Answer:
Area = 42.43 square cm
Step-by-step explanation:
Using Heron's Formula :
![Area = \sqrt{s\times (s-a) \times (s-b) \times(s -c)} \ , \ where \ s \ =\ \frac{a+ b+ c}{2}](https://tex.z-dn.net/?f=Area%20%3D%20%5Csqrt%7Bs%5Ctimes%20%28s-a%29%20%5Ctimes%20%28s-b%29%20%5Ctimes%28s%20-c%29%7D%20%5C%20%2C%20%5C%20where%20%20%5C%20s%20%5C%20%3D%5C%20%5Cfrac%7Ba%2B%20b%2B%20c%7D%7B2%7D)
<em><u>Find Area :</u></em>
<em><u></u></em>
![s = \frac{11 + 9 + 10 }{2} = \frac{30}{2} = 15](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B11%20%2B%209%20%2B%2010%20%7D%7B2%7D%20%3D%20%5Cfrac%7B30%7D%7B2%7D%20%3D%2015)
![Area = \sqrt{15 \times ( 15 - 11 ) \times { (15 - 9 ) \times ( 15 -10)}\\\\](https://tex.z-dn.net/?f=Area%20%3D%20%5Csqrt%7B15%20%5Ctimes%20%28%2015%20-%2011%20%29%20%5Ctimes%20%7B%20%2815%20-%209%20%29%20%5Ctimes%20%28%2015%20-10%29%7D%5C%5C%5C%5C)
![= \sqrt{15 \times 4 \times 6 \times 5}\\\\=\sqrt{1800}](https://tex.z-dn.net/?f=%3D%20%5Csqrt%7B15%20%5Ctimes%204%20%5Ctimes%206%20%5Ctimes%205%7D%5C%5C%5C%5C%3D%5Csqrt%7B1800%7D)
![= 42 . 43 \ cm^2](https://tex.z-dn.net/?f=%3D%2042%20.%2043%20%5C%20cm%5E2)
Agree with the one at the top it’s 36
Answer:
296.89 m
Step-by-step explanation:
assuming that both the buildings are on level ground (i.e their bases are at the same elevation), see attached.
Answer:
![\large\boxed{A_\triangle=25\sqrt{15}\ cm^2}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7BA_%5Ctriangle%3D25%5Csqrt%7B15%7D%5C%20cm%5E2%7D)
Step-by-step explanation:
Look at the picture.
The formula of an area of a triangle:
![A_\triangle=\dfrac{bh}{2}](https://tex.z-dn.net/?f=A_%5Ctriangle%3D%5Cdfrac%7Bbh%7D%7B2%7D)
<em>b</em><em> - base</em>
<em>h</em><em> - height</em>
<em />
We need a length of a height.
Use the Pythagorean theorem:
![leg^2+leg^2=hypotenuse^2](https://tex.z-dn.net/?f=leg%5E2%2Bleg%5E2%3Dhypotenuse%5E2)
We have:
![leg=5,\ leg=h,\ hypotenuse=20](https://tex.z-dn.net/?f=leg%3D5%2C%5C%20leg%3Dh%2C%5C%20hypotenuse%3D20)
Substitute:
![5^2+h^2=20^2](https://tex.z-dn.net/?f=5%5E2%2Bh%5E2%3D20%5E2)
<em>subtract 25 from both sides</em>
![h^2=375\to h=\sqrt{375}\\\\h=\sqrt{(25)(15)}\\\\h=\sqrt{25}\cdot\sqrt{15}\\\\h=5\sqrt{15}\ cm](https://tex.z-dn.net/?f=h%5E2%3D375%5Cto%20h%3D%5Csqrt%7B375%7D%5C%5C%5C%5Ch%3D%5Csqrt%7B%2825%29%2815%29%7D%5C%5C%5C%5Ch%3D%5Csqrt%7B25%7D%5Ccdot%5Csqrt%7B15%7D%5C%5C%5C%5Ch%3D5%5Csqrt%7B15%7D%5C%20cm)
Calculate the area:
![A_\triangle=\dfrac{(10)(5\sqrt{15})}{2}=\dfrac{50\sqrt{15}}{2}=25\sqrt{15}\ cm^2](https://tex.z-dn.net/?f=A_%5Ctriangle%3D%5Cdfrac%7B%2810%29%285%5Csqrt%7B15%7D%29%7D%7B2%7D%3D%5Cdfrac%7B50%5Csqrt%7B15%7D%7D%7B2%7D%3D25%5Csqrt%7B15%7D%5C%20cm%5E2)