\:\sf \sqrt{36\:+\:36}$

$:\implies\:\sf \sqrt{72}$

$:\implies\:\bf 8.5\:units$

Diagonal AC :

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

$:\implies\:\sf \sqrt{[8\:-\:(-4)]^2\:+\:[5\:-\:(-7)]^2}$

$:\implies\:\sf \sqrt{(8\:+\:4)^2\:+\:(5\:+\:7)^2 }$

$:\implies\:\sf \sqrt{(12)^2\:+\:(12)^2 }$

$:\implies\:\sf \sqrt{144\:+\:144}$

$:\implies\:\sf \sqrt{288}$

$:\implies\:\bf 16.9\:units$

RESULT :

• AB = BC = CD = DA
• BD ≠ AC

• Hence it is a rhombus! •

NOW,

♣ Using formula of area of rhombus :-

ㅤ➱ㅤ½ × Product of it's diagonals

ㅤ➱ㅤ½ × 8.5 × 16.9

ㅤ➱ㅤ½ × 143.65

ㅤ➱ㅤ71.8

∴ Area of rhombus ≈ 71.8 sq.units.

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NOTE :

• Dear user if you are not able to see the diagram from app. Kindly please see it from the the site (brainly.in). It will be correctly displayed there.

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Answer 2

Given :

Show that the points (– 4, – 7), (– 1, 2), (8, 5) and (5, – 4) taken in order are the vertices of a rhombus.

To find :

• Area of rhombus

Solution :

• As we know that according to the distance formula

→ √(x2 - x1)² + (y2 - y1)²

According to the question

• A = (– 4, – 7)
• B = (– 1, 2)
• C = (8, 5)
• D = (5, – 4)

• Distance of AB

→ AB = √[-1 - (-4)]² + [2 - (-7)]²

→ AB = √[-1 + 4]² + [2 + 7]²

→AB = √(-3)² + (9)²

→ AB = √9 + 81

→ AB = √90 unit

• Distance of BC

→ BC = √[8 - (-1)]² + [5 - 2]²

→ BC = √(9)² + (3)²

→ BC = √81 + 9

→ BC = √90 unit

• Distance of CD

→ CD = √[5 - 8]² + [-4 - 5]²

→ CD = √(-3)² + (-9)²

→ CD = √9 + 81

→ CD = √90 unit

• Distance of DA

→ DA = √[-4 - 5]² + [-7 - (-4)]²

→ DA = √(-9)² + (-3)²

→ DA = √81 + 9

→ DA = √90 unit

• Distance of AC

→ AC = √[8 - (-4)]² + [5 - (-7)]²

→ AC = √(12)² + (12)²

→ AC = √144 + 144

→ AC = √288

• Distance of DB

→ DB = √[-1 - 5]² + [2 - (-4)]²

→ DB = √(-6)² + (6)²

→ DB = √36 + 36

→ DB = √72

• Results

AB = BC = CD = DA, AC ≠ DB

Hence, it's a rhombus

• Area of rhombus

→ ½ × product of diagonals

→ ½ × AC × DB

→ ½ × √288 × √72

→ ½ × 12√2 × 6√2

→ 12 × 6

→ 72 sq.unit

•°• Area of rhombus is 72 sq.unit