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SSAT Upper Level Math : How to find if right triangles are congruent

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Example Questions

Example Question #1 : How To Find If Right Triangles Are Congruent

Given:

$\bigtriangleup ABC$ , where $\angle B$ is a right angle; AB = 10; BC = 24 ;

$\bigtriangleup DEF$ , where $\angle E$ is a right angle and DF = 28 ;

$\bigtriangleup GHI$ , where $\angle H$ is a right angle and $\bigtriangleup GHI$ has perimeter 60;

$\bigtriangleup JKL$ , where $\angle K$ is a right angle and $\bigtriangleup JKL$ has area 120;

$\bigtriangleup MNO$ , where $\angle N$ is a right triangle and $\angle M \cong \angle A$

Which of the following must be a false statement?

Possible Answers:

$\bigtriangleup JKL \cong \bigtriangleup ABC$

All of the statements given in the other responses are possible

$\bigtriangleup DEF \cong \bigtriangleup ABC$

$\bigtriangleup GHI \cong \bigtriangleup ABC$

$\bigtriangleup MNO \cong \bigtriangleup ABC$

Correct answer:

$\bigtriangleup DEF \cong \bigtriangleup ABC$

Explanation:

$\bigtriangleup ABC$ has as its leg lengths 10 and 24, so the length of its hypotenuse, $\overline{AC}$ , is

$AC = \sqrt{(AB)^{2}+(BC)^{2}} = \sqrt{10^{2}+24^{2}} = \sqrt{100+576} = \sqrt{676 } = 26$

Its perimeter is the sum of its sidelengths:

P= 10+24+26 = 60

Its area is half the product of the lengths of its legs:

$A = \frac{1}{2} \cdot 10 \cdot 24 = 120$

$\bigtriangleup GHI$ and $\bigtriangleup JKL$ have the same perimeter and area, respectively, as $\bigtriangleup ABC$ ; also, between $\bigtriangleup MNO$ and $\bigtriangleup ABC$ , corresponding angles are congruent. In the absence of other information, none of these three triangles can be eliminated as being congruent to $\bigtriangleup ABC$ .

However, DF = 28 and AC = 26 . Therefore, $\overline{AC} \ncong \overline{DF}$ . Since a pair of corresponding sides is noncongruent, it follows that $\bigtriangleup DEF \ncong \bigtriangleup ABC$ .

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Example Question #2 : How To Find If Right Triangles Are Congruent

Given: $\bigtriangleup ABC$ and $\bigtriangleup DEF$ with right angles $\angle B$ and $\angle E$ ; $\angle A \cong \angle D$ .

Which of the following statements alone, along with this given information, would prove that $\bigtriangleup ABC \cong \bigtriangleup DEF$ ?

I) $\overline{AB } \cong \overline{DE}$

II) $\overline{BC }\cong \overline{EF}$

III) $\overline{AC } \cong \overline{DF}$

Possible Answers:

I or III only

III only

II or III only

Any of I, II, or III

I or II only

Correct answer:

Any of I, II, or III

Explanation:

$\angle A \cong \angle D$ ; $\angle B \cong \angle E$ since both are right angles.

Given that two pairs of corresponding angles are congruent and any one side of corresponding sides is congruent, it follows that the triangles are congruent. In the case of Statement I, the included sides are congruent, so by the Angle-Side-Angle Congruence Postulate, $\bigtriangleup ABC \cong \bigtriangleup DEF$ . In the case of the other two statements, a pair of nonincluded sides are congruent, so by the Angle-Angle-Side Congruence Theorem, $\bigtriangleup ABC \cong \bigtriangleup DEF$ . Therefore, the correct choice is I, II, or III.

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Example Question #51 : Properties Of Triangles

$\bigtriangleup ABC \cong \bigtriangleup DEF$ , where $\angle B$ is a right angle, AC = 20 , and AB = BC .

Which of the following is true?

Possible Answers:

$\bigtriangleup DEF$ has area 100

$\bigtriangleup DEF$ has perimeter 40

None of the statements given in the other choices is true.

$m \angle F = 60^{\circ }$

$DE = 20\sqrt{2}$

Correct answer:

$\bigtriangleup DEF$ has area 100

Explanation:

$\bigtriangleup ABC \cong \bigtriangleup DEF$ , and corresponding parts of congruent triangles are congruent.

Since $\angle B$ is a right angle, so is $\angle E$ . DE = AB and EF= BC ; since AB = BC , it follows that DE = EF . $\bigtriangleup DEF$ is an isosceles right triangle; consequently, $m \angle A = m \angle C = 45^{\circ }$ .

$\bigtriangleup DEF$ is a 45-45-90 triangle with hypotenuse of length DF= AC = 20 . By the 45-45-90 Triangle Theorem, the length of each leg is equal to that of the hypotenuse divided by $\sqrt{2}$ ; therefore,

$DE=EF = \frac{DF}{\sqrt{2}} = \frac{20}{\sqrt{2}} =\frac{20 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{20 \sqrt{2}}{2} = 10 \sqrt{2}$

$DE = 20\sqrt{2}$ is eliminated as the correct choice.

Also, the perimeter of $\bigtriangleup DEF$ is

$P = DE+EF+DF = 10 \sqrt{2} + 10 \sqrt{2} + 20 = 20+ 20 \sqrt{2} \ne 40$ .

This eliminates the perimeter of $\bigtriangleup DEF$ being 40 as the correct choice.

Also, $m \angle F = 60^{\circ }$ is eliminated as the correct choice, since the triangle is 45-45-90.

The area of $\bigtriangleup DEF$ is half the product of the lengths of its legs:

$A = \frac{1}{2} \cdot DE \cdot EF$

$= \frac{1}{2} \cdot 10 \sqrt{2}\cdot 10 \sqrt{2}$

$= \frac{1}{2} \cdot 10\cdot 10 \cdot \sqrt{2} \cdot \sqrt{2}$

$= \frac{1}{2} \cdot 10\cdot 10 \cdot 2$

= 100

The correct choice is the statement that $\bigtriangleup DEF$ has area 100.

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SSAT Upper Level Math : How to find if right triangles are congruent

Study concepts, example questions & explanations for SSAT Upper Level Math

All SSAT Upper Level Math Resources

Example Questions

Example Question #1 : How To Find If Right Triangles Are Congruent

Example Question #2 : How To Find If Right Triangles Are Congruent

Example Question #51 : Properties Of Triangles

All SSAT Upper Level Math Resources

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