Determine the area of the following special quadrilaterals.
The area of the rectangle equals:
AB \times BC
Using the Pythagoras theorem, we have:
AB=\sqrt{\left(AC\right)^2 - \left(BC\right)^2}= \sqrt{15^2 - 12^2}= 9
The area of the rectangle equals:
12 \times 9 = 108
The area is 108.
The area of a square with side x equals:
x^2
Using the Pythagoras theorem, we have:
x^2+x^2=10^2
2x^2=100
x^2=50
x=\sqrt{50}
The area of the square is:
x^2=\sqrt{50}^2=50
The area is 50.
The area of a trapezoid is the average of the bases times the height:
\dfrac{1}{2}\left(AB + CE\right) \times AD
Using the Pythagoras theorem, we have:
AD=\sqrt{\left(AE\right)^2 - \left(DE\right)^2}= \sqrt{13^2-5^2}= 12
The area of the trapezoid equals:
\dfrac{1}{2}\left(15 + 20\right) \times 12=210
The area is 210.
The area of a trapezoid is the average of the bases times the height:
\dfrac{1}{2}\left(AB + CD\right) \times BH
Using the Pythagoras theorem, we have:
BH=\sqrt{\left(BC\right)^2 - \left(CH\right)^2}= \sqrt{5^2-3^2}= 4
The area of the trapezoid equals:
\dfrac{1}{2}\left(13 + 10\right) \times 4= 46
The area is 46.
The area of a kite is half of the product of the diagonals:
\dfrac{1}{2}\left(AC \times BD\right)
Using the Pythagoras theorem, we have:
AO=\sqrt{\left(AB\right)^2 - \left(OB\right)^2}= \sqrt{50 - 25}=5
And:
CO=\sqrt{\left(BC\right)^2 - \left(OB\right)^2}= \sqrt{169 - 25}=12
Therefore:
- AC = 5+12=17
- OB=2\times OB = 10
The area of the kite equals:
\dfrac{1}{2}\left(17 \times 10\right) = 85
The area is 85.
ABCD is a parallelogram.
The area of the parallelogram equals the product of the base by the height:
CD \times BH
Using the Pythagoras theorem, we have:
BH=\sqrt{\left(10\right)^2 - \left(6\right)^2}= \sqrt{100 - 36}= 8
The area of the parallelogram equals:
11 \times 8 = 88
The area is 88.
The area of the rhombus equals the half of the products of the diagonals:
\dfrac{1}{2} \times AC \times BD
Using the Pythagoras theorem, we have:
OB=\sqrt{\left(BC\right)^2 - \left(OC\right)^2}= \sqrt{13^2 - 5^2}= 12
Therefore:
- AC = 2 \times OC =10
- BD=2 \times OB = 24
The area of the rhombus equals:
\dfrac{1}{2}\times 24 \times 10 = 120
The area is 120.