<h2>
The area of a triangle is =54 square units</h2><h2>
The perpendicular distance from B to AC is = 
</h2>
Step-by-step explanation:
Given a triangle ABC with vertices A(2,1),B(12,2) and C(12,8)

The area of a triangle is= ![\frac{1}{2} [x_1(y_2-y_3) +x_2 (y_3- y_1)+x_3(y_1-y_2)]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%5Bx_1%28y_2-y_3%29%20%2Bx_2%20%28y_3-%20y_1%29%2Bx_3%28y_1-y_2%29%5D)
=![|\frac{1}{2} [2(2-8+12(8-1)+12(1-2)]|](https://tex.z-dn.net/?f=%7C%5Cfrac%7B1%7D%7B2%7D%20%5B2%282-8%2B12%288-1%29%2B12%281-2%29%5D%7C)
=
= 54 square units
The length of AC = 
= 
=
units
Let the perpendicular distance from B to AC be = x
According To Problem

⇔
units
Therefore the perpendicular distance from B to AC is = 
Answer: $12000 was invested at 7% while $8000 was invested at 9%.
Step-by-step explanation:
Based on the information given in the question, let the amount that was invested at 7% be n.
Therefore,
0.07n + 0.09(20000 - n) = 1560
0.07n + 1800 - 0.09n = 1560
-0.02n = 1560 - 1800
-0.02n = -240
n = 240/0.02
n = 120
Therefore, $12000 was invested at 7% while $8000 was invested at 9%.
Answer: a) Mean = 5, Median = 5
b) Mean = 15, Median = 5
c) Due to presence of outlier i.e. 99.
Step-by-step explanation:
Since we have given that
1,2,3,4,5,6,7,8,9
Here, n = 9 which is odd
So, Mean would be

Median = 
If 9 is replaced by 99,
1,2,3,4,5,6,7,8,99
So, mean would be

Median would be same as before i.e. 5
The mean is neither central nor typical for the data due to outlier i.e. 99
Hence, a) Mean = 5, Median = 5
b) Mean = 15, Median = 5
c) Due to presence of outlier i.e. 99.
A right triangle. Hope this helps.
There are 100 cm(centimeters) in a m(meter).
So, to solve this, we divide 250 by 100. this gives us 2.5meters.
1500÷100=15meters
The rule is m=cm÷100.