2021 AMC 10 B Fall

Complete problem set with solutions and individual problem pages

Problem 13 Medium

A square with side length 3is inscribed in an isosceles triangle with one side of the square along the base of the triangle. A square with side length 2 has two vertices on the other square and the other two on sides of the triangle, as shown. What is the area of the triangle?(2021 AMC Fall 10B, Question #13)

  • A.

    19 \frac{1}{4}

  • B.

    20 \frac{1}{4}

  • C.

    21 \frac{3}{4}

  • D.

    22 \frac{1}{2}

  • E.

    23 \frac{3}{4}

Answer:B

Solution 1:

Let's split the triangle down the middle and label it:

We see that \triangle A D G \sim \triangle B E G \sim \triangle C F G by AA similarity. B E=\frac{3}{2} because A K cuts the side length of the square in half; similarly, C F=1. Let C G=h : then by side ratios, \frac{h+2}{h}=\frac{\frac{3}{2}}{1} \Longrightarrow 2(h+2)=3 h \Longrightarrow h=4 Now the height of the triangle is A G=4+2+3=9. By side ratios, \frac{9}{4}=\frac{A D}{1} \Longrightarrow A D=\frac{9}{4} The area of the triangle is A G \cdot A D=9 \cdot \frac{9}{4}=\frac{81}{4}=B

Solution 2:

By similarity, the height is 3+\frac{3}{1} \cdot 2=9 and the base is \frac{9}{2} \cdot 1=4.5. Thus the area is \frac{9 \cdot 4.5}{2}=20.25=20 \frac{1}{4}, or (\text{B})

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