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Side of Triangle : AMC 10B, 2011 - Problem 9

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. Here we have to find the side of triangle.

Try the problem

The area of $\triangle (EBD)$ is one third of the area of $\triangle (ABC)$. Segment  is perpendicular to segment $(AB)$. What is $(BD)$?

2011 AMC 10B-Problem 9

Geometry

6 out of 10

Challenges and Thrills in Pre College Mathematics

Use some hints

Notice that here $\triangle ABC$ and $\triangle BDE$ are similar.

Therefore $DE$=$\frac{3}{4} BD$

Now we have to find the area.

We know this is = $\frac{1}{2}$ $\times$ base $\times$ height.

Now using this formula can you find the area of $\triangle ABC$ and $\triangle BDE$?

Now $\triangle ABC$ = $\frac{1}{2} \times 3 \times 4$ =$6$ and $\triangle BDE$= $\frac{1}{2} \times DE \times BD$ = $\frac{1}{2}$* $\frac{3}{4} \times BD$ $\times$ $BD$ = $\frac{3}{8} BD^2$

Again we know the area of $\triangle BDE$ is one third of the area of $\triangle ABC$ .

Therefore, $\frac{3}{8} BD^2$ = $6$ $\times$ $\frac{1}{3}$

$9*BD^2$=$48$ , or,$BD^2$=$\frac{48}{9}$ , or, $BD^2$=$\frac{16}{3}$

so, the answer is= $BD$=$\frac{4 \sqrt3}{3}$

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. Here we have to find the side of triangle.

Try the problem

The area of $\triangle (EBD)$ is one third of the area of $\triangle (ABC)$. Segment  is perpendicular to segment $(AB)$. What is $(BD)$?

2011 AMC 10B-Problem 9

Geometry

6 out of 10

Challenges and Thrills in Pre College Mathematics

Use some hints

Notice that here $\triangle ABC$ and $\triangle BDE$ are similar.

Therefore $DE$=$\frac{3}{4} BD$

Now we have to find the area.

We know this is = $\frac{1}{2}$ $\times$ base $\times$ height.

Now using this formula can you find the area of $\triangle ABC$ and $\triangle BDE$?

Now $\triangle ABC$ = $\frac{1}{2} \times 3 \times 4$ =$6$ and $\triangle BDE$= $\frac{1}{2} \times DE \times BD$ = $\frac{1}{2}$* $\frac{3}{4} \times BD$ $\times$ $BD$ = $\frac{3}{8} BD^2$

Again we know the area of $\triangle BDE$ is one third of the area of $\triangle ABC$ .

Therefore, $\frac{3}{8} BD^2$ = $6$ $\times$ $\frac{1}{3}$

$9*BD^2$=$48$ , or,$BD^2$=$\frac{48}{9}$ , or, $BD^2$=$\frac{16}{3}$

so, the answer is= $BD$=$\frac{4 \sqrt3}{3}$

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