In quadrilateral ABCD, AB = 17 cm, BC = 8 cm, CD = 9 cm, AD = 12 cm, and AC = 15 cm. What is the area (in cm²) of the quadrilateral?​

Correct option is B

Given:

AB = 17 cm, BC = 8 cm, CD = 9 cm,

AD = 12 cm, AC = 15 cm

Formula Used:

Heron's formula for the area of a triangle with sides a, b, and c is:

Area = $$\sqrt{s(s-a)(s-b)(s-c)}$$​

where s is the semi-perimeter, given by:

s = $$\frac{a + b + c}{2}$$​​

Solution: ​

Area of Triangle ABC

Sides: AB = 17, BC = 8, and AC = 15

Semi-perimeter:

s =$$\frac{17 + 8 + 15}{2}$$​ = 20 cm

Area of ABC = $$\sqrt{20(20 - 17)(20 - 8)(20 - 15)}$$​​

$$= \sqrt{20 \times 3 \times 12 \times 5}$$​ 

​$$= \sqrt{3600} = 60 \text{ cm}^2$$​​

Area of Triangle ACD

Semi-perimeter  = $$\frac{12 + 9 + 15}{2} = 18 \text{ cm}$$​​

Area of ACD = $$\sqrt{18(18 - 12)(18 - 9)(18 - 15)}$$​​

​$$= \sqrt{18 \times 6 \times 9 \times 3}$$​​

​$$= \sqrt{2916} = 54 \text{ cm}^2$$​​

Total Area  of quadrilateral ABCD = 60 + 54 = 114 cm² 

Alternate Solution: ​

In triangle ABC 

Sides: AB = 17, BC = 8, and AC = 15 

Form triplet , 

So, Area of triangle ABC = $$\frac{1}2\times 8 \times 15 = 60 \ cm^2$$ 

​Area of Triangle ACD:

Sides: AC = 15,DC = 9, and AD = 12 (triplet)

Area of ACD = $$\frac12\times 9\times 12 = 54 \ cm^2$$​  

Area of ABCD = 60 + 54 = 114 cm²​