QuestionIn rectangle $ABCD$ , with $AB = 20$ and $BC = 15$ :
(a) Find $AC$ . (b) Prove $\triangle ABC \sim \triangle CMD$ . (c) Find $MC$ . (d) Find ratio $AM : MC$ .

Studdy Solution

STEP 1

Assumptions1. $ABCD$ is a rectangle. . $M$ is a point lying on $AC$ .
3. $DM \perp AC$ .
4. The lengths of $AB$ and $BC$ are20 and15 units respectively.

STEP 2

We know that in a rectangle, the diagonals are equal. Therefore, the length of $AC$ is equal to the length of $BD$ . Since $ABCD$ is a rectangle, $BD = AB =20$ units and $BC = AD =15$ units.We can use the Pythagorean theorem to find the length of $AC$ .
$AC = \sqrt{AB^2 + BC^2}$

STEP 3

Now, plug in the given values for $AB$ and $BC$ to calculate the length of $AC$ .
$AC = \sqrt{20^2 +15^2}$

STEP 4

Calculate the length of $AC$ .
$AC = \sqrt{400 +225} = \sqrt{625} =25$ So, the length of $AC$ is25 units.

STEP 5

To prove that $\triangle ABC \sim \triangle CMD$ , we need to show that the corresponding angles are equal and the corresponding sides are proportional.
In $\triangle ABC$ and $\triangle CMD$ , we have $\angle ABC = \angle CMD$ (each90 degrees as $ABCD$ is a rectangle) $\angle BAC = \angle MCD$ (each equal to angle $AC$ as $DM \perp AC$ )
So, by AA (Angle-Angle) criterion, $\triangle ABC \sim \triangle CMD$ .

STEP 6

Since $\triangle ABC \sim \triangle CMD$ , the sides are proportional. Therefore, we have $\frac{AB}{CM} = \frac{BC}{MD}$ We know that $AB =20$ units, $BC =15$ units, and $MD = AB =20$ units. We can solve for $CM$ .
$CM = \frac{AB \times MD}{BC}$

STEP 7

Plug in the values for $AB$ , $MD$ , and $BC$ to calculate the length of $CM$ .
$CM = \frac{20 \times20}{15}$

STEP 8

Calculate the length of $CM$ .
$CM = \frac{400}{15} = \frac{80}{3} \approx26.67$ So, the length of $MC$ is approximately26.67 units.

STEP 9

To find the ratio $AM MC$ , we subtract $MC$ from $AC$ to get $AM$ .
$AM = AC - MC$

STEP 10

Plug in the values for $AC$ and $MC$ to calculate the length of $AM$ .
$AM =25 - \frac{80}{3}$

STEP 11

Calculate the length of $AM$ .
$AM = \frac{75}{3} - \frac{80}{3} = -\frac{5}{3} \approx -.67$ The length of $AM$ cannot be negative, so there must be a mistake in our calculations. Let's go back to6 and re-evaluate the proportionality of sides.

STEP 12

Since $\triangle ABC \sim \triangle CMD$ , the sides are proportional. Therefore, we have $\frac{AB}{BC} = \frac{CM}{MD}$ We know that $AB =20$ units, $BC =15$ units, and $MD = AB =20$ units. We can solve for $CM$ .
$CM = \frac{AB \times MD}{BC}$

STEP 13

Plug in the values for $AB$ , $MD$ , and $BC$ to calculate the length of $CM$ .
$CM = \frac{20 \times20}{15}$

STEP 14

Calculate the length of $CM$ .
$CM = \frac{400}{} = \frac{80}{3} \approx26.67$ So, the length of $MC$ is approximately26.67 units.

STEP 15

To find the ratio $AM MC$ , we subtract $MC$ from $AC$ to get $AM$ .
$AM = AC - MC$

STEP 16

Plug in the values for $AC$ and $MC$ to calculate the length of $AM$ .
$AM =25 - \frac{80}{3}$

STEP 17

Calculate the length of $AM$ .
$AM = \frac{75}{3} - \frac{80}{3} = -\frac{5}{3} \approx -.67$ The length of $AM$ cannot be negative, so there must be a mistake in our calculations. Let's go back to6 and re-evaluate the proportionality of sides.

STEP 18

Since $\triangle ABC \sim \triangle CMD$ , the sides are proportional. Therefore, we have $\frac{AB}{BC} = \frac{MD}{CM}$ We know that $AB =20$ units, $BC =15$ units, and $MD = AB =20$ units. We can solve for $CM$ .
$CM = \frac{MD \times BC}{AB}$

STEP 19

Plug in the values for $MD$ , $BC$ , and $AB$ to calculate the length of $CM$ .
$CM = \frac{ \times15}{}$

STEP 20

Calculate the length of $CM$ .
$CM = \frac{300}{20} =15$ So, the length of $MC$ is15 units.

STEP 21

To find the ratio $AM MC$ , we subtract $MC$ from $AC$ to get $AM$ .
$AM = AC - MC$

STEP 22

Plug in the values for $AC$ and $MC$ to calculate the length of $AM$ .
$AM =25 -15$

STEP 23

Calculate the length of $AM$ .
$AM =10$ So, the length of $AM$ is10 units.

STEP 24

The ratio $AM MC$ is then $1015$ , which simplifies to $3$ .
(a) The length of $AC$ is units. (b) $\triangle ABC \sim \triangle CMD$ . (c) The length of $MC$ is15 units. (d) The ratio $AM MC$ is $3$ .

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