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solid geometry - Length, Pythagoras theorem

SOURCE:COMPETITION
Number of Problems: 21.
FOR PRINT ::: (Book)

Problem Num : 1

Type: None

Topic:solid geometry

Theme:None
Adjustment# :

Difficulty: 1

Category Length, Pythagoras theorem

Analysis

Solution/Hint

Problem Num : 2

Type: None

Topic:solid geometry

Theme:None
Adjustment# :

Difficulty: 1

Category Length, Pythagoras theorem

Analysis

Solution/Hint

Problem Num : 3

Type: None

Topic:solid geometry

Theme:None
Adjustment# :

Difficulty: 1

Category Length, Pythagoras theorem

Analysis

Solution/Hint

Problem Num : 4

Type: None

Topic:solid geometry

Theme:None
Adjustment# :

Difficulty: 1

Category Length, Pythagoras theorem

Analysis

Solution/Hint

Problem Num : 5

Type: Complex

Topic:solid geometry

Theme:Euqivalence
Adjustment# :

Difficulty: 2

Category Length, Pythagoras theorem

Analysis

Solution/Hint

Problem Num : 6

Type: None

Topic:solid geometry

Theme:None
Adjustment# :

Difficulty: 1

Category Length, Pythagoras theorem

Analysis

Solution/Hint

Problem Num : 7

Type: Complex

Topic:solid geometry

Theme:None
Adjustment# :

Difficulty: 2

Category Length, Pythagoras theorem

Analysis

Solution/Hint

Problem Num : 8

Type: Complex

Topic:solid geometry

Theme:None
Adjustment# :

Difficulty: 2

Category Length, Pythagoras theorem

Analysis

Solution/Hint

Problem Num : 9

Type:

Topic:solid geometry

Adjustment# : 0

Difficulty: 1

'
In trapezoid $ABCD$ , $overline{AB}$ and $overline{CD}$ are perpendicular to $overline{AD}$ , with $AB+CD=BC$ , $AB<CD$ , and $AD=7$ . What is $ABcdot CD$ ?
$extbf{(A)} 12 qquad extbf{(B)} 12.25 qquad extbf{(C)} 12.5 qquad extbf{(D)} 12.75 qquad extbf{(E)} 13$
'

Category Length, Pythagoras theorem

Analysis

Solution/Hint
If $AB=x$ and $CD=y$ , we have $BC=x+y$ .
By the Pythagorean theorem, we have $(x+y)^2=(y-x)^2+49$
Solving the equation, we get $4xy=49 implies xy = oxed{ extbf{(B)} 12.25}$ .

Problem Num : 10

Type:

Topic:solid geometry

Adjustment# : 0

Difficulty: 1

'
In $riangle ABC$ , we have $AC=BC=7$ and $AB=2$ . Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD=8$ . What is $BD$ ?
$mathrm{(A)} 3 qquadmathrm{(B)} 2sqrt{3} qquadmathrm{(C)} 4 qquadmathrm{(D)} 5 qquadmathrm{(...$
'

Category Length, Pythagoras theorem

Analysis

Solution/Hint
Draw height $CH$ . We have that $BH=1$ . From the Pythagorean Theorem, $CH=sqrt{48}$ . Since $CD=8$ , $HD=sqrt{8^2-48}=sqrt{16}=4$ , and $BD=HD-1$ , so $BD=3 ightarrowoxed{A}$ .

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