Right Triangles - ISEE Upper Level Quantitative Reasoning

Card 0 of 740

Question

Examine the above diagram. If $l \parallel m$ , give in terms of .

Answer

The two marked angles are same-side exterior angles of parallel lines, which are supplementary - that is, their measures have sum 180. We can solve for in this equation:

Compare your answer with the correct one above

Question

Examine the above diagram. If $l \parallel m$ , give in terms of .

Answer

The two marked angles are same-side interior angles of parallel lines, which are supplementary - that is, their measures have sum 180. We can solve for in this equation:

$2y \div 2 = \left ( 184 -x \right )\div 2$

$y= 92 - \frac{1}{2}x$

Compare your answer with the correct one above

Question

Examine the above diagram. Which of the following statements must be true whether or not and are parallel?

Answer

Four statements can be eliminated by the various parallel theorems and postulates. Congruence of alternate interior angles or corresponding angles forces the lines to be parallel, so

$\angle 5 \cong \angle 4 \Rightarrow l \parallel m$ and

$\angle 2 \cong \angle 6 \Rightarrow l \parallel m$ .

Also, if same-side interior angles or same-side exterior angles are supplementary, the lines are parallel, so

$m \angle 4 + m \angle 7 = 180 ^{\circ } \Rightarrow l \parallel m$ and

$m \angle 1 + m \angle 6 = 180 ^{\circ } \Rightarrow l \parallel m$ .

However, $\angle 5 \cong \angle 8$ whether or not $l \parallel m$ since they are vertical angles, which are always congruent.

Compare your answer with the correct one above

Question

Examine the above diagram. What is ?

Answer

By angle addition,

$\left ( x - 13\right )+ 100 + (y + 25) = 180$

$x - 13\right+ 100 + y + 25 = 180$

$x + y - 13\right+ 100+ 25 = 180$

Compare your answer with the correct one above

Question

$\angle 1$ and $\angle 2$ are supplementary; $\angle 1$ and $\angle 3$ are complementary.

$m \angle 2 = 100 ^{\circ }$ .

What is $m \angle 3$ ?

Answer

Supplementary angles and complementary angles have measures totaling $180^{\circ }$ and $90^{\circ }$ , respectively.

$m \angle 2 = 100 ^{\circ }$ , so its supplement $\angle 1$ has measure

$m \angle 1 = 180^{\circ } - m \angle 2 = 180^{\circ } - 100 ^{\circ } = 80^{\circ }$

$\angle 3$ , the complement of $\angle 1$ , has measure

$m \angle 3 = 90^{\circ } - m \angle 1 = 90^{\circ } - 80 ^{\circ } = 10^{\circ }$

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

In the above figure, $m \angle 1=\left ( x+17 \right )^{\circ}$ and $m \angle 2= \left (7x+11 \right )^{\circ}$ . Which of the following is equal to $m \angle 1$ ?

Answer

$\angle 1$ and $\angle 2$ form a linear pair, so their angle measures total $180^{\circ}$ . Set up and solve the following equation:

$m \angle 1+ m \angle 2 = 180^{\circ}$

$\left ( x+17 \right ) + \left (7x+11 \right ) = 180$

$m \angle 1=\left ( x+17 \right )^{\circ} = \left ( 19+17 \right )^{\circ} = 36^{\circ}$

Compare your answer with the correct one above

Question

Two angles which form a linear pair have measures $(2x+72)^{\circ}$ and $(5x-125)^{\circ}$ . Which is the lesser of the measures (or the common measure) of the two angles?

Answer

Two angles that form a linear pair are supplementary - that is, they have measures that total $180^{\circ}$ . Therefore, we set and solve for in this equation:

$x = \frac{233}{7}= 33 \frac{2}{7}$

The two angles have measure

$2 \cdot 33 \frac{2}{7} +72= 138 \frac{4}{7} ^{\circ}$

and

$5 \cdot 33 \frac{2}{7} -125 = 41\frac{3}{7}^{\circ}$

$41\frac{3}{7}^{\circ}$ is the lesser of the two measures and is the correct choice.

Compare your answer with the correct one above

Question

Two vertical angles have measures $(3x+14)^{\circ }$ and $\left ( 5x-17\right )^{\circ }$ . Which is the lesser of the measures (or the common measure) of the two angles?

