Math / Geometry

QuestionQ1. $\sqrt{10^{2}-6^{2}}=A D B=8$
ADE and AEC are straight lines DE s paralle to $B C$ . $\begin{array}{l} \text { Angle } A B C=90^{\circ} \\ A C=10 \mathrm{~cm} . \\ B C=6 \mathrm{~cm} . \end{array}$ $D$ is the midpoint of $A B$ . Work out the area of trapez um BCED.

Studdy Solution

STEP 1

What is this asking? We need to find the area of the trapezoid BCED, given a right triangle ABC with AC = $10$ cm, BC = $6$ cm, and D being the midpoint of AB. Watch out! Don't forget that the area of a trapezoid is different from the area of a triangle!

STEP 2

1. Find AB
2. Find AD and DB
3. Find AE
4. Find the area of triangle ABC
5. Find the area of triangle ADE
6. Find the area of trapezoid BCED

STEP 3

We're given a right triangle ABC, so we can use the Pythagorean theorem: $a^2 + b^2 = c^2$ , where a and b are the legs, and c is the hypotenuse.
In our case, $AC = 10$ cm (hypotenuse), $BC = 6$ cm (one leg), and we want to find $AB$ (the other leg).

STEP 4

Let's plug in the values: $AB^2 + 6^2 = 10^2$ .
This simplifies to $AB^2 + 36 = 100$ .

STEP 5

Subtracting 36 from both sides gives us $AB^2 = 64$ .
Taking the square root of both sides, we get $AB = 8$ cm.
So, $AB = \mathbf{8}$ cm!

STEP 6

We know that D is the midpoint of AB.
Since $AB = 8$ cm, then $AD = DB = \frac{1}{2} \cdot AB = \frac{1}{2} \cdot 8 = \mathbf{4}$ cm.

STEP 7

Since DE is parallel to BC, triangles ADE and ABC are similar.
This means their corresponding sides are proportional.

STEP 8

We can set up a proportion: $\frac{AD}{AB} = \frac{AE}{AC}$ .
Plugging in the values we know, we get $\frac{4}{8} = \frac{AE}{10}$ .

STEP 9

Simplifying the fraction gives us $\frac{1}{2} = \frac{AE}{10}$ .
Multiplying both sides by 10, we find $AE = \frac{1}{2} \cdot 10 = \mathbf{5}$ cm.

STEP 10

The area of a triangle is given by $\frac{1}{2} \cdot \text{base} \cdot \text{height}$ .
For triangle ABC, the base is BC and the height is AB.

STEP 11

So, the area of triangle ABC is $\frac{1}{2} \cdot 6 \cdot 8 = \frac{1}{2} \cdot 48 = \mathbf{24}$ square cm.

STEP 12

Similarly, the area of triangle ADE is $\frac{1}{2} \cdot AD \cdot AE = \frac{1}{2} \cdot 4 \cdot 5 = \frac{1}{2} \cdot 20 = \mathbf{10}$ square cm.

STEP 13

The area of trapezoid BCED is simply the area of the large triangle ABC minus the area of the small triangle ADE.

STEP 14

Therefore, the area of trapezoid BCED is $24 - 10 = \mathbf{14}$ square cm.

STEP 15

The area of trapezoid BCED is 14 square cm.

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QuestionQ1. $\sqrt{10^{2}-6^{2}}=A D B=8$
ADE and AEC are straight lines DE s paralle to $B C$ . $\begin{array}{l} \text { Angle } A B C=90^{\circ} \\ A C=10 \mathrm{~cm} . \\ B C=6 \mathrm{~cm} . \end{array}$ $D$ is the midpoint of $A B$ . Work out the area of trapez um BCED.

Studdy Solution

STEP 1

What is this asking? We need to find the area of the trapezoid BCED, given a right triangle ABC with AC = $10$ cm, BC = $6$ cm, and D being the midpoint of AB. Watch out! Don't forget that the area of a trapezoid is different from the area of a triangle!

