Triangles

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ISEE Upper Level Quantitative Reasoning › Triangles

Questions 1 - 10

In a certain quadrilateral, three of the angles are $65^{\circ}$ , $120^{\circ}$ , and $34^{\circ}$ . What is the measure of the fourth angle?

$141^{\circ}$

$139^{\circ}$

$219^{\circ}$

$41^{\circ}$

$29^{\circ}$

Explanation

A quadrilateral has four angles totalling $360^{\circ}$ . So, first add up the three angles given. The sum is $219^{\circ}$ . Then, subtract that from 360. This gives you the missing angle, which is $141^{\circ}$ .

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

Which is the greater quantity?

(a)

(b)

(a) is greater.

(b) is greater.

(a) and (b) are equal.

It is impossible to tell from the information given.

Explanation

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

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

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

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

The lengths of the hypotenuses of ten similar right triangles form an arithmetic sequence. The smallest triangle has legs of lengths 5 and 12 inches; the second-smallest triangle has a hypotenuse of length one and one half feet.

Which of the following responses comes closest to the area of the largest triangle?

4 square feet

3 square feet

5 square feet

6 square feet

7 square feet

Explanation

The hypotenuse of the smallest triangle can be calculated using the Pythagorean Theorem:

$c = \sqrt{a^{2}+b^{2}} = \sqrt{5^{2}+12^{2}} = \sqrt{25+144} = \sqrt{169}=13$ inches.

Let $a_{n}$ be the lengths of the hypotenuses of the triangles in inches. $a_{1} = 13$ and $a_{2} = 1 \frac{1}{2} \cdot 12 = 18$ , so their common difference is

$d = a_{2} - a_{1} = 18- 13= 5$

The arithmetic sequence formula is

$a_{n} = a_{1} + (n-1)d$

The length of the hypotenuse of the largest triangle - the tenth triangle - can be found by substituting $a_{1} = 13, n= 10, d = 5$ :

$a_{10} = a_{1} + (10-1)5 = 13+ 9 \cdot 5= 13 + 45 = 58$ inches.

The largest triangle has hypotenuse of length 58 inches. Since the triangles are similar, corresponding sides are in proportion. If we let and be the lengths of the legs of the largest triangle, then

$\frac{a}{5} = \frac{58}{13}$

$\frac{a}{5}\cdot 5= \frac{58}{13} \cdot 5$

$a \approx 22.3$

Similarly,

$\frac{b}{12} = \frac{58}{13}$

$\frac{b}{12} \cdot 12= \frac{58}{13} \cdot 12$

$b \approx 53.5$

The area of a right triangle is half the product of its legs:

$A = \frac{ab}{2}= \frac{22.3 \cdot 53.5}{2} \approx 596.5$ square inches.

Divide this by 144 to convert to square feet:

$596.5 \div 144 \approx 4.1$

Of the given responses, 4 square feet is the closest, and is the correct choice.

One side of a regular hexagon is 20% shorter than one side of a regular pentagon. Which is the greater quantity?

(A) The perimeter of the pentagon

(B) The perimeter of the hexagon

(A) is greater

(A) and (B) are equal

It is impossible to determine which is greater from the information given

(B) is greater

Explanation

Let be the length of one side of the pentagon. Then its perimeter is .

Each side of the hexagon is 20% less than this length, or

The perimeter is five times this, or $6 \cdot 0.8s = 4.8s$ .

Since and is positive, , so the pentagon has greater perimeter, and (A) is greater.

The sum of the lengths of three sides of a regular pentagon is one foot. Give the perimeter of the pentagon in inches.

$20 ; \textrm{in}$

$25; \textrm{in}$

$30 ; \textrm{in}$

$15; \textrm{in}$

It is impossible to determine the perimeter from the information given.

Explanation

A regular pentagon has five sides of the same length.

One foot is equal to twelve inches; since the sum of the lengths of three of the congruent sides is twelve inches, each side measures

$12 \div 3 = 4$ inches.

The perimeter is

$4 \times 5 = 20$ inches.

One side of a regular pentagon is 20% longer than one side of a regular hexagon. Which is the greater quantity?

(A) The perimeter of the pentagon

(B) The perimeter of the hexagon

(A) and (B) are equal

(B) is greater

(A) is greater

It is impossible to determine which is greater from the information given

Explanation

Let be the length of one side of the hexagon. Then its perimeter is .

Each side of the pentagon is 20% greater than this length, or

The perimeter is five times this, or $5 \cdot 1.2s = 6s$ .

The perimeters are the same.

$\Delta ABC$ and $\Delta DEF$ are right triangles, with right angles $\angle B , \angle E$ , respectively.

Which is the greater quantity?

(a) The perimeter of $\Delta ABC$

(b) The perimeter of $\Delta DEF$

It is impossible to tell from the information given.

(a) and (b) are equal.

(a) is greater.

(b) is greater.

Explanation

No information is given about the legs of either triangle; therefore, no information about their perimeters can be deduced.

$\Delta ABC$ and $\Delta DEF$ are right triangles, with right angles $\angle B , \angle E$ , respectively.

Which is the greater quantity?

(a) The perimeter of $\Delta ABC$

(b) The perimeter of $\Delta DEF$

It is impossible to tell from the information given.

(a) and (b) are equal.

(a) is greater.

(b) is greater.

Explanation

No information is given about the legs of either triangle; therefore, no information about their perimeters can be deduced.

Icecreamcone

Refer to the above figure. The shaded region is a semicircle with area $16 \pi$ . Give the perimeter of $\bigtriangleup ABC$ .

$24 \sqrt{2}$

$24 \sqrt{3}$

$12 \sqrt{2}$

$12 \sqrt{3}$

Explanation

Given the radius of a semicircle, its area can be calculated using the formula

$A = \frac{1}{2} \pi r ^{2}$ .

Substituting $A = 16 \pi$ :

$\frac{1}{2} \pi r ^{2} = 16 \pi$

$\frac{2}{\pi} \cdot \frac{1}{2} \pi r ^{2} = \frac{2}{\pi} \cdot 16 \pi$

$r ^{2} = 32$

$r = \sqrt{32} = \sqrt{16} \cdot \sqrt{2} = 4\sqrt{2}$

The diameter of this semicircle is twice this, which is $d = 2 \cdot 4\sqrt{2} = 8 \sqrt{2}$ ; this is also the length of $\overline{BC}$ .

$\bigtriangleup ABC$ has two angles of degree measure 60; its third angle must also have measure 60, making $\bigtriangleup ABC$ an equilateral triangle with sidelength $s = 8 \sqrt{2}$ . Its perimeter is three times this, or $3 \cdot 8 \sqrt{2} = 24 \sqrt{2}$