AMC 10

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

The simplest example of Area of Triangle from 2011 AMC 10B-Problem 9. Learn in this self-learning module for math olympiad.We may use sequential hints.

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) $?

[asy] unitsize(10mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4;  pair A=(0,0), B=(5,0), C=(1.8,2.4), D=(5-4sqrt(3)/3,0), E=(5-4sqrt(3)/3,sqrt(3)); pair[] ps={A,B,C,D,E};  draw(A--B--C--cycle); draw(E--D); draw(rightanglemark(E,D,B));  dot(ps); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,S); label("$E$",E,NE); label("$3$",midpoint(A--C),NW); label("$4$",midpoint(C--B),NE); label("$5$",midpoint(A--B),SW); [/asy]
$textbf{(A)} frac{4}{3} qquadtextbf{(B)} sqrt{5} qquadtextbf{(C)} frac{9}{4} qquadtextbf{(D)} frac{4sqrt{3}}{3} qquadtextbf{(E)} frac{5}{2}$

2011 AMC 10B-Problem 9


6 out of 10

Challenges and Thrills in Pre College Mathematics

Knowledge Graph

area of Triangle-Knowledge Graph

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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