Email:
password:
Sign Up

.solid geometry.
Angle : 14

Area : 1

Area of a rectangle, a triangle : 26

Circle Related : 19

Geometric visualization : 5

Length, Pythagoras theorem : 21

Ratio : 14

special triagle : 2

Triangle Inequality : 5

Volume of a cone, cube, and cylinder : 8


Imagination is more important than knowledge... Albert Einstein

                                                                                   Guess is more important than calculation --- Knowhowacademy.com

 

solid geometry - special triagle

For COMPETITION
Number of Total Problems: 2.
FOR PRINT ::: (Book)

Problem Num : 1
From : AMC10B
Type:
Section:solid geometry 
Theme:
Adjustment# : 0
Difficulty: 1
'

Six points are equally spaced around a circle of radius 1. Three of these points are the vertices of a triangle that is neither equilateral nor isosceles. What is the area of this triangle? extbf{(A)} frac{sqrt{3}}{3}qquad extbf{(B)} frac{sqrt{3}}{2}qquad extbf{(C)} extbf{1}qquad extbf{(D)} sqrt...

'
Category special triagle
Analysis

Solution/Answer

If there are no two points on the circle that are adjacent, then the triangle would be equilateral. If the three points are all adjacent, it would be isosceles. Thus, the only possibility is two adjacent points and one point two away. Because one of the sides of this triangle is the diameter, the opposite angle is a right angle. Also, because the two adjacent angles are one sixth of the circle apart, the angle opposite them is thirty degrees. This is a 30-60-90 triangle. If the original six points are connected, a regular hexagon is created. This hexagon consists of six equilateral triangles, so the radius is equal to one of its side lengths. The radius is 1, so the side opposite the thirty degree angle in the triangle is also 1. From rules with 30-60-90 triangles, the area is 1*sqrt3/2=oxed{ extbf{(B) } sqrt3/2}

Answer:


Problem Num : 2
From : AMC10B
Type:
Section:solid geometry 
Theme:
Adjustment# : 0
Difficulty: 1
'

In rectangle ABCD, AD=1, P is on overline{AB}, and overline{DB} and overline{DP} trisect angle ADC. What is the perimeter of riangle BDP?

draw((0,2)--(3.4,2)--(3.4,0)--(0,0)--cycle);draw((0,0)--(1.3,2));draw((0,0)--(3.4,2));dot((0,0));dot((0,2));dot((3.4,2));dot(...

mathrm{(A)} 3+frac{sqrt{3}}{3} qquadmathrm{(B)} 2+frac{4sqrt{3}}{3} qquadmathrm{(C)} 2+2sqrt{2} qquadmathrm{(D...

'
Category special triagle
Analysis

Solution/Answer

draw((0,2)--(3.4,2)--(3.4,0)--(0,0)--cycle);draw((0,0)--(1.3,2));draw((0,0)--(3.4,2));dot((0,0));dot((0,2));dot((3.4,2));dot(...

AD=1.

Since angle ADC is trisected, angle ADP= angle PDB= angle BDC=30^circ.

Thus, PD=frac{2sqrt{3}}{3}

DB=2

BP=sqrt{3}-frac{sqrt{3}}{3}=frac{2sqrt{3}}{3}.

Adding, 2+frac{4sqrt{3}}{3}.

oxed{ ext{B}}

Answer:


Array ( [0] => 8194 [1] => 8213 ) 2