Math / Geometry

Question $\triangle \mathrm{DEF}$ is shown in the sketch below. G is a point on DF and EG is drawn. $\mathrm{DG}=\mathrm{DE}=x . \quad \mathrm{FG}=\frac{3 x}{2}$ and $\mathrm{EG}=\sqrt{3} x$ . 2.1 Calculate the size of $\dot{D}$ . (4) 2.2 Calculate the area of $\triangle G E F$ in terms of $x$ , in its simplest form. (5)

Studdy Solution

STEP 1

Assumptions
1. $\triangle DEF$ is a triangle with point $G$ on $DF$ .
2. $DG = DE = x$ .
3. $FG = \frac{3x}{2}$ .
4. $EG = \sqrt{3}x$ .
5. We need to calculate the size of $\angle D$ .
6. We need to calculate the area of $\triangle GEF$ in terms of $x$ .

STEP 2

To find the size of $\angle D$ , we will use the Law of Cosines in $\triangle DGE$ .
According to the Law of Cosines: $EG^2 = DE^2 + DG^2 - 2 \cdot DE \cdot DG \cdot \cos(\angle D)$

STEP 3

Substitute the given lengths into the Law of Cosines formula: $(\sqrt{3}x)^2 = x^2 + x^2 - 2 \cdot x \cdot x \cdot \cos(\angle D)$

STEP 4

Simplify the equation: $3x^2 = x^2 + x^2 - 2x^2 \cos(\angle D)$

STEP 5

Combine like terms: $3x^2 = 2x^2 - 2x^2 \cos(\angle D)$

STEP 6

Isolate $\cos(\angle D)$ : $3x^2 - 2x^2 = -2x^2 \cos(\angle D)$ $x^2 = -2x^2 \cos(\angle D)$

STEP 7

Divide both sides by $-2x^2$ : $\cos(\angle D) = -\frac{1}{2}$

STEP 8

Find $\angle D$ using the inverse cosine function: $\angle D = \cos^{-1}\left(-\frac{1}{2}\right)$

STEP 9

Recognize that $\cos^{-1}\left(-\frac{1}{2}\right)$ corresponds to $120^\circ$ : $\angle D = 120^\circ$

STEP 10

Now, we need to calculate the area of $\triangle GEF$ . We will use the formula for the area of a triangle: $\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height}$

STEP 11

In $\triangle GEF$ , the base is $FG = \frac{3x}{2}$ and the height can be considered as the perpendicular distance from $E$ to $FG$ , which is $EG = \sqrt{3}x$ .

STEP 12

Substitute the values into the area formula: $\text{Area} = \frac{1}{2} \cdot \frac{3x}{2} \cdot \sqrt{3}x$

STEP 13

Simplify the expression: $\text{Area} = \frac{1}{2} \cdot \frac{3x \cdot \sqrt{3}x}{2}$ $\text{Area} = \frac{1}{2} \cdot \frac{3 \sqrt{3} x^2}{2}$ $\text{Area} = \frac{3 \sqrt{3} x^2}{4}$

STEP 14

Thus, the area of $\triangle GEF$ in terms of $x$ is: $\text{Area} = \frac{3 \sqrt{3} x^2}{4}$
Solution:
1. The size of $\angle D$ is $120^\circ$ .
2. The area of $\triangle GEF$ in terms of $x$ is $\frac{3 \sqrt{3} x^2}{4}$ .

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Studdy Solution

STEP 1

Assumptions
1. $\triangle DEF$ is a triangle with point $G$ on $DF$ .
2. $DG = DE = x$ .
3. $FG = \frac{3x}{2}$ .
4. $EG = \sqrt{3}x$ .
5. We need to calculate the size of $\angle D$ .
6. We need to calculate the area of $\triangle GEF$ in terms of $x$ .

