How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?
Learn More

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) $?

[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= $BD$=$\frac{4 \sqrt3}{3}$

Subscribe to Cheenta at Youtube

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.