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ISEE Upper Level Quantitative : How to find if triangles are similar

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Example Questions

Example Question #1 : How To Find If Triangles Are Similar

$\Delta ABC$ $\displaystyle \Delta ABC$ and $\Delta DEF$ $\displaystyle \Delta DEF$ are right triangles, with right angles $\angle B , \angle E$ $\displaystyle \angle B , \angle E$ , respectively. $m \angle A = m\angle D = 45^{\circ }$ $\displaystyle m \angle A = m\angle D = 45^{\circ }$ and AC > DF $\displaystyle AC > DF$ .

Which is the greater quantity?

(a) $\displaystyle AB$

(b) $\displaystyle EF$

Possible Answers:

(a) is greater.

(a) and (b) are equal.

It is impossible to tell from the information given.

(b) is greater.

Correct answer:

(a) is greater.

Explanation:

Each right triangle is a $45 ^{\circ }-45 ^{\circ }-90 ^{\circ }$ $\displaystyle 45 ^{\circ }-45 ^{\circ }-90 ^{\circ }$ triangle, making each triangle isosceles by the Converse of the Isosceles Triangle Theorem.

Since $\angle B$ $\displaystyle \angle B$ and $\angle E$ $\displaystyle \angle E$ are the right triangles, the legs are $\overline{AB}, \overline{BC}, \overline{DE}, \overline{EF}$ $\displaystyle \overline{AB}, \overline{BC}, \overline{DE}, \overline{EF}$ , and the hypotenuses are $\overline{AC}, \overline{DF}$ $\displaystyle \overline{AC}, \overline{DF}$ .

By the $45 ^{\circ }-45 ^{\circ }-90 ^{\circ }$ $\displaystyle 45 ^{\circ }-45 ^{\circ }-90 ^{\circ }$ Theorem, $AB \sqrt{2} = AC$ $\displaystyle AB \sqrt{2} = AC$ and $EF \sqrt{2} = DF$ $\displaystyle EF \sqrt{2} = DF$ .

AC > DF $\displaystyle AC > DF$ , so $AB \sqrt{2} > EF \sqrt{2}$ $\displaystyle AB \sqrt{2} > EF \sqrt{2}$ and subsequently, AB > EF $\displaystyle AB > EF$ .

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Example Question #2 : How To Find If Triangles Are Similar

Untitled

Figure NOT drawn to scale.

In the above figure, $\angle ABC$ $\displaystyle \angle ABC$ is a right angle.

What is the ratio of the area of $\bigtriangleup AXB$ $\displaystyle \bigtriangleup AXB$ to that of $\bigtriangleup BXC$ $\displaystyle \bigtriangleup BXC$ ?

Possible Answers:

169 to 25

12 to 5

144 to 25

13 to 5

Correct answer:

144 to 25

Explanation:

The altitude of a right triangle from the vertex of its right angle divides the triangle into two smaller, similar triangles.

The similarity ratio of $\bigtriangleup AXB$ $\displaystyle \bigtriangleup AXB$ to $\bigtriangleup BXC$ $\displaystyle \bigtriangleup BXC$ can be found by determining the ratio of one pair of corresponding sides; we will use the short leg of each, $\overline{XB}$ $\displaystyle \overline{XB}$ and $\overline{XC}$ $\displaystyle \overline{XC}$ .

$\overline{XB}$ $\displaystyle \overline{XB}$ is also the long leg of $\bigtriangleup BXC$ $\displaystyle \bigtriangleup BXC$ , so its length can be found using the Pythagorean Theorem:

$(BX)^{2} = (BC)^{2} - (CX)^{2}$ $\displaystyle (BX)^{2} = (BC)^{2} - (CX)^{2}$

$(BX)^{2} = 13^{2} - 5^{2} = 169 - 25 = 144$ $\displaystyle (BX)^{2} = 13^{2} - 5^{2} = 169 - 25 = 144$

$BX = \sqrt{144} = 12$ $\displaystyle BX = \sqrt{144} = 12$

The similarity ratio is therefore

$\frac{XB}{XC} = \frac{12}{5}$ $\displaystyle \frac{XB}{XC} = \frac{12}{5}$ .

The ratio of the areas is the square of this ratio:

$\left ( \frac{12}{5} \right )^{2} = \frac{12^{2}}{5^{2}} = \frac{144}{25}$ $\displaystyle \left ( \frac{12}{5} \right )^{2} = \frac{12^{2}}{5^{2}} = \frac{144}{25}$ - that is, 144 to 25.

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ISEE Upper Level Quantitative : How to find if triangles are similar

Study concepts, example questions & explanations for ISEE Upper Level Quantitative

All ISEE Upper Level Quantitative Resources

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Example Question #1 : How To Find If Triangles Are Similar

Example Question #2 : How To Find If Triangles Are Similar

All ISEE Upper Level Quantitative Resources

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