1970 AHSME Problems/Problem 30

Problem

In the accompanying figure, segments $AB$ and $CD$ are parallel, the measure of angle $D$ is twice that of angle $B$ , and the measures of segments $AD$ and $CD$ are $a$ and $b$ respectively. Then the measure of $AB$ is equal to

$\text{(A) } \tfrac{1}{2}a+2b\quad \text{(B) } \tfrac{3}{2}b+\tfrac{3}{4}a\quad \text{(C) } 2a-b\quad \text{(D) } 4b-\tfrac{1}{2}a\quad \text{(E) } a+b$

Solution

$\fbox{E}$

1970 AHSC (Problems • Answer Key • Resources)
Preceded by [[1970 AHSME Problems/Problem {{{num-b}}}\|Problem {{{num-b}}}]]	Followed by Problem 31
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1970 AHSME Problems/Problem 30

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