Question

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

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Answer 1

`6+2√(21)\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}`

Explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs
`AB=3`,
`BC=4`, and
`AC=5`. The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base
`b` and height
`h` is given by
`A=(1)/(2)bh`. Therefore, the area of this right triangle is:

`A=(1)/(2)\cdot 3\cdot 4=(1)/(2)\cdot 12=6\:\mathrm{cm^2}`

The other triangle is a bit trickier. Triangle
`\triangle ADC` is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

`A=√(s(s-a)(s-b)(s-c))`, where
`a`,
`b`, and
`c` are three sides of the triangle and
`s` is the semi-perimeter (
`s=(a+b+c)/(2)`).

The semi-perimeter,
`s`, is:

`s=(5+5+4)/(2)=(14)/(2)=7`

Therefore, the area of the isosceles triangle is:

`A=√(7(7-5)(7-5)(7-4)),\\A=√(7\cdot 2\cdot 2\cdot 3),\\A=√(84), \\A=2√(21)\:\mathrm{cm^2}`

Thus, the area of the quadrilateral is:

`6\:\mathrm{cm^2}+2√(21)\:\mathrm{cm^2}=\boxed{6+2√(21)\:\mathrm{cm^2}}`