QuestionIn isosceles triangle $\triangle A B E$ , if $m \angle B = 2 m \angle A$ and $A E = 5 \sqrt{2}$ , find the area of $\triangle A B E$ .

Studdy Solution

STEP 1

Assumptions1. Triangle ABE is an isosceles triangle. . Angle A and angle are the base angles.
3. The measure of angle B is twice the measure of angle A.
4. The length of AE is $5\sqrt{}$ .
5. We are asked to find the area of the triangle.

STEP 2

In an isosceles triangle, the base angles are equal. Since $\angle B$ is twice $\angle A$ , then $\angle A = \angle =60^\circ$ and $\angle B =120^\circ$ .

STEP 3

The formula for the area of a triangle is $\frac{1}{2} \times base \times height$ . In this case, the base of the triangle is AE.

STEP 4

We can find the height by drawing a line from B to the midpoint of AE, creating two30-60-90 triangles.

STEP 5

In a30-60-90 triangle, the ratio of the sides opposite the30,60, and90 degree angles is1 $\sqrt{3}$ 2. Therefore, if the hypotenuse (AE) is $5\sqrt{2}$ , the side opposite the60 degree angle (the height of $\triangle ABE$ ) is $\frac{5\sqrt{2}}{2} \times \sqrt{3}$ .

STEP 6

Calculate the height of the triangle.
$Height = \frac{5\sqrt{2}}{2} \times \sqrt{3} = \frac{5\sqrt{6}}{2}$

STEP 7

Now we can calculate the area of the triangle using the formula $\frac{1}{2} \times base \times height$ .

STEP 8

Plug in the values for the base and the height to calculate the area.
$Area = \frac{1}{2} \times5\sqrt{2} \times \frac{5\sqrt{6}}{2}$

STEP 9

Calculate the area of the triangle.
$Area = \frac{}{2} \times5\sqrt{2} \times \frac{5\sqrt{6}}{2} =25$ So, the area of $\triangle ABE$ is25 square units.

Was this helpful?

QuestionIn isosceles triangle $\triangle A B E$ , if $m \angle B = 2 m \angle A$ and $A E = 5 \sqrt{2}$ , find the area of $\triangle A B E$ .

Studdy Solution

STEP 1

Assumptions1. Triangle ABE is an isosceles triangle. . Angle A and angle are the base angles.
3. The measure of angle B is twice the measure of angle A.
4. The length of AE is $5\sqrt{}$ .
5. We are asked to find the area of the triangle.

STEP 2

In an isosceles triangle, the base angles are equal. Since $\angle B$ is twice $\angle A$ , then $\angle A = \angle =60^\circ$ and $\angle B =120^\circ$ .

STEP 3

The formula for the area of a triangle is $\frac{1}{2} \times base \times height$ . In this case, the base of the triangle is AE.

STEP 4

We can find the height by drawing a line from B to the midpoint of AE, creating two30-60-90 triangles.

STEP 5

In a30-60-90 triangle, the ratio of the sides opposite the30,60, and90 degree angles is1 $\sqrt{3}$ 2. Therefore, if the hypotenuse (AE) is $5\sqrt{2}$ , the side opposite the60 degree angle (the height of $\triangle ABE$ ) is $\frac{5\sqrt{2}}{2} \times \sqrt{3}$ .

STEP 6

Calculate the height of the triangle.
$Height = \frac{5\sqrt{2}}{2} \times \sqrt{3} = \frac{5\sqrt{6}}{2}$

STEP 7

Now we can calculate the area of the triangle using the formula $\frac{1}{2} \times base \times height$ .

STEP 8

Plug in the values for the base and the height to calculate the area.
$Area = \frac{1}{2} \times5\sqrt{2} \times \frac{5\sqrt{6}}{2}$

STEP 9

Calculate the area of the triangle.
$Area = \frac{}{2} \times5\sqrt{2} \times \frac{5\sqrt{6}}{2} =25$ So, the area of $\triangle ABE$ is25 square units.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionIn isosceles triangle △ABE\triangle A B E△ABE, if m∠B=2m∠Am \angle B = 2 m \angle Am∠B=2m∠A and AE=52A E = 5 \sqrt{2}AE=52​, find the area of △ABE\triangle A B E△ABE.

