AMC 10 Daily Practice Round 4

Complete problem set with solutions and individual problem pages

Problem 4 Easy

A rectangle A B C D has A B=8 and B C=4. Points P and Q lie on sides \overline{A B} and \overline{B C}, respectively, such that A P=C Q and the area of \triangle B P Q is 6. What is P Q^2?

  • A.

    32

  • B.

    34

  • C.

    36

  • D.

    38

  • E.

    40

Answer:E

Since the area of \triangle B P Q is 6, we get that \frac{B P \cdot B Q}{2}=6. Thus, B P \cdot B Q=12. Let A P=Q C=x. Then B P=4-x and B Q=8-x, so (4-x)(8-x)=12. Expanding and factoring gives (x-2)(x-10)=0, so either x=2 or x=10.

If x=10, then B P=-6 and B Q=-2, which is impossible, so thus x=2. This gives B P=2 and B Q=6. Since A B C D is a rectangle, \angle B=90^{\circ}, so applying the Pythagorean Theorem on \triangle B P Q gives 2^2+6^2=P Q^2. Thus, P Q^2=4+36=40.

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Problem 2 AMC 10 Daily Practice Round 4
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