STEP 2

To find the size of $\angle D$ , we will use the Law of Cosines in $\triangle DGE$ .
According to the Law of Cosines: $EG^2 = DE^2 + DG^2 - 2 \cdot DE \cdot DG \cdot \cos(\angle D)$

STEP 3

Substitute the given lengths into the Law of Cosines formula: $(\sqrt{3}x)^2 = x^2 + x^2 - 2 \cdot x \cdot x \cdot \cos(\angle D)$

STEP 4

Simplify the equation: $3x^2 = x^2 + x^2 - 2x^2 \cos(\angle D)$

STEP 5

Combine like terms: $3x^2 = 2x^2 - 2x^2 \cos(\angle D)$

STEP 6

Isolate $\cos(\angle D)$ : $3x^2 - 2x^2 = -2x^2 \cos(\angle D)$ $x^2 = -2x^2 \cos(\angle D)$

STEP 7

Divide both sides by $-2x^2$ : $\cos(\angle D) = -\frac{1}{2}$

STEP 8

Find $\angle D$ using the inverse cosine function: $\angle D = \cos^{-1}\left(-\frac{1}{2}\right)$

STEP 9

Recognize that $\cos^{-1}\left(-\frac{1}{2}\right)$ corresponds to $120^\circ$ : $\angle D = 120^\circ$

STEP 10

Now, we need to calculate the area of $\triangle GEF$ . We will use the formula for the area of a triangle: $\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height}$

STEP 11

In $\triangle GEF$ , the base is $FG = \frac{3x}{2}$ and the height can be considered as the perpendicular distance from $E$ to $FG$ , which is $EG = \sqrt{3}x$ .

STEP 12

Substitute the values into the area formula: $\text{Area} = \frac{1}{2} \cdot \frac{3x}{2} \cdot \sqrt{3}x$

STEP 13

Simplify the expression: $\text{Area} = \frac{1}{2} \cdot \frac{3x \cdot \sqrt{3}x}{2}$ $\text{Area} = \frac{1}{2} \cdot \frac{3 \sqrt{3} x^2}{2}$ $\text{Area} = \frac{3 \sqrt{3} x^2}{4}$

STEP 14

Thus, the area of $\triangle GEF$ in terms of $x$ is: $\text{Area} = \frac{3 \sqrt{3} x^2}{4}$
Solution:
1. The size of $\angle D$ is $120^\circ$ .
2. The area of $\triangle GEF$ in terms of $x$ is $\frac{3 \sqrt{3} x^2}{4}$ .

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Studdy Solution

STEP 1

STEP 2

To find the size of ∠D\angle D∠D, we will use the Law of Cosines in △DGE\triangle DGE△DGE.According to the Law of Cosines: EG2=DE2+DG2−2⋅DE⋅DG⋅cos⁡(∠D) EG^2 = DE^2 + DG^2 - 2 \cdot DE \cdot DG \cdot \cos(\angle D) EG2=DE2+DG2−2⋅DE⋅DG⋅cos(∠D)

STEP 3

Substitute the given lengths into the Law of Cosines formula: (3x)2=x2+x2−2⋅x⋅x⋅cos⁡(∠D) (\sqrt{3}x)^2 = x^2 + x^2 - 2 \cdot x \cdot x \cdot \cos(\angle D) (3​x)2=x2+x2−2⋅x⋅x⋅cos(∠D)

STEP 4

Simplify the equation: 3x2=x2+x2−2x2cos⁡(∠D) 3x^2 = x^2 + x^2 - 2x^2 \cos(\angle D) 3x2=x2+x2−2x2cos(∠D)

STEP 5

Combine like terms: 3x2=2x2−2x2cos⁡(∠D) 3x^2 = 2x^2 - 2x^2 \cos(\angle D) 3x2=2x2−2x2cos(∠D)

STEP 6

Isolate cos⁡(∠D)\cos(\angle D)cos(∠D): 3x2−2x2=−2x2cos⁡(∠D) 3x^2 - 2x^2 = -2x^2 \cos(\angle D) 3x2−2x2=−2x2cos(∠D) x2=−2x2cos⁡(∠D) x^2 = -2x^2 \cos(\angle D) x2=−2x2cos(∠D)

STEP 7

Divide both sides by −2x2-2x^2−2x2: cos⁡(∠D)=−12 \cos(\angle D) = -\frac{1}{2} cos(∠D)=−21​

STEP 8

Find ∠D\angle D∠D using the inverse cosine function: ∠D=cos⁡−1(−12) \angle D = \cos^{-1}\left(-\frac{1}{2}\right) ∠D=cos−1(−21​)

