4 in rectangle abcd on the right, point o is the intersection point of the diagonals. answer the questions below.
(1) given $ \angle bdc = 55 ^ { \circ } $, find the measure of $ \angle cad $.
(2) given $ oc = 5 \mathrm { cm } $, find the measure of diagonal bd.
(3) given that quadrilateral abcd is a square, find the measure of $ \angle bac $.
5 in rhombus abcd on the right, given points e and f are placed on bc and cd so that $ be = df $. prove that $ ae = af $. fill the appropriate words in the blanks.
proof
in $ \triangle abe $ and $ \triangle \square $,
given: $ \square = \square \cdots \cdots $ (1)
since quadrilateral abcd is a rhombus,
$ \angle b = \angle d \cdots \cdots $ (2)
$ \square = \square \cdots \cdots $ (3)
from (1), (2), and (3), since the two sides and the included angle are equal respectively,
$ \triangle \square \cong \triangle \square $
therefore, $ \square = \square $.
6 the figure on the right is a trapezoid with $ ad = 5 \mathrm { cm } $, $ bc = 8 \mathrm { cm } $ and the distance between lines ad and bc is $ 10 \mathrm { cm } $. answer the questions below.
(1) find the area of $ \triangle abc $.
(2) find the area of $ \triangle abd $.
(3) what is the area of trapezoid abcd?

Question

Accepted Answer

### Problem 4 (1)
# Explanation:
## Step1: Recall rectangle properