QuestionGiven $AB=36 \mathrm{~cm}, BC=48 \mathrm{~cm}, AD=109 \mathrm{~cm}$ , find $AC$ , $CD$ , and the area of quadrilateral $ABCD$ .

Studdy Solution

STEP 1

Assumptions1. The lengths of the sides are given as $AB=36$ cm, $BC=48$ cm, and $AD=109$ cm. . The angles $ABC$ and $AC$ are right angles, i.e., $90^{\circ}$ .
3. We are asked to find the lengths of $AC$ and $CD$ .
4. We are also asked to find the area of the quadrilateral $ABCD$ .

STEP 2

Since $\triangle ABC$ and $\triangle ACD$ are right triangles, we can use the Pythagorean theorem to find the lengths of $AC$ and $CD$ .
The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. This can be written as $c^2 = a^2 + b^2$ where $c$ is the length of the hypotenuse and $a$ and $b$ are the lengths of the other two sides.

STEP 3

First, let's find $AC$ in $\triangle ABC$ .
$AC^2 = BC^2 + AB^2$

STEP 4

Plug in the given values for $BC$ and $AB$ to calculate $AC$ .
$AC^2 =48^2 +36^2$

STEP 5

Calculate the value of $AC^2$ .
$AC^2 =48^2 +36^2 =2304 +129 =3600$

STEP 6

Take the square root of both sides to find $AC$ .
$AC = \sqrt{3600} =60 \text{ cm}$

STEP 7

Now, let's find $CD$ in $\triangle ACD$ .
$CD^2 = AD^2 - AC^2$

STEP 8

Plug in the given values for $AD$ and the calculated value for $AC$ to calculate $CD$ .
$CD^2 =109^2 -60^2$

STEP 9

Calculate the value of $CD^2$ .
$CD^2 =109^2 -60^2 =11881 -360 =828$

STEP 10

Take the square root of both sides to find $CD$ .
$CD = \sqrt{828} =91 \text{ cm}$

STEP 11

Now that we have the lengths of $AC$ and $CD$ , we can calculate the area of the quadrilateral $ABCD$ .
The area of a right triangle is given by the formula $Area = \frac{}{} \times base \times height$

STEP 12

The area of quadrilateral $ABCD$ is the sum of the areas of $\triangle ABC$ and $\triangle ACD$ .
$Area_{ABCD} = Area_{ABC} + Area_{AC}$

STEP 13

Plug in the values for the base and height for $\triangle ABC$ and $\triangle ACD$ to calculate the areas.
$Area_{ABC} = \frac{}{2} \times AB \times BC = \frac{}{2} \times36 \times48$ $Area_{AC} = \frac{}{2} \times AC \times CD = \frac{}{2} \times60 \times91$

STEP 14

Calculate the areas of $\triangle ABC$ and $\triangle ACD$ .
$Area_{ABC} = \frac{}{2} \times36 \times48 =864 \text{ cm}^2$ $Area_{AC} = \frac{}{2} \times60 \times91 =2730 \text{ cm}^2$

STEP 15

Add the areas of $\triangle ABC$ and $\triangle ACD$ to find the area of quadrilateral $ABCD$ .
$Area_{ABCD} =864 +2730 =3594 \text{ cm}^2$ So, $AC =60$ cm, $CD =91$ cm, and the area of quadrilateral $ABCD$ is $3594$ cm $^2$ .

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