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Work out the area of ABCD.
Give your answer to 1 decimal place.

Work Out The Area Of ABCD Give Your Answer To 1 Decimal Place class=

Sagot :

Area of the quadrilateral ABCD = (21.5 + 23.4) cm² = 44.9 cm²

How to calculate the area of ABCD?

  • Quadrilateral ABCD is formed from two triangles ΔABD and ΔBDC.

So, Area of quadrilateral ABCD= area of ΔABD and Δ BDC.

How to find area of triangle ABD?

ΔABD is a right angled triangle, right angled at A, BD is the hypotenuse.

  • Area of ΔABD =  AD x AB/2
  • We know that in a right angled triangle, cosФ= [tex]\frac{base}{hypotenuse}[/tex] and    sinФ=  [tex]\frac{perpendicular}{hypotenuse}[/tex], where Ф is the angle between base and hypotenuse.

Cos(55°) = [tex]\frac{AD}{10}[/tex]

⇒ AD = 5.7 cm

Sin(55°)=[tex]\frac{AB}{10}[/tex]

⇒AB=8.2 cm

Area of Δ ABD = ( 8.2 x 5.7/2) cm² =23.4 cm²

How to find area of triangle BDC?

  • We can find the area of the triangle by the formula                           BD x CD x sin(44°)/2
  • We know that sum of all the angles of a triangle is 180°

∠BDC+∠BCD+∠DBC=180°

⇒∠BCD=180°-44°-38°=99°

  • We know according to sine rule,

in a triangle ABC,   [tex]\frac{sinA}{a}=\frac{sinB}{b} =\frac{sinc}{c}[/tex], where a,b,c are corresponding sides opposite to angle A,B,C

In ΔBDC

[tex]\frac{sin99}{10}=\frac{sin38}{CD}[/tex]

⇒CD =6.2 cm

Area of Δ BDC = {6.2 x 10 x sin(44°)/2} cm² = 21.5 cm²

∴ Area of the quadrilateral ABCD = (21.5 + 23.4) cm² = 44.9 cm²

Find out more about 'Quadrilateral' here: https://brainly.com/question/9338914

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