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. Also the area of Quadrilateral is defined as the half of the product of the length of the diagonals
Triangle AMC is isosceles with AM = AC. Medians \(\overline {MV}\) and \(\overline {CU}\) are perpendicular to each other, and \(MV = CU =12 \) . What is the area of \(\triangle {AMC}\) ?
American Mathematics Competition 10 (AMC 10A), {2020}, {12}
Geometry - Area of Triangle
4 out of 10
Challenges and Thrills of Pre - College Mathematics
We can imagine the portion \(UVCM\) to be a quadrilateral having perpendicular diagonals .So its area can be found as half of the product of the length of the diagonals .
Again : - \(\triangle AUV \) has \(\frac {1}{4}\) of the triangle
\(AMC \) by similarity.
So, \(UVVM = \frac {3}{4} AMC \)
\(\frac {1}{2} .12.12 = \frac {3}{4} AMC \)
\(72 = \frac {3}{4} AMC \)
\(AMC = 96 \)
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You havee mentioned above:
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. Also the area of Quadrilateral is defined as the half of the product of the length of the diagonals
You should correct : last line that is :Also the area of Quadrilateral is defined as the half of the product of the length of the diagonals, It should be (1/2)* the product of the length of the diagonals* sin of angle between Two diagonols;
In the example it clicked because Angle between Two Diagonals=90°& Sin 90°=1;
Thank you for going through my post. In this sum as i am using the area of quadrilateral as =1/2 of the product of the length of the diagonals so in competency i have mentioned whatever i have used in the problem.I got your point but as triangle is the main focus here so i haven't mentioned the angle part in are of quadrilateral. Once again thank you.