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solid geometry - Ratio

For COMPETITION
Number of Total Problems: 14.
FOR PRINT ::: (Book)

Problem Num : 11

From : AMC10

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
Trapezoid $ABCD$ has bases $overline{AB}$ and $overline{CD}$ and diagonals intersecting at $K$ . Suppose that $AB = 9$ , $DC = 12$ , and the area of $riangle AKD$ is $24$ . What is the area of trapezoid $ABCD$ ?
$mathrm{(A)} 92qquadmathrm{(B)} 94qquadmathrm{(C)} 96qquadmathrm{(D)} 98 qquadmathrm{(E)} 100$
'

Category Ratio

Analysis

Solution/Answer
Since $overline{AB} parallel overline{DC}$ it follows that $riangle ABK sim riangle CDK$ . Thus $frac{KA}{KC} = frac{KB}{KD} = frac{AB}{DC} = frac{3}{4}$ .
We now introduce the concept of area ratios: given two triangles that share the same height, the ratio of the areas is equal to the ratio of their bases. Since $riangle AKB, riangle AKD$ share a common altitude to $overline{BD}$ , it follows that (we let $[ riangle ldots]$ denote the area of the triangle) $frac{[ riangle AKB]}{[ riangle AKD]} = frac{KB}{KD} = frac{3}{4}$ , so $[ riangle AKB] = frac{3}{4}(24) = 18$ . Similarly, we find $[ riangle DKC] = frac{4}{3}(24) = 32$ and $[ riangle BKC] = 24$ .
Therefore, the area of $ABCD = [AKD] + [AKB] + [BKC] + [CKD] = 24 + 18 + 24 + 32 = 98 mathrm{(D)}$ .

Answer:

Problem Num : 12

From : AMC10

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
Triangle $ABC$ has a right angle at $B$ . Point $D$ is the foot of the altitude from $B$ , $AD=3$ , and $DC=4$ . What is the area of $riangle ABC$ ?

$mathrm{(A)} 4sqrt3qquadmathrm{(B)} 7sqrt3qquadmathrm{(C)} 21qquadmathrm{(D)} 14sqrt3 qquadmathrm{(E)} 42$
'

Category Ratio

Analysis

Solution/Answer
It is a well-known fact that in any right triangle $ABC$ with the right angle at $B$ and $D$ the foot of the altitude from $B$ onto $AC$ we have $BD^2 = ADcdot CD$ . (See below for a proof.) Then $BD = sqrt{ 3cdot 4 } = 2sqrt 3$ , and the area of the triangle $ABC$ is $frac{ACcdot BD}2 = 7sqrt3Rightarrowoxed{ ext{(B)}}$ .
Proof: Consider the Pythagorean theorem for each of the triangles $ABC$ , $ABD$ , and $CBD$ . We get:

$AB^2 + BC^2 = AC^2 = (AD+DC)^2 = AD^2 + DC^2 + 2 cdot AD cdot DC$ .
$AB^2 = AD^2 + BD^2$
$BC^2 = BD^2 + CD^2$

Substituting equations 2 and 3 into the left hand side of equation 1, we get $BD^2 = AD cdot DC$ .
Alternatively, note that $riangle ABD sim riangle BCD Longrightarrow frac{AD}{BD} = frac{BD}{CD}$ . $lacksquare$

Answer:

Problem Num : 13

From : AMC10B

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

' The area of $riangle$ $EBD$ is one third of the area of $3-4-5$ $riangle$ $ABC$ . Segment $DE$ is perpendicular to segment $AB$ . What is $BD$ ?

$extbf{(A)} frac{4}{3} qquad extbf{(B)} sqrt{5} qquad extbf{(C)} frac{9}{4} qquad extbf{(D)} frac{4sqrt{3}}{3...$
'

Category Ratio

Analysis

Solution/Answer
Answer:

Problem Num : 14

From : AMC10

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
Externally tangent circles with centers at points A and B have radii of lengths 5 and 3, respectively. A line externally tangent to both circles intersects ray AB at point C. What is BC?
$extbf{(A)} 4qquad extbf{(B)} 4.8qquad extbf{(C)} 10.2qquad extbf{(D)} 12qquad extbf{(E)} 14.4$
'

Category Ratio

Analysis

Solution/Answer
Let $D$ and $E$ be the points of tangency on circles $A$ and $B$ with line $CD$ . $AB=8$ . Also, let $BC=x$ . As $angle ADC$ and $angle BEC$ are right angles (a radius is perpendicular to a tangent line at the point of tangency) and both triangles share $angle ACD$ , $riangle ADC sim riangle BCE$ . From this we can get a proportion.
$frac{BC}{AC}=frac{BE}{AD} ightarrow frac{x}{x+8}=frac{3}{5} ightarrow 5x=3x+24 ightarrow x=oxed{ extbf{(D)} 12}$

Answer:

Array ( [0] => 7878 [1] => 7890 [2] => 8148 [3] => 7929 ) 141 2