Question

find the area of quadrilateral ABCD in which ab=18,bc=20,cd=24,da=26,ac=28 cm​

Answer 1

Given:-

• AB = 18 cm.
• BC = 20 cm.
• CD = 24 cm.
• DA = 26 cm.
• AC = 28 cm.

To find:-

• Area of quadrilateral ABCD.

Solution:-

• Two triangles are forming . One is ∆ABC and second is ∆ADC.

In ∆ABC

S = AB + BC + AC/2

S = 18 + 20 + 28/2

S = 66/2

S = 33

Semi-perimeter = 33 cm.

Area of ∆ABC :

$\implies \sf \sqrt{33 (33 - 18)(33 - 20)(33 - 28)}$

$\implies \sf \sqrt{ 33 \times 15 \times 13 \times 5}$

$\implies \sf \sqrt{3 \times 11 \times 5 \times 3 \times 13 \times 5}$

$\implies \sf 3 \times 5 \sqrt{11 \times 13}$

$\implies \sf 15 \times \sqrt{143}$

$\implies \sf 178.5$

Area of ∆ABC is 178.5 cm².

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In ∆ADC

S = CD + DA + AC/2

S = 24 + 26 + 28/2

S = 78/2

S = 39

Semi-perimeter = 39 cm.

Area of ∆ADC :

$\implies \sf \sqrt{39 (39 - 24)(39 - 26)(39 - 28)}$

$\implies \sf \sqrt{ 39 \times 15 \times 13 \times 11}$

$\implies \sf \sqrt{3 \times 13 \times 5 \times 3 \times 13 \times 11}$

$\implies \sf 3 \times 13 \sqrt{11 \times 5}$

$\implies \sf 39 \times \sqrt{55}$

$\implies \sf 288.6$

Area of ∆ADC is 288.6 cm².

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Area of ABCD = Area of ∆ABC + Area of ∆ADC

= 178.5 + 288.6 cm²

= 467.1 cm²

Therefore,

Area of quadrilateral ABCD is 467.1 cm².

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Formula used:

$\large \boxed{\sf Area\:of\: triangle= \sqrt{s(s-a)(s-b)(s-c)}}$

In which,

• S is semi-perimeter of triangle.
• a , b and c are sides of triangle