## What is Area of Triangle ?

*The area of a triangle is defined as the total space that is enclosed by any particular triangle. The basic formula to find the area of a given triangle is A = 1/2 Ã— b Ã— h, where b is the base and h is the height of the given triangle, whether it is scalene, isosceles or equilateral. *

## Try This Problem from AMC 10A – 2019 -Problem No.7

** Two lines with slopes **\(\frac{1}{2}\)

**\(x + y = 10 \) ?**

*and 2 intersect at (2,2) . What is the area of the triangle enclosed by these two lines and the line**A) 4 B) \(4\sqrt 2\) C) 6 D) 8 E) \(6 \sqrt 2\)*

*American Mathematics Competition 10 (AMC 10A), 2019, Problem Number – 7*

*Area of Triangle*

*6 out of 10*

*Problems in Plane Geometry by Sharygin*

## Knowledge Graph

## Use some hints

First Hint

*If you need a hint to start this sum use this *

*Lets try to find the slop – intercept form of all three lines : (x,y) = (2,2) and y =*

* *\(\frac{x}{2}+b\)** implies **\(2 = \frac{2}{2}+b = 1+b\)

*. So, b = 1 . While y = 2x + c implies 2 = 2.2 + c So, c = -2 And again x+y = 10 implies y = -x + 10.*Second Hint

*Thus the lines are \( y = \frac {x}{2} + 1 \) , y = 2x – 2 and y = -x + 10 . Now we find the intersection points between each of the lines with y = -x + 10 , which are (6,4) and (4,6) .*

Final Hint

** In the last hint we can apply the distance formula and then the Pythagorean Theorem, we see that we have an isosceles triangle where the base is \(2\sqrt 2\) and the height \(3 \sqrt 2\), whose area is 6 .The answer is 6** (c) .

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