Studdy Solution

STEP 1

Assumptions1. Triangle ABE is an isosceles triangle. . Angle A and angle are the base angles.3. The measure of angle B is twice the measure of angle A.4. The length of AE is 55\sqrt{}5​.5. We are asked to find the area of the triangle.

STEP 2

In an isosceles triangle, the base angles are equal. Since ∠B\angle B∠B is twice ∠A\angle A∠A, then ∠A=∠=60∘\angle A = \angle =60^\circ∠A=∠=60∘ and ∠B=120∘\angle B =120^\circ∠B=120∘.

STEP 3

The formula for the area of a triangle is 12×base×height\frac{1}{2} \times base \times height21​×base×height. In this case, the base of the triangle is AE.

STEP 4

We can find the height by drawing a line from B to the midpoint of AE, creating two30-60-90 triangles.

STEP 5

In a30-60-90 triangle, the ratio of the sides opposite the30,60, and90 degree angles is1 3\sqrt{3}3​2. Therefore, if the hypotenuse (AE) is 525\sqrt{2}52​, the side opposite the60 degree angle (the height of △ABE\triangle ABE△ABE) is 522×3\frac{5\sqrt{2}}{2} \times \sqrt{3}252​​×3​.

STEP 6

Calculate the height of the triangle.Height=522×3=562Height = \frac{5\sqrt{2}}{2} \times \sqrt{3} = \frac{5\sqrt{6}}{2}Height=252​​×3​=256​​

STEP 7

Now we can calculate the area of the triangle using the formula 12×base×height\frac{1}{2} \times base \times height21​×base×height.

STEP 8

Plug in the values for the base and the height to calculate the area.Area=12×52×562Area = \frac{1}{2} \times5\sqrt{2} \times \frac{5\sqrt{6}}{2}Area=21​×52​×256​​

STEP 9

Calculate the area of the triangle.Area=2×52×562=25Area = \frac{}{2} \times5\sqrt{2} \times \frac{5\sqrt{6}}{2} =25Area=2​×52​×256​​=25So, the area of △ABE\triangle ABE△ABE is25 square units.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionIn isosceles triangle $\triangle A B E$ , if $m \angle B = 2 m \angle A$ and $A E = 5 \sqrt{2}$ , find the area of $\triangle A B E$ .

Assumptions1. Triangle ABE is an isosceles triangle. . Angle A and angle are the base angles.
3. The measure of angle B is twice the measure of angle A.
4. The length of AE is $5\sqrt{}$ .
5. We are asked to find the area of the triangle.

In an isosceles triangle, the base angles are equal. Since $\angle B$ is twice $\angle A$ , then $\angle A = \angle =60^\circ$ and $\angle B =120^\circ$ .

The formula for the area of a triangle is $\frac{1}{2} \times base \times height$ . In this case, the base of the triangle is AE.

In a30-60-90 triangle, the ratio of the sides opposite the30,60, and90 degree angles is1 $\sqrt{3}$ 2. Therefore, if the hypotenuse (AE) is $5\sqrt{2}$ , the side opposite the60 degree angle (the height of $\triangle ABE$ ) is $\frac{5\sqrt{2}}{2} \times \sqrt{3}$ .

Calculate the height of the triangle.
$Height = \frac{5\sqrt{2}}{2} \times \sqrt{3} = \frac{5\sqrt{6}}{2}$

Now we can calculate the area of the triangle using the formula $\frac{1}{2} \times base \times height$ .

Plug in the values for the base and the height to calculate the area.
$Area = \frac{1}{2} \times5\sqrt{2} \times \frac{5\sqrt{6}}{2}$

Calculate the area of the triangle.
$Area = \frac{}{2} \times5\sqrt{2} \times \frac{5\sqrt{6}}{2} =25$ So, the area of $\triangle ABE$ is25 square units.