Question

find the area of quadrilateral ABCD in which ANGLE BCA =90degree AB=13cm and ACD is an equilateral triangle of side 12 cm

Answer 1

Take a (5, 12, 13) Pythagorean triple and multiply by 2. 
You get (10, 24, 26) 
Since ABD is a right triangle with BD = 24 and AD = 26, 
then we must have AB = 10 

The area of triangle ABD = (1/2)AB BD = (1/2)(10)(24) = 120 

The area of an equilateral triangle with sides s is (√3 / 4)s^2. 

The area of triangle BCD = (√3 / 4)24^2 = 144√3 

So the area of ABCD = 120 + 144√3

Answer 2

It is given that

ABCD is a quadrilateral in which ∠BCA = 900 and AB = 13 cm

ABCD is an equilateral triangle in which AC = CD = AD = 12 cm

In right angled △ABC

Using Pythagoras theorem,

AB2 = AC2 + BC2

Substituting the values

132 = 122 + BC2

By further calculation

BC2 = 132 – 122

BC2 = 169 – 144 = 25

So we get

BC = √25 = 5 cm

We know that

Area of quadrilateral ABCD = Area of △ABC + Area of △ACD

It can be written as

= ½ × base × height + √3/4 × side2

= ½ × AC × BC + √3/4 × 122

Substituting the values

= ½ × 12 × 5 + √3/4 × 12 × 12

So we get

= 6 × 5 + √3 × 3 × 12

= 30 + 36√3

Substituting the value of √3

= 30 + 36 × 1.732

= 30 + 62.28

= 92.28 cm2

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