Determine the area of the following triangles.
The area of the triangle is:
\dfrac{1}{2}\times\left(BC \times AH\right)
Or:
\dfrac{1}{2}\times\left[\left(BH+HC\right) \times AH\right]
Using the Pythagoras theorem, we have:
BH = \sqrt{5^2-3^2}= 4
Similarly, we have:
HC=4
The area of the triangle is:
\dfrac{1}{2}\times\left[\left(BH+HC\right) \times AH\right]=\dfrac{1}{2}\times\left[\left(4+4\right) \times 3\right]=12
The area of the triangle is 12.
The area of the triangle is:
\dfrac{1}{2}\times\left(BC \times AB\right)
We know that:
\tan{\widehat{C}} = \dfrac{AB}{BC}
And:
- \widehat{C}= 30^\circ
- \tan{30^\circ} = \dfrac{\sqrt{3}}{3}
Therefore:
\dfrac{\sqrt{3}}{3} = \dfrac{3}{BC}
BC = \dfrac{9}{\sqrt{3}} = 3\sqrt{3}
The area of the triangle is:
\dfrac{1}{2} \times AB \times BC = \dfrac{1}{2} \times3 \times 3{\sqrt{3}} = \dfrac{9{\sqrt{3}} }{2}
The area of the triangle is \dfrac{9{\sqrt{3}} }{2}.
The area of the triangle is:
\dfrac{1}{2}\times\left(BC \times AH\right)
Using the Pythagoras theorem, we have:
AH = \sqrt{5^2 - 3^2} = 4
Therefore:
S = \dfrac{1}{2} \times 6 \times 4 = 12
The area of this triangle is 12.
The area of the triangle is:
S = \dfrac{1}{2} \times BC \times AH
Using the Pythagoras theorem, we have:
AH = \sqrt{4^2 - 2^2} = \sqrt{12}= 2\sqrt{3}
Therefore:
S = \dfrac{1}{2} \times 4 \times 2\sqrt{3} = 4\sqrt{3}
The area of this triangle is 4\sqrt{3}.
The area of the triangle is:
\dfrac{1}{2}\times\left(BC \times AH\right)
Or:
\dfrac{1}{2}\times\left[\left(BH+HC\right) \times AH\right]
Using the Pythagoras theorem, we have:
BH = \sqrt{5^2-3^2}= 4
Also:
HC=\sqrt{6^2-3^2} = \sqrt{27}
The area of the triangle is:
\dfrac{1}{2}\times\left[\left(BH+HC\right) \times AH\right]=\dfrac{1}{2}\times\left[\left(4+\sqrt{27}\right) \times 3\right]=\dfrac{3}{2}\left(4+\sqrt{27}\right)
The area of the triangle is \dfrac{3}{2}\left(4+\sqrt{27}\right).
The area of the triangle with sides a and b, and the included angle \theta, can be computed using the following formula:
A=\dfrac{1}{2}a.b \sin \left(\theta\right)
Therefore:
A=\dfrac{1}{2} \times 2 \times\sqrt2 \times \sin \left(45^\circ\right)
\dfrac{1}{2} \times 2 \times\sqrt2 \times \dfrac{\sqrt{2}}{2}
A=1
The area of the triangle is 1.
The area of the triangle is:
\dfrac{1}{2}\times\left(BC \times AH\right) or \dfrac{1}{2}\times\left[\left(BH+HC\right) \times AH\right]
We know that:
- \tan{\widehat{C}} = \dfrac{AH}{BH}
- \tan{\widehat{B}} = \dfrac{AH}{CH}
We also have:
- \widehat{C}= 45^\circ
- \widehat{B}= 45^\circ
- \tan{45^\circ}= 1
We can write:
1 = \dfrac{2}{BH} \Rightarrow BH = 2
1 = \dfrac{2}{CH} \Rightarrow CH=2
The area of the triangle is:
\dfrac{1}{2} \times 2 \times \left(2+2\right)=4
The area of the triangle is 4.