STEP 2

1. Find AB
2. Find AD and DB
3. Find AE
4. Find the area of triangle ABC
5. Find the area of triangle ADE
6. Find the area of trapezoid BCED

STEP 3

We're given a right triangle ABC, so we can use the Pythagorean theorem: $a^2 + b^2 = c^2$ , where a and b are the legs, and c is the hypotenuse.
In our case, $AC = 10$ cm (hypotenuse), $BC = 6$ cm (one leg), and we want to find $AB$ (the other leg).

STEP 4

Let's plug in the values: $AB^2 + 6^2 = 10^2$ .
This simplifies to $AB^2 + 36 = 100$ .

STEP 5

Subtracting 36 from both sides gives us $AB^2 = 64$ .
Taking the square root of both sides, we get $AB = 8$ cm.
So, $AB = \mathbf{8}$ cm!

STEP 6

We know that D is the midpoint of AB.
Since $AB = 8$ cm, then $AD = DB = \frac{1}{2} \cdot AB = \frac{1}{2} \cdot 8 = \mathbf{4}$ cm.

STEP 7

Since DE is parallel to BC, triangles ADE and ABC are similar.
This means their corresponding sides are proportional.

STEP 8

We can set up a proportion: $\frac{AD}{AB} = \frac{AE}{AC}$ .
Plugging in the values we know, we get $\frac{4}{8} = \frac{AE}{10}$ .

STEP 9

Simplifying the fraction gives us $\frac{1}{2} = \frac{AE}{10}$ .
Multiplying both sides by 10, we find $AE = \frac{1}{2} \cdot 10 = \mathbf{5}$ cm.

STEP 10

The area of a triangle is given by $\frac{1}{2} \cdot \text{base} \cdot \text{height}$ .
For triangle ABC, the base is BC and the height is AB.

STEP 11

So, the area of triangle ABC is $\frac{1}{2} \cdot 6 \cdot 8 = \frac{1}{2} \cdot 48 = \mathbf{24}$ square cm.

STEP 12

Similarly, the area of triangle ADE is $\frac{1}{2} \cdot AD \cdot AE = \frac{1}{2} \cdot 4 \cdot 5 = \frac{1}{2} \cdot 20 = \mathbf{10}$ square cm.

STEP 13

The area of trapezoid BCED is simply the area of the large triangle ABC minus the area of the small triangle ADE.

STEP 14

Therefore, the area of trapezoid BCED is $24 - 10 = \mathbf{14}$ square cm.

STEP 15

The area of trapezoid BCED is 14 square cm.

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Studdy Solution

STEP 1

What is this asking? We need to find the area of the trapezoid BCED, given a right triangle ABC with AC = 101010 cm, BC = 666 cm, and D being the midpoint of AB. Watch out! Don't forget that the area of a trapezoid is different from the area of a triangle!

STEP 2

1. Find AB2. Find AD and DB3. Find AE4. Find the area of triangle ABC5. Find the area of triangle ADE6. Find the area of trapezoid BCED

STEP 3

We're given a right triangle ABC, so we can use the **Pythagorean theorem**: a2+b2=c2a^2 + b^2 = c^2a2+b2=c2, where a and b are the legs, and c is the hypotenuse.In our case, AC=10AC = 10AC=10 cm (hypotenuse), BC=6BC = 6BC=6 cm (one leg), and we want to find ABABAB (the other leg).

STEP 4

Let's plug in the values: AB2+62=102AB^2 + 6^2 = 10^2AB2+62=102.This simplifies to AB2+36=100AB^2 + 36 = 100AB2+36=100.

STEP 5

Subtracting 36 from both sides gives us AB2=64AB^2 = 64AB2=64.Taking the square root of both sides, we get AB=8AB = 8AB=8 cm.So, AB=8AB = \mathbf{8}AB=8 **cm**!

STEP 6

We know that D is the **midpoint** of AB.Since AB=8AB = 8AB=8 cm, then AD=DB=12⋅AB=12⋅8=4AD = DB = \frac{1}{2} \cdot AB = \frac{1}{2} \cdot 8 = \mathbf{4}AD=DB=21​⋅AB=21​⋅8=4 **cm**.

STEP 7

Since DE is parallel to BC, triangles ADE and ABC are **similar**.This means their corresponding sides are proportional.

