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2021 AMC Fall 10B Problem 13

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2021 AMC Fall 10B Problem 13AMC 10 Step by Step
Name:                             Date: September 13, 2026
Similar & Congruent Triangles · difficulty 3 of 5 · about 4 min · position P11-15

A square with side length 33 is inscribed in an isosceles triangle with one side of the square along the base of the triangle. A square with side length 22 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?

figure

  1. A)

    191419\frac14

  2. B)

    201420\frac14

  3. C)

    213421 \frac34

  4. D)

    221222\frac12

  5. E)

    233423\frac34

Answer

B

Key insight

The triangle's width shrinks linearly with height: it is 3 at height 3 and 2 at height 5, so it is 9/2 at height 0 and 0 at height 9.

Solution

Measure heights from the base. A horizontal slice of the triangle at height hh has some width w(h)w(h); because the two slanted sides are straight lines, ww is a linear function of hh.

The big square's top corners lie on the triangle's sides at height 33, so w(3)=3w(3) = 3. The small square sits on top of the big one, so its top corners are at height 3+2=53 + 2 = 5 and lie on the sides: w(5)=2w(5) = 2.

Going up 22 units shrinks the width by 11, so the width shrinks 12\frac12 per unit of height. Therefore
w(0)=3+3⋅12=92,w=0 at height 3+3⋅2=9.w(0) = 3 + 3 \cdot \tfrac12 = \tfrac92, \qquad w = 0 \text{ at height } 3 + 3 \cdot 2 = 9.
The triangle has base 92\frac92 and height 99, so its area is
12⋅92⋅9=814=2014.\frac12 \cdot \frac92 \cdot 9 = \frac{81}{4} = 20\tfrac14.

The answer is (B) 2014\boxed{\textbf{(B)}\ 20\frac14}.

Why this works

Every horizontal cross-section of a triangle is similar to its base, so width is linear in height; two measured widths determine the whole triangle. This "linear taper" viewpoint packages the similar-triangles argument into a single slope computation and works for any stack of inscribed rectangles.

Alternative approach

Similar triangles explicitly: the small triangle above height 55 has base 22; the triangle above height 33 has base 33. Their heights are in ratio 2:32 : 3 and differ by 22, so they are 44 and 66. The apex is at height 3+6=93 + 6 = 9, and the full triangle is similar with height 99 and base 3⋅96=923 \cdot \frac{9}{6} = \frac92.

The trap

Assuming the small square's top corners are at height 6 or that the apex sits directly above the big square's top corners, instead of using the two known widths.

Common mistakes

  • Assuming the small square's top corners are at height 6 or that the apex sits directly above the big square's top corners, instead of using the two known widths.
  • Computing the base as 3+2⋅12=43 + 2 \cdot \frac12 = 4 by tapering over the wrong height interval.

Techniques

Add construction lines/points (drop altitudes, extend segments, connect centers) · Set up the equation/formula and compute; no special trick needed