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Intermediate Geometry : How to find the area of an acute / obtuse isosceles triangle

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Example Question #21 : Triangles

An isosceles triangle is placed in a circle as shown by the figure below.

If the diameter of the circle is , find the area of the shaded region.

Possible Answers:

1145.52

1190.02

1098.98

1132.90

Correct answer:

1132.90

Explanation:

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

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

$\text{Area of Triangle}=\frac{1}{2}(46 \times 23)=529$

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

$\text{Area of Circle}=\pi\times\text{radius}^2$

$\text{Area of Circle}=\pi\times 23^2= 529\pi$

To find the area of the shaded region, subtract the two areas.

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

$\text{Area of Shaded Region}=529\pi-529=1132.90$

Make sure to round to places after the decimal.

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Example Question #22 : Triangles

An isosceles triangle is placed in a circle as shown by the figure below.

If the diameter of the circle is , find the area of the shaded region.

Possible Answers:

498.20

402.08

481.86

441.12

Correct answer:

481.86

Explanation:

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

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

$\text{Area of Triangle}=\frac{1}{2}(30 \times 15)=225$

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

$\text{Area of Circle}=\pi\times\text{radius}^2$

$\text{Area of Circle}=\pi\times 15^2= 225\pi$

To find the area of the shaded region, subtract the two areas.

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

$\text{Area of Shaded Region}=225\pi-225=481.86$

Make sure to round to places after the decimal.

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Example Question #23 : Triangles

An isosceles triangle is placed in a circle as shown by the figure below.

If the diameter of the circle is , find the area of the shaded region.

Possible Answers:

1190.21

1181.21

1447.72

1098.21

Correct answer:

1447.72

Explanation:

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

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

$\text{Area of Triangle}=\frac{1}{2}(52 \times 26)=676$

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

$\text{Area of Circle}=\pi\times\text{radius}^2$

$\text{Area of Circle}=\pi\times 26^2= 676\pi$

To find the area of the shaded region, subtract the two areas.

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

$\text{Area of Shaded Region}=676\pi-676=1447.72$

Make sure to round to places after the decimal.

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Example Question #24 : Triangles

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

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

Possible Answers:

1.25

0.75

1.5

Correct answer:

1.5

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.

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

$\text{Area of Parallelogram}=2\times 1 = 2$

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.

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

$\text{Height of Triangle}=\frac{1}{2}$

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

$\text{Area of Triangle}=\frac{1}{2}(2\times \frac{1}{2})=\frac{1}{2}=0.5$

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

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

$\text{Area of Shaded Region}=2-0.5=1.5$

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Intermediate Geometry : How to find the area of an acute / obtuse isosceles triangle

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #21 : Triangles

Example Question #22 : Triangles

Example Question #23 : Triangles

Example Question #24 : Triangles

All Intermediate Geometry Resources

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