STEP 8

We can set up a proportion: ADAB=AEAC\frac{AD}{AB} = \frac{AE}{AC}ABAD​=ACAE​.Plugging in the values we know, we get 48=AE10\frac{4}{8} = \frac{AE}{10}84​=10AE​.

STEP 9

Simplifying the fraction gives us 12=AE10\frac{1}{2} = \frac{AE}{10}21​=10AE​.Multiplying both sides by 10, we find AE=12⋅10=5AE = \frac{1}{2} \cdot 10 = \mathbf{5}AE=21​⋅10=5 **cm**.

STEP 10

The area of a triangle is given by 12⋅base⋅height\frac{1}{2} \cdot \text{base} \cdot \text{height}21​⋅base⋅height.For triangle ABC, the base is BC and the height is AB.

STEP 11

So, the area of triangle ABC is 12⋅6⋅8=12⋅48=24\frac{1}{2} \cdot 6 \cdot 8 = \frac{1}{2} \cdot 48 = \mathbf{24}21​⋅6⋅8=21​⋅48=24 **square cm**.

STEP 12

Similarly, the area of triangle ADE is 12⋅AD⋅AE=12⋅4⋅5=12⋅20=10\frac{1}{2} \cdot AD \cdot AE = \frac{1}{2} \cdot 4 \cdot 5 = \frac{1}{2} \cdot 20 = \mathbf{10}21​⋅AD⋅AE=21​⋅4⋅5=21​⋅20=10 **square cm**.

STEP 13

The area of trapezoid BCED is simply the area of the large triangle ABC minus the area of the small triangle ADE.

STEP 14

Therefore, the area of trapezoid BCED is 24−10=1424 - 10 = \mathbf{14}24−10=14 **square cm**.

STEP 15

The area of trapezoid BCED is **14 square cm**.

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What is this asking? We need to find the area of the trapezoid BCED, given a right triangle ABC with AC = $10$ cm, BC = $6$ cm, and D being the midpoint of AB. Watch out! Don't forget that the area of a trapezoid is different from the area of a triangle!

1. Find AB
2. Find AD and DB
3. Find AE
4. Find the area of triangle ABC
5. Find the area of triangle ADE
6. Find the area of trapezoid BCED

We're given a right triangle ABC, so we can use the Pythagorean theorem: $a^2 + b^2 = c^2$ , where a and b are the legs, and c is the hypotenuse.
In our case, $AC = 10$ cm (hypotenuse), $BC = 6$ cm (one leg), and we want to find $AB$ (the other leg).

Let's plug in the values: $AB^2 + 6^2 = 10^2$ .
This simplifies to $AB^2 + 36 = 100$ .

Subtracting 36 from both sides gives us $AB^2 = 64$ .
Taking the square root of both sides, we get $AB = 8$ cm.
So, $AB = \mathbf{8}$ cm!

We know that D is the midpoint of AB.
Since $AB = 8$ cm, then $AD = DB = \frac{1}{2} \cdot AB = \frac{1}{2} \cdot 8 = \mathbf{4}$ cm.

Since DE is parallel to BC, triangles ADE and ABC are similar.
This means their corresponding sides are proportional.

We can set up a proportion: $\frac{AD}{AB} = \frac{AE}{AC}$ .
Plugging in the values we know, we get $\frac{4}{8} = \frac{AE}{10}$ .

Simplifying the fraction gives us $\frac{1}{2} = \frac{AE}{10}$ .
Multiplying both sides by 10, we find $AE = \frac{1}{2} \cdot 10 = \mathbf{5}$ cm.

The area of a triangle is given by $\frac{1}{2} \cdot \text{base} \cdot \text{height}$ .
For triangle ABC, the base is BC and the height is AB.

So, the area of triangle ABC is $\frac{1}{2} \cdot 6 \cdot 8 = \frac{1}{2} \cdot 48 = \mathbf{24}$ square cm.

Similarly, the area of triangle ADE is $\frac{1}{2} \cdot AD \cdot AE = \frac{1}{2} \cdot 4 \cdot 5 = \frac{1}{2} \cdot 20 = \mathbf{10}$ square cm.

Therefore, the area of trapezoid BCED is $24 - 10 = \mathbf{14}$ square cm.

The area of trapezoid BCED is 14 square cm.