Question 15 Marks
In a quadrilateral ABCD, AB = 28cm, BC = 26cm, CD = 50cm, DA = 40cm and diagonal AC = 30cm. Find the area of the quadrilateral.
Answer
View full question & answer→In quad. ABCD,
AB = 28cm, BC = 26cm, CD = 50cm, DA = 40cm
and diagonal AC = 30cm
In $\triangle\text{ABC},$
$\text{s}=\frac{\text{sum of sides}}{2}$
$=\frac{26+28+30}{2}$
$=\frac{84}{2}=42$
$\therefore\text{Area of}\triangle\text{ABC}$
$=\sqrt{\text{s(s}-\text{a})\text{(s}-\text{b})\text{(s}-\text{c})}$
$=\sqrt{42(42-26)(42-28)(42-30)}$
$=\sqrt{42\times16\times14\times12}$
$\sqrt{7\times3\times2\times2\times2\times2\times2\times2\times7\times2\times2\times3}$
$=2\times2\times2\times3\times7=336\text{cm}^2$
In $\triangle\text{ADC},$
$\text{s}=\frac{30+40+50}{2}=\frac{120}{2}=60$
$\therefore\text{Area of }\triangle\text{ADC}$
$=\sqrt{\text{s(s}-\text{a})\text{(s}-\text{b})\text{(s}-\text{c})}$
$=\sqrt{60(60-30)(60-40)(60-50)}$
$=\sqrt{60\times30\times20\times10}$
$=\sqrt{360000}=600\text{cm}^2$
$\therefore$ Area of quad. ABCD $= 336 + 600$
$=936\text{cm}^2$
AB = 28cm, BC = 26cm, CD = 50cm, DA = 40cm
and diagonal AC = 30cm
In $\triangle\text{ABC},$
$\text{s}=\frac{\text{sum of sides}}{2}$
$=\frac{26+28+30}{2}$
$=\frac{84}{2}=42$
$\therefore\text{Area of}\triangle\text{ABC}$
$=\sqrt{\text{s(s}-\text{a})\text{(s}-\text{b})\text{(s}-\text{c})}$
$=\sqrt{42(42-26)(42-28)(42-30)}$
$=\sqrt{42\times16\times14\times12}$
$\sqrt{7\times3\times2\times2\times2\times2\times2\times2\times7\times2\times2\times3}$
$=2\times2\times2\times3\times7=336\text{cm}^2$
In $\triangle\text{ADC},$
$\text{s}=\frac{30+40+50}{2}=\frac{120}{2}=60$
$\therefore\text{Area of }\triangle\text{ADC}$
$=\sqrt{\text{s(s}-\text{a})\text{(s}-\text{b})\text{(s}-\text{c})}$
$=\sqrt{60(60-30)(60-40)(60-50)}$
$=\sqrt{60\times30\times20\times10}$
$=\sqrt{360000}=600\text{cm}^2$
$\therefore$ Area of quad. ABCD $= 336 + 600$
$=936\text{cm}^2$