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# Area of Triangle - AMC 10A - 2019 - Problem No. - 7

## 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 × b × 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 $\frac{1}{2}$ and 2 intersect at (2,2) . What is the area of the triangle enclosed by these two lines and the line $x + y = 10$ ?

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

## Use some hints

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.

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) .

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) .

## 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 × b × 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 $\frac{1}{2}$ and 2 intersect at (2,2) . What is the area of the triangle enclosed by these two lines and the line $x + y = 10$ ?

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

## Use some hints

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.

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) .

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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