STEP 9

Recognize that cos⁡−1(−12)\cos^{-1}\left(-\frac{1}{2}\right)cos−1(−21​) corresponds to 120∘120^\circ120∘: ∠D=120∘ \angle D = 120^\circ ∠D=120∘

STEP 10

Now, we need to calculate the area of △GEF\triangle GEF△GEF. We will use the formula for the area of a triangle: Area=12⋅base⋅height \text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} Area=21​⋅base⋅height

STEP 11

In △GEF\triangle GEF△GEF, the base is FG=3x2FG = \frac{3x}{2}FG=23x​ and the height can be considered as the perpendicular distance from EEE to FGFGFG, which is EG=3xEG = \sqrt{3}xEG=3​x.

STEP 12

Substitute the values into the area formula: Area=12⋅3x2⋅3x \text{Area} = \frac{1}{2} \cdot \frac{3x}{2} \cdot \sqrt{3}x Area=21​⋅23x​⋅3​x

STEP 13

Simplify the expression: Area=12⋅3x⋅3x2 \text{Area} = \frac{1}{2} \cdot \frac{3x \cdot \sqrt{3}x}{2} Area=21​⋅23x⋅3​x​ Area=12⋅33x22 \text{Area} = \frac{1}{2} \cdot \frac{3 \sqrt{3} x^2}{2} Area=21​⋅233​x2​ Area=33x24 \text{Area} = \frac{3 \sqrt{3} x^2}{4} Area=433​x2​

STEP 14

Thus, the area of △GEF\triangle GEF△GEF in terms of xxx is: Area=33x24 \text{Area} = \frac{3 \sqrt{3} x^2}{4} Area=433​x2​Solution:1. The size of ∠D\angle D∠D is 120∘120^\circ120∘.2. The area of △GEF\triangle GEF△GEF in terms of xxx is 33x24\frac{3 \sqrt{3} x^2}{4}433​x2​.

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To find the size of $\angle D$ , we will use the Law of Cosines in $\triangle DGE$ .
According to the Law of Cosines: $EG^2 = DE^2 + DG^2 - 2 \cdot DE \cdot DG \cdot \cos(\angle D)$

Substitute the given lengths into the Law of Cosines formula: $(\sqrt{3}x)^2 = x^2 + x^2 - 2 \cdot x \cdot x \cdot \cos(\angle D)$

Simplify the equation: $3x^2 = x^2 + x^2 - 2x^2 \cos(\angle D)$

Combine like terms: $3x^2 = 2x^2 - 2x^2 \cos(\angle D)$

Isolate $\cos(\angle D)$ : $3x^2 - 2x^2 = -2x^2 \cos(\angle D)$ $x^2 = -2x^2 \cos(\angle D)$

Divide both sides by $-2x^2$ : $\cos(\angle D) = -\frac{1}{2}$

Find $\angle D$ using the inverse cosine function: $\angle D = \cos^{-1}\left(-\frac{1}{2}\right)$

Recognize that $\cos^{-1}\left(-\frac{1}{2}\right)$ corresponds to $120^\circ$ : $\angle D = 120^\circ$

Now, we need to calculate the area of $\triangle GEF$ . We will use the formula for the area of a triangle: $\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height}$

In $\triangle GEF$ , the base is $FG = \frac{3x}{2}$ and the height can be considered as the perpendicular distance from $E$ to $FG$ , which is $EG = \sqrt{3}x$ .

Substitute the values into the area formula: $\text{Area} = \frac{1}{2} \cdot \frac{3x}{2} \cdot \sqrt{3}x$

Simplify the expression: $\text{Area} = \frac{1}{2} \cdot \frac{3x \cdot \sqrt{3}x}{2}$ $\text{Area} = \frac{1}{2} \cdot \frac{3 \sqrt{3} x^2}{2}$ $\text{Area} = \frac{3 \sqrt{3} x^2}{4}$

Thus, the area of $\triangle GEF$ in terms of $x$ is: $\text{Area} = \frac{3 \sqrt{3} x^2}{4}$
Solution:
1. The size of $\angle D$ is $120^\circ$ .
2. The area of $\triangle GEF$ in terms of $x$ is $\frac{3 \sqrt{3} x^2}{4}$ .