Categories

## Largest Area of Triangle | AIME I, 1992 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Largest Area of Triangle.

## Area of Triangle – AIME I, 1992

Triangle ABC has AB=9 and BC:AC=40:41, find the largest area that this triangle can have.

• is 107
• is 820
• is 840
• cannot be determined from the given information

### Key Concepts

Ratio

Area

Triangle

AIME I, 1992, Question 13

Coordinate Geometry by Loney

## Try with Hints

Let the three sides be 9, 40x, 41x

area = $\frac{1}{4}\sqrt{(81^2-81x^2)(81x^2-1)} \leq \frac{1}{4}\frac{81^2-1}{2}$

or, $\frac{1}{4}\frac{81^2-1}{2}=\frac{1}{8}(81-1)(81+1)$

=(10)(82)

=820.

Categories

## What is Ratio ?

In mathematics, a ratio indicates how many times one number contains another. For example, if there are eight oranges and six lemons in a bowl of fruit, then the ratio of oranges to lemons is eight to six which is equivalent to 4:3.

## Try this Problem from AMC 10B – 2020 – Problem No.-3

The ratio of w to x, y to z and z to x are 4:3, 3:2 and 1:6 respectively . What is the ratio of w to y ?

A) 4:3 B) 3:2 C) 8:3 D) 4:1 E) 16 :3

American Mathematics Competition 10 (AMC 10B), 2020, Problem Number 3

Ratio

3 out of 10

Mathematics can be fun

## Use some hints

This one is an easy sum to solve but those who are confused right now can use the first hint :

The ratio of w : x = 4 : 3 ; so we can write it $\frac {w}{x} = \frac {4}{3}$. Similarly we can do it for the other given ratios. Try to do it……

I think you have already got the answer . If not then try this out …..

z : x = 1 : 6 i.e $\frac {z}{x} = \frac {1}{6}$

y : z = 3 : 2 i.e $\frac {y}{z} = \frac {3}{2}$

This hint is the final hint as already mentioned in header :

Lets multiply each ratios :

$\frac {w}{x} \times \frac {x}{z} \times \frac {z}{y} = \frac {4}{3} \times 6 \times \frac {2}{3}$

After canceling out all the similar terms $\frac {w}{y} = \frac {16}{3}$

Categories

## Ratio and Proportion

The given problem is based upon calculating the number of marbles in jars of specific color, to do so we have to use ratios of the marbles of different colors and use the ratio to calculate the actual number of marbles of required color.

## Try the problem

Two jars each contain the same number of marbles, and every marble is either blue or green. In Jar $1$ the ratio of blue to green marbles is $9:1$, and the ratio of blue to green marbles in Jar $2$ is $8:1$. There are $95$ green marbles in all. How many more blue marbles are in Jar $1$ than in Jar $2$?

$\textbf{(A) } 5\qquad\textbf{(B) } 10 \qquad\textbf{(C) }25 \qquad\textbf{(D) } 45 \qquad \textbf{(E) } 50$

2019 AMC 10B Problem 11

Ratio and proportion

6 out of 10

Secrets in Inequalities.

## Use some hints

Let $2x$ is the total no of marbles in both the jars. so each of the jar have $x$ marbles.

Thus, $\frac{x}{10}$ is the number of green marbles in Jar $1$, and $\frac{x}{9}$ is the number of green marbles in Jar $2$.

Since $\frac{x}{9}+\frac{x}{10}=\frac{19x}{90}$, we have $\frac{19x}{90}=95$, so there are $x=450$ marbles in each jar.

Since $\frac{9}{10}th$ of the jar 1 marbles are of blue color and $\frac{8}{9}th$ of the jar 2 marbles are of blue color.

Now we can easily find the no of blue marbles in both the jars and then we can subtract them to get the amount by which one exceed the other.