Pre-Calculus - Solve a Right Triangle

Solve the right triangle.

Pcq1

C=90°

B=45°

a=5

c= $\sqrt{50}$

A=45°

b=5

A=135°

b=5

A=135°

b=2.07

A=45°

b= $\sqrt{50}$

None of these answers are correct.

Explanation

Pcq1

Given that:

C=90°

B=45°

a=5

c= $\sqrt{50}$

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

Therefore...

$5^{2}+b^{2}=(\sqrt{50})^{2}$

$25+b^{2}=50$

$b^{2}=50-25$

$b^{2}=25$

$\sqrt{b^{2}} =\sqrt{25}$

All angles of a triangle add up to 180°.

$A+45^{\circ}+90^{\circ}=180^{\circ}$

$A+135^{\circ}=180^{\circ}$

$A=180^{\circ}-135^{\circ}$

$A=45^{\circ}$

A right triangle has a base of 10 and a hypotenuse of 20. What is the length of the other leg?

$10\sqrt3$

$15\sqrt2$

$10\sqrt2$

$5\sqrt3$

Explanation

Write the Pythagorean Theorem.

Substitute the values of the leg and hypotenuse. The hypotenuse is the longest side of the right triangle. Solve for the unknown variable.

$a=\sqrt{300} = \sqrt{100\times3} = \sqrt{100}\times \sqrt3 = 10\sqrt3$

In the right triangle ABC, side AB is cm long, side AC is cm long, and side BC is the hypotenuse. How long is side BC?

$\sqrt{106}$ cm

$\sqrt{56}$ cm

Explanation

Given that ABC is a right triangle, the length of hypotenuse BC is the root of the sum of the squares of the two other sides (in other words, . Since AB is cm long and AC is cm long, we get that , and so $BC = \sqrt{106}$ .

Pcq1

Solve the right triangle given that a=5, b=12, and A=22.620°

B=67.380°

C=90°

c=13

B=90°

C=67.380°

c=13

B=67.380°

C=90°

c=17

B=90°

C=67.380°

c=169

None of these answers are correct.

Explanation

Pcq1

C is given as 90°.

A is given as 22.620°

a is given as 5

b is given as 12

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

Therefore...

$5^{2}+12^{2}=c^{2}$

$169=c^{2}$

$\sqrt{169}=\sqrt{c^{2}}$

All angles of a triangle add up to equal 180°.

$90^{\circ}+22.620^{\circ}+B=180^{\circ}$

$112.620^{\circ}+B=180^{\circ}$

$B=180^{\circ}-112.620^{\circ}$

$B=67.380^{\circ}$

Find the area of the given isosceles triangle:

Varsity log graph

$6.25 : cm^{2}$

$3.11 : cm^{2}$

$3.34 : cm^{2}$

$6.71 : cm^{2}$

$25.55: cm^{2}$

Explanation

The first step toward finding the area is to divide this isosceles triangle into two right triangles:

Varsity log graph

Trigonometric ratios can be used to find both the height and the base, which are needed to calculate area:

$sin(75^{\circ})=\frac{height}{5: cm}$

$0.9659=\frac{height}{5: cm}$

$sin(15^{\circ})=\frac{1/2base}{5: cm}$

$0.2588=\frac{1/2base}{5: cm}$

With both of those values calculated, we can now calculate the area of the triangle:

$Area=\frac{1}{2}\cdot2.59: cm\cdot4.83: cm=6.25: cm^2$

Find the area of the given isosceles triangle:

Varsity log graph

$17.85; cm^{2}$

$49 ; cm^{2}$

$67.32 ; cm^{2}$

$35.7 ; cm^{2}$

$24.5; cm^{2}$

Explanation

The first step is to divide this isosceles triangle into 2 right triangles, making it easier to solve:

Varsity log graph

The equation for area is $A=\frac{1}{2}base\cdot height$

We already know the base, so we need to solve for height to get the area.

$tan20^{\circ}=\frac{height}{7; cm}$

$0.364=\frac{height}{7; cm}$

Then we plug in all values for the equation:

$Area=\frac{1}{2}\cdot14; cm\cdot2.55; cm=17.85; cm^{2}$

Find the area of the given isosceles triangle and round all values to the nearest tenth:

Varsity log graph

$51.1 ; cm^{2}$

$66.6 ; cm^{2}$

$133.2 ; cm^{2}$

$102.2 ; cm^{2}$

$46.2 ; cm^{2}$

Explanation

The first step to solve for area is to divide the isosceles into two right triangles:

Varsity log graph

From there, we can determine the height and base needed for our area equation $A=\frac{1}{2}\cdot b \cdot h$

$sin(67.5^{\circ})=\frac{1/2Base}{12; cm}$

$0.924=\frac{1/2Base}{12; cm}$

From there, height can be easily determined using the Pathegorean Theorem:

$(12; cm)^{2}=(11.1; cm)^{2}+h^{2}$

$h^{2}=144; cm^{2}-123.2; cm^{2}=20.8cm^{2}$

Now both values can be plugged into the Area formula:

$A=\frac{1}{2}\cdot22.2; cm\cdot4.6; cm=51.1cm^{2}$

In a right triangle, if the hypotenuse is and a leg is , what is the area of the triangle?

$\frac{3\sqrt7}{2}$

$3\sqrt{7}$

$\frac{15}{2}$

Explanation

Use the Pythagorean Theorem to find the other leg.

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

$a=\sqrt7$

The length of the given leg is 3, and the unknown leg is $\sqrt7$ .

Use the area of a triangle formula and solve.

$A=\frac{bh}{2}= \frac{3\sqrt7}{2}$

An isosceles right triangle has a hypotenuse of 1. What is the area of this triangle?

$\frac{1}{4}$

$\frac{1}{2}$

$\frac{\sqrt2}{2}$

$\sqrt2$

Explanation

Write the formula for the Pythagorean theorem.

In an isosceles right triangle, both legs of the right triangle are equal.

Substitute the either variable and the known hypotheuse and determine the side length.

$b^2=\frac{1}{2}$

$b=\sqrt\frac{1}{2}$

This length represents both the base and the height of the triangle. Write the area of a triangle and substitute to solve for the area.

$A=\frac{bh}{2}=\frac{\sqrt\frac{1}{2}\sqrt\frac{1}{2}}{2}=\frac{\frac{1}{2}}{2}= \frac{1}{4}$

The side lengths of right triangle ABC are such that AC > BC > AB. AC = 25 and AB = 9. What is the length of BC?

$4\sqrt{34}$

$45\sqrt{2}$

$6\sqrt{17}$

Explanation

When you are using Pythagorean Theorem to calculate the missing side of a right triangle, it is crucial that you identify which side is the hypotenuse, in the Pythagorean equation . Here you're told that side AC is the longest of the three sides, so 25 will serve as the length of the hypotenuse and the value of . This allows you to set up the equation: