How to find if two acute / obtuse triangles are similar - ISEE Upper Level Quantitative Reasoning

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Question

NOTE: Figures NOT drawn to scale.

Refer to the above two triangles.

$\Delta ABC \sim \Delta DEF ,AB = 25, DE = 20, AC = 20$

What is ?

Answer

Corresponding sides of similar triangle are proportional, so if

$\Delta ABC \sim \Delta DEF$ , then

$\frac{DF}{AC} = \frac{DE}{AB}$

Substitute the known sidelengths, then solve for :

$\frac{DF}{20} = \frac{20}{25}$

$\frac{DF}{20}\cdot 20= \frac{20}{25}\cdot 20$

$DF = \frac{400}{25}$

Compare your answer with the correct one above

Question

$\Delta ABC \sim \Delta DEF$

What is the perimeter of $\Delta DEF$ ?

Answer

By definition, since, $\Delta ABC \sim \Delta DEF$ , side lengths are in proportion.

So,

$\frac{DE}{AB}= \frac{DF}{AC}$

$\frac{DE}{15}= \frac{36}{12} = 3$

$\frac{DE}{15} \cdot 15 = 3\cdot 15$

$\frac{EF}{BC}= \frac{DF}{AC}$

$\frac{EF}{21}= \frac{36}{12} = 3$

$\frac{EF}{21} \cdot 21 = 3\cdot 21$

The perimeter of $\Delta DEF$ is

.

Compare your answer with the correct one above

Question

$\Delta ABC \sim \Delta DEF$

What is ?

Answer

By definition, since $\Delta ABC \sim \Delta DEF$ , all side lengths are in proportion.

$\frac{AC}{DF} = \frac{AB}{DE}$

$\frac{AC}{66} = \frac{15}{45}$

$\frac{AC}{66} = \frac{1}{3}$

$AC = \frac{1}{3} \cdot 66 = 22$

Compare your answer with the correct one above

Question

NOTE: Figures NOT drawn to scale.

Refer to the above two triangles.

$\Delta ABC \sim \Delta DEF ,AB = 25, DE = 20, AC = 20$

What is ?

Answer

Corresponding sides of similar triangle are proportional, so if

$\Delta ABC \sim \Delta DEF$ , then

$\frac{DF}{AC} = \frac{DE}{AB}$

Substitute the known sidelengths, then solve for :

$\frac{DF}{20} = \frac{20}{25}$

$\frac{DF}{20}\cdot 20= \frac{20}{25}\cdot 20$

$DF = \frac{400}{25}$

Compare your answer with the correct one above

Question

$\Delta ABC \sim \Delta DEF$

What is the perimeter of $\Delta DEF$ ?

Answer

By definition, since, $\Delta ABC \sim \Delta DEF$ , side lengths are in proportion.

So,

$\frac{DE}{AB}= \frac{DF}{AC}$

$\frac{DE}{15}= \frac{36}{12} = 3$

$\frac{DE}{15} \cdot 15 = 3\cdot 15$

$\frac{EF}{BC}= \frac{DF}{AC}$

$\frac{EF}{21}= \frac{36}{12} = 3$

$\frac{EF}{21} \cdot 21 = 3\cdot 21$

The perimeter of $\Delta DEF$ is

.

Compare your answer with the correct one above

Question

$\Delta ABC \sim \Delta DEF$

What is ?

Answer

By definition, since $\Delta ABC \sim \Delta DEF$ , all side lengths are in proportion.

$\frac{AC}{DF} = \frac{AB}{DE}$

$\frac{AC}{66} = \frac{15}{45}$

$\frac{AC}{66} = \frac{1}{3}$

$AC = \frac{1}{3} \cdot 66 = 22$

Compare your answer with the correct one above

Question

NOTE: Figures NOT drawn to scale.

Refer to the above two triangles.

$\Delta ABC \sim \Delta DEF ,AB = 25, DE = 20, AC = 20$

What is ?

Answer

Corresponding sides of similar triangle are proportional, so if

$\Delta ABC \sim \Delta DEF$ , then

$\frac{DF}{AC} = \frac{DE}{AB}$

Substitute the known sidelengths, then solve for :

$\frac{DF}{20} = \frac{20}{25}$

$\frac{DF}{20}\cdot 20= \frac{20}{25}\cdot 20$

$DF = \frac{400}{25}$

Compare your answer with the correct one above

Question

$\Delta ABC \sim \Delta DEF$

What is the perimeter of $\Delta DEF$ ?

Answer

By definition, since, $\Delta ABC \sim \Delta DEF$ , side lengths are in proportion.

So,

$\frac{DE}{AB}= \frac{DF}{AC}$

$\frac{DE}{15}= \frac{36}{12} = 3$

$\frac{DE}{15} \cdot 15 = 3\cdot 15$

$\frac{EF}{BC}= \frac{DF}{AC}$

$\frac{EF}{21}= \frac{36}{12} = 3$

$\frac{EF}{21} \cdot 21 = 3\cdot 21$

The perimeter of $\Delta DEF$ is

.

Compare your answer with the correct one above

Question

$\Delta ABC \sim \Delta DEF$

What is ?

Answer

By definition, since $\Delta ABC \sim \Delta DEF$ , all side lengths are in proportion.

$\frac{AC}{DF} = \frac{AB}{DE}$

$\frac{AC}{66} = \frac{15}{45}$

$\frac{AC}{66} = \frac{1}{3}$

$AC = \frac{1}{3} \cdot 66 = 22$

Compare your answer with the correct one above

Question

NOTE: Figures NOT drawn to scale.

Refer to the above two triangles.

$\Delta ABC \sim \Delta DEF ,AB = 25, DE = 20, AC = 20$

What is ?

Answer

Corresponding sides of similar triangle are proportional, so if

$\Delta ABC \sim \Delta DEF$ , then

$\frac{DF}{AC} = \frac{DE}{AB}$

Substitute the known sidelengths, then solve for :

$\frac{DF}{20} = \frac{20}{25}$

$\frac{DF}{20}\cdot 20= \frac{20}{25}\cdot 20$

$DF = \frac{400}{25}$

Compare your answer with the correct one above

Question

$\Delta ABC \sim \Delta DEF$

What is the perimeter of $\Delta DEF$ ?

Answer

By definition, since, $\Delta ABC \sim \Delta DEF$ , side lengths are in proportion.

So,

$\frac{DE}{AB}= \frac{DF}{AC}$

$\frac{DE}{15}= \frac{36}{12} = 3$

$\frac{DE}{15} \cdot 15 = 3\cdot 15$

$\frac{EF}{BC}= \frac{DF}{AC}$

$\frac{EF}{21}= \frac{36}{12} = 3$

$\frac{EF}{21} \cdot 21 = 3\cdot 21$

The perimeter of $\Delta DEF$ is

.

Compare your answer with the correct one above

Question

$\Delta ABC \sim \Delta DEF$

What is ?

Answer

By definition, since $\Delta ABC \sim \Delta DEF$ , all side lengths are in proportion.

$\frac{AC}{DF} = \frac{AB}{DE}$

$\frac{AC}{66} = \frac{15}{45}$

$\frac{AC}{66} = \frac{1}{3}$

$AC = \frac{1}{3} \cdot 66 = 22$

Compare your answer with the correct one above

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