Acute / Obtuse Triangles

Help Questions

Math › Acute / Obtuse Triangles

Questions 1 - 10

An acute isosceles triangle has an area of square units and a base with length . Find the perimeter of this triangle.

$2\sqrt{73} +6$

$\sqrt{73}+6$

$2\sqrt{73}+8$

$\sqrt{73}+8$

Explanation

To solve this problem, first work backwards using the formula:

$Area=\frac{b\cdot h}{2}$

Plugging in the given information and solving for height.

$24=\frac{6h}{2}$

$24\times2=6h$

$h=\frac{48}{6}=8$

Now that you've found the height of the triangle, use the Pythagorean Theorem to find the length of one of the equivalent sides of the triangle.

(Note, when applying the formula divide the measurement of the base in half--so the triangle will have sides of and .

Thus, the solution is:

$c=\sqrt{73}$

In this problem, the isosceles triangle will have two sides with the length of $\sqrt{73}$ and one side length of . Therefore, the perimeter is: $2\sqrt{73}+6$

The largest angle in an obtuse scalene triangle is degrees. The second largest angle in the triangle is $\frac{1}{3}$ the measurement of the largest angle. What is the measurement of the smallest angle in the obtuse scalene triangle?

$20^\circ$

$40^\circ$

$30^\circ$

$25^\circ$

$15^\circ$

Explanation

Since this is a scalene triangle, all of the interior angles will have different measures. However, it's fundemental to note that in any triangle the sum of the measurements of the three interior angles must equal degrees.

The largest angle is equal to degrees and second interior angle must equal:

$120\div3=40$

Therefore, the final angle must equal:

An acute isosceles triangle has an area of square units and a base with length . Find the perimeter of this triangle.

$2\sqrt{73} +6$

$\sqrt{73}+6$

$2\sqrt{73}+8$

$\sqrt{73}+8$

Explanation

To solve this problem, first work backwards using the formula:

$Area=\frac{b\cdot h}{2}$

Plugging in the given information and solving for height.

$24=\frac{6h}{2}$

$24\times2=6h$

$h=\frac{48}{6}=8$

Now that you've found the height of the triangle, use the Pythagorean Theorem to find the length of one of the equivalent sides of the triangle.

(Note, when applying the formula divide the measurement of the base in half--so the triangle will have sides of and .

Thus, the solution is:

$c=\sqrt{73}$

In this problem, the isosceles triangle will have two sides with the length of $\sqrt{73}$ and one side length of . Therefore, the perimeter is: $2\sqrt{73}+6$

For the triangle below, the perimeter is . Find the value of .

Explanation

The dashes on the two sides of the triangle indicate that those two sides are congruent. Thus, the length of the missing side is also.

Now, use the information given about the perimeter to set up an equation to solve for . The sum of all the sides will give the perimeter.

A triangle is placed in a parallelogram so they share a base.

If the height of the triangle is half that of the parallelogram, find the area of the shaded region.

Explanation

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=12\times8=96$

Next, recall how to find the area of a triangle.

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{2}$

$\text{Height of Triangle}=\frac{8}{2}=4$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(12 \times 4)=24$

Subtract the two areas to find the area of the shaded region.

$\text{Area of Shaded Region}=\text{Area of Parallelogram}-\text{Area of Triangle}$

$\text{Area of Shaded Region}=96-24=72$

In the figure, a triangle that shares its base with the width of the rectangle has a height that is half the length of the rectangle. Find the area of the shaded region.

Explanation

In order to find the area of the shaded region, we must first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle:

$\text{Area of Rectangle}=\text{length}\times\text{width}$

Substitute in the given length and width to find the area.

$\text{Area of Rectangle}=8 \times 12$

$\text{Area of Rectangle}= 96$

Next, recall how to find the area of a triangle:

$\text{Area of Triangle}=\frac{1}{2}\times\text{base}\times\text{height}$

From the question, we know that the height of the triangle is half the length of the rectangle. Use the length of the rectangle to find the height of the triangle:

$\text{height}=\frac{\text{length}}{2}$

$\text{height}=\frac{8}{2}$

$\text{height}=4$

Since the base of the triangle and the width of the rectangle are the same, we can find the area of the triangle:

$\text{Area of Triangle}=\frac{1}{2}\times 12\times 4$

$\text{Area of Triangle}=24$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle.

$\text{Area of Shaded Region}=\text{Area of Rectangle}-\text{Area of Triangle}$

$\text{Area of Shaded Region}=96-24$

Solve.

$\text{Area of Shaded Region}=72$

In ΔABC: a = 10, c = 15, and B = 50°.

Find b (to the nearest tenth).

11.5

9.7

12.1

11.9

14.3

Explanation

This problem requires us to use either the Law of Sines or the Law of Cosines. To figure out which one we should use, let's write down all the information we have in this format:

A = ? a = 10

B = 50° b = ?

C = ? c = 15

Now we can easily see that we do not have any complete pairs (such as A and a, or B and b) but we do have one term from each pair (a, B, and c). This tells us that we should use the Law of Cosines. (We use the Law of Sines when we have one complete pair, such as B and b).

Law of Cosines:
$a^2 = b^2 + c^2 - 2bc\cos(A)$
$b^2 = a^2 + c^2 - 2ac\cos(B)$
$c^2 = a^2 + b^2 - 2ab\cos(C)$

Since we are solving for b, let's use the second version of the Law of Cosines.
This gives us:

$b^2 = 10^2 + 15^2 - 2(10)(15)\cos(50)$

Exterior_angle

If the measure of $\angle A=52^{\circ}$ and the measure of $\angle B= 43^{\circ}$ then what is the meausre of $\angle\theta$ ?

$95^{\circ}$

$85^{\circ}$

$82^{\circ}$

$98^{\circ}$

Not enough information to solve

Explanation

The key to solving this problem lies in the geometric fact that a triangle possesses a total of $180^{\circ}$ between its interior angles. Therefore, one can calculate the measure of $\angle C$ and then find the measure of its supplementary angle, $\angle\theta$ .

$180^{\circ}=\angle A + \angle B + \angle C$

$\angle C= 180^{\circ}- 52^{\circ} - 43^{\circ}$

$\angle C= 85^{\circ}$

$\angle C$ and $\angle\theta$ are supplementary, meaning they form a line with a measure of $180^{\circ}$ .

$\angle\theta =180^{\circ} - 85^{\circ}$

$\rightarrow 95^{\circ}$

One could also solve this problem with the knowledge that the sum of the exterior angle of a triangle is equal to the sum of the two interior angles opposite of it.

For the triangle below, the perimeter is . Find the value of .