Answer

Two vertical angles - angles which share a vertex and whose union is a pair of lines - have the same measure. Therefore, we set up and solve the equation

$x= 15\frac{1}{2}$

$3x+14 = 3 \cdot 15\frac{1}{2} + 14 = 46\frac{1}{2} + 14= 60\frac{1}{2} ^{\circ }$

Compare your answer with the correct one above

Question

A line intersects parallel lines and . $\angle 1$ and $\angle 2$ are corresponding angles; $\angle 1$ and $\angle 3$ are same side interior angles.

$m \angle 1 = \left (3x+2y \right )^{\circ }$

$m \angle 2 = \left ( 4x+21\right )^{\circ }$

$m \angle 3 = \left ( 2y-27\right )^{\circ}$

Evaluate .

Answer

When a transversal such as crosses two parallel lines, two corresponding angles - angles in the same relative position to their respective lines - are congruent. Therefore,

Two same-side interior angles are supplementary - that is, their angle measures total 180 - so

We can solve this system by the substitution method as follows:

Backsolve:

, which is the correct response.

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. Give the measure of $\angle 1$ .

Answer

The top and bottom angles, being vertical angles - angles which share a vertex and whose union is a pair of lines - have the same measure, so

,

or, simplified,

The right and bottom angles form a linear pair, so their degree measures total 180. That is,

Substitute for :

The left and right angles, being vertical angles, have the same measure, so, since the right angle measures $\left (8x \right )^{\circ } =\left (8 \cdot 15 \right )^{\circ } = 120 ^{\circ }$ , this is also the measure of the left angle, $\angle 1$ .

Compare your answer with the correct one above

Question

Figure NOT drawn to scale

The above figure shows Trapezoid , with $\overline{RT}$ and $\overline{RA }$ tangent to the circle. $m \angle T = 106 ^{\circ }$ ; evaluate $m \angle A$ .

Answer

By the Same-Side Interior Angle Theorem, since $\overline{TR} || \overline{AP}$ , $\angle T$ and $\angle P$ are supplementary - that is, their degree measures total $180 ^{\circ }$ . Therefore,

$m \angle P + m \angle T = 180 ^{\circ }$

$m \angle P = 180 ^{\circ } - m \angle T$

$m \angle P = 180 ^{\circ } - 106 ^{\circ } = 74^{\circ }$

$\angle P$ is an inscribed angle, so the arc it intercepts, $\overarc{TA}$ , has twice its degree measure;

$m \overarc{TA} = 2 \cdot m \angle P =2 \cdot 74^{\circ } = 148 ^{\circ }$ .

The corresponding major arc, $\overarc{TPA}$ , has as its measure

$m \overarc{TPA} = 360 ^{\circ }- m \overarc{TA} = 360 ^{\circ }- 148 ^{\circ }= 212 ^{\circ }$

The measure of an angle formed by two tangents to a circle is equal to half the difference of those of its intercepted arcs:

$m \angle R = \frac{1}{2} \left (m \overarc{TPA} - m \overarc{TA} \right )$

$m \angle R = \frac{1}{2} \left (212 ^{\circ } -148 ^{\circ } \right )= \frac{1}{2} \cdot 64 ^{\circ } = 32^{\circ }$

Again, by the Same-Side Interior Angles Theorem, $\angle R$ and $\angle A$ are supplementary, so

$m \angle A + m \angle R= 180 ^{\circ }$

$m \angle A = 180 ^{\circ } - m \angle R$

$m \angle A = 180 ^{\circ } - 32 ^{\circ } = 148^{\circ }$

Compare your answer with the correct one above

Question

Which of the following is true about a triangle with two angles that measure $45^{\circ}$ and $45^{\circ}$ ?

Answer

The sum of the two given angles is 90 degrees, which means that the third angle should be a right angle (90 degrees). We also know that two of the angles are equal. Therefore, the triangle is right and isosceles.

Compare your answer with the correct one above

Question

What is the value of x in a right triangle if the two acute angles are equal to 5x and 25x?

Answer

In a right triangle, there is one right angle of 90 degrees, while the two acute angles add up to 90 degrees.

Given that the two acute angles are equal to 5x and 25x, the value of x can be calculated with the equation below:

Compare your answer with the correct one above

Question

$\Delta ABC \sim \Delta DEF$

$\angle B$ is a right angle; , .

Which is the greater quantity?

(a) $m \angle A$

(b) $60 ^{\circ }$

Answer

$\Delta ABC \sim \Delta DEF$ . Corresponding angles of similar triangles are congruent, so since $\angle B$ is a right angle, so is $\angle E$ .

The hypotenuse $\overline{DF}$ of $\Delta DEF$ is twice as long as leg $\overline{DE}$ ; by the $30 ^{\circ } - 60^{\circ } -90^{\circ }$ Theorem, $m \angle D = 60 ^{\circ }$ . Again, by similiarity,

$m \angle A = m \angle D = 60 ^{\circ }$ .

Compare your answer with the correct one above

Question

One angle of a right triangle measures 45 degrees, and the hypotenuse measures 8 centimeters. Give the area of the triangle.

Answer

This triangle has two angles of 45 and 90 degrees, so the third angle must measure 45 degrees; this is therefore an isosceles right triangle.

By the Pythagorean Theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Let hypotenuse and side length.

$c^2=s^2+s^2\Rightarrow c^2=2s^2\Rightarrow 8^2=2s^2\Rightarrow s^2=32$

$\Rightarrow s=\sqrt{32}=\sqrt{16\times 2}=4\sqrt{2}\ cm$

We can then plug this side length into the formula for area.

$area=\frac{s^2}{2}=\frac{(4\sqrt{2})^2}{2}=\frac{32}{2}=16\ cm^2$

Compare your answer with the correct one above

Question

The legs of a right triangle measure $2 \frac{2}{5}$ and $3 \frac{1}{5}$ . What is its perimeter?

Answer

The hypotenuse of the triangle can be calculated using the Pythagorean Theorem. Set $a = 2 \frac{2}{5} = \frac{12}{5} , b=3 \frac{1}{5}= \frac{16}{5}$ :

$c ^{2} = a ^{2} + b ^{2}$

$c ^{2} = \left ( \frac{12}{5} \right ) ^{2} + \left ( \frac{16}{5} \right ) ^{2}$

$c ^{2} = \frac{144}{25} + \frac{256}{25}$

$c ^{2} = \frac{400}{25} = 16$

$c = \sqrt{16} = 4$

Add the three sidelengths:

$P =2 \frac{2}{5} + 3 \frac{1}{5} + 4 = 9 \frac{3}{5}$

Compare your answer with the correct one above

Question

Figure NOT drawn to scale

$\bigtriangleup ABC$ is a right triangle with altitude $\overline{BX}$ . Give the ratio of the area of $\bigtriangleup BXC$ to that of $\bigtriangleup AXB$ .

Answer

The altitude of a right triangle from the vertex of its right angle - which, here, is $\overline{BX}$ - divides the triangle into two triangles similar to each other. The ratio of the hypotenuse of the white triangle to that of the gray triangle (which are corresponding sides) is

$\frac{BC}{AB} = \frac{40}{30} =\frac{4}{3}$ ,

making this the similarity ratio. The ratio of the areas of two similar triangles is the square of their similarity ratio, which here is

$\left ( \frac{4}{3} \right )^{2} = \frac{4^{2}}{3^{2}} = \frac{16}{9}$ , or .

Compare your answer with the correct one above

Question

Find the area of a right triangle with a base of 7cm and a height of 20cm.

Answer

To find the area of a right triangle, we will use the following formula:

$A = \frac{1}{2} \cdot b \cdot h$

where b is the base and h is the height of the triangle.

We know the base of the triangle is 7cm. We know the height of the triangle is 20cm. Knowing this, we can substitute into the formula. We get

$A = \frac{1}{2} \cdot 7\text{cm} \cdot 20\text{cm}$

$A = 7\text{cm} \cdot 10\text{cm}$

$A = 70\text{cm}^2$

Compare your answer with the correct one above

Question

Refer to the above figure. Evaluate the length of $\overline{BD}$ in terms of .

Answer

The altitude of a right triangle to its hypotenuse divides the triangle into two smaller trangles similar to each other and to the large triangle.

Therefore,

$\Delta ABC \sim \Delta ADB$

and, consequently,

$\frac{DB}{BC} = \frac{AB}{AC}$ ,

or, equivalently,

$DB = \frac{AB \cdot BC}{AC}$

$AC = \sqrt{ n^{2}+36}$ by the Pythagorean Theorem, so

$DB = \frac{6n }{\sqrt{ n^{2}+36}}$ .

Compare your answer with the correct one above

Question

Refer to the above figure. Evaluate the length of $\overline{BD}$ in terms of .

Answer

The height of a right triangle from the vertex of its right angle is the geometric mean - in this case, the square root of the product - of the lengths of the two segments of the hypotenuse that it forms. Therefore,

$BD = \sqrt{(AD)(CD)} = \sqrt{6n}$

Compare your answer with the correct one above

Tap the card to reveal the answer