Answer: The area of ABC is 56 m².
Explanation:
It is given that in △ABC, point P∈ AB is so that AP:BP=1:3 and point M is the midpoint of segment CP.
Since point P divides the line AB in 1:3, therefore the area of triangle APC and BPC is also in ratio 1:3. To prove this draw a perpendicular h on AB from C.
![\frac{\text{Area of } \triangle BCP}{\text{Area of } \triangle ABC} =\frac{\frac{1}{2}\times BP\times CH}{\frac{1}{2}\times AB\times CH} =\frac{BP}{AB}= \frac{3}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BArea%20of%20%7D%20%5Ctriangle%20BCP%7D%7B%5Ctext%7BArea%20of%20%7D%20%5Ctriangle%20ABC%7D%20%3D%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20BP%5Ctimes%20CH%7D%7B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20AB%5Ctimes%20CH%7D%20%3D%5Cfrac%7BBP%7D%7BAB%7D%3D%20%5Cfrac%7B3%7D%7B4%7D)
Since the area of BPC is
th part of total area, therefore area of APC is
th part of total area.
The point M is the midpoint of CP, therefore the area of BMP and BMC is equal by midpoint theorem.
![\text{Area of } \triangle BMP=\text{Area of } \triangle BMC](https://tex.z-dn.net/?f=%5Ctext%7BArea%20of%20%7D%20%5Ctriangle%20BMP%3D%5Ctext%7BArea%20of%20%7D%20%5Ctriangle%20BMC)
![21=\text{Area of } \triangle BMC](https://tex.z-dn.net/?f=21%3D%5Ctext%7BArea%20of%20%7D%20%5Ctriangle%20BMC)
Area of BPC is,
![\text{Area of } \triangle BPC=\text{Area of } \triangle BMP+\text{Area of } \triangle BMC](https://tex.z-dn.net/?f=%5Ctext%7BArea%20of%20%7D%20%5Ctriangle%20BPC%3D%5Ctext%7BArea%20of%20%7D%20%5Ctriangle%20BMP%2B%5Ctext%7BArea%20of%20%7D%20%5Ctriangle%20BMC)
![\text{Area of } \triangle BPC=21+21](https://tex.z-dn.net/?f=%5Ctext%7BArea%20of%20%7D%20%5Ctriangle%20BPC%3D21%2B21)
![\text{Area of } \triangle BPC=42](https://tex.z-dn.net/?f=%5Ctext%7BArea%20of%20%7D%20%5Ctriangle%20BPC%3D42)
Area of APC is,
![\text{Area of } \triangle APC=\frac{1}{3}\times \text{Area of } \triangle BPC](https://tex.z-dn.net/?f=%5Ctext%7BArea%20of%20%7D%20%5Ctriangle%20APC%3D%5Cfrac%7B1%7D%7B3%7D%5Ctimes%20%5Ctext%7BArea%20of%20%7D%20%5Ctriangle%20BPC)
![\text{Area of } \triangle APC=\frac{1}{3}\times 42](https://tex.z-dn.net/?f=%5Ctext%7BArea%20of%20%7D%20%5Ctriangle%20APC%3D%5Cfrac%7B1%7D%7B3%7D%5Ctimes%2042)
![\text{Area of } \triangle APC=14](https://tex.z-dn.net/?f=%5Ctext%7BArea%20of%20%7D%20%5Ctriangle%20APC%3D14)
Area of ABC is,
![\text{Area of } \triangle ABC=\text{Area of } \triangle APC+\text{Area of } \triangle BPC](https://tex.z-dn.net/?f=%5Ctext%7BArea%20of%20%7D%20%5Ctriangle%20ABC%3D%5Ctext%7BArea%20of%20%7D%20%5Ctriangle%20APC%2B%5Ctext%7BArea%20of%20%7D%20%5Ctriangle%20BPC)
![\text{Area of } \triangle ABC=14+42=56](https://tex.z-dn.net/?f=%5Ctext%7BArea%20of%20%7D%20%5Ctriangle%20ABC%3D14%2B42%3D56)
Therefore, the area of ABC is 56 m².
Answer:
t = 54 seconds
Step-by-step explanation:
The equation that models the path of a rocket into the air is given by :
![h(t)= -16t^2+864t](https://tex.z-dn.net/?f=h%28t%29%3D%20-16t%5E2%2B864t)
It is required to find the time after which the rocket will hit the ground. When it hits the ground, the height of the rocket will becomes zero. It means,
h(t) = 0
i.e.
![-16t^2+864t=0\\\\16t(-t+54)=0\\\\16t=0, -t+54=0\\\\t=0\ and\ t=54\ s](https://tex.z-dn.net/?f=-16t%5E2%2B864t%3D0%5C%5C%5C%5C16t%28-t%2B54%29%3D0%5C%5C%5C%5C16t%3D0%2C%20-t%2B54%3D0%5C%5C%5C%5Ct%3D0%5C%20and%5C%20t%3D54%5C%20s)
It means after 54 seconds, the rocket will hit the ground.
3x^2 + 6x - 10 = 0
x = [-6 +/- sqrt(^2 - 3*3*-10)] / 2*3
= 1.08 and -3.08
Other x -intercept is (-3.08,0)
Answer:
40%
Step-by-step explanation:
18+12+20=50
20/50×100=
40%