Question

Question 2
Find the area of a quadrilateral ABCD in which AB=3cm,BC=4cm,CD=4cm, DA = 5cm and AC= 5cm.

Answer 1

For $\Delta \text{ABC}$,
${\text{AC}}^2={\text{AB}}^2+{\text{BC}}^2(5{)}^2=(3{)}^2+(4{)}^2\text{Therefore,}\Delta \text{ABC}\text{is a right triangle, right-angled at point~B}\text{Area of}\Delta \text{ABC =}\frac{1}{2}\times \text{AB}\times \text{BC =}\frac{1}{2}\times 3\times 4=6c{m}^2\text{For}\Delta \text{ADC},\text{Perimeter = 2s = AC+CD+DA =}(5+4+5)cm=14cmS=7cm\text{By, Heron's formula,}\text{Area of triangle =}\sqrt{s(s-a)(s-b)(s-c)}\text{Area of}\Delta \text{ADC =}\lfloor \sqrt{7(7-5)(7-5)(7-4)}\rfloor c{m}^2=(\sqrt{7\times 2\times 2\times 3})c{m}^2=2\sqrt{21}c{m}^2=9.166c{m}^2\text{Area of ABCD = Area of}\Delta \text{ABC + Area of}\Delta \text{ACD}=(6+9.166)c{m}^2=15.166c{m}^2=15.2c{m}^2\text{(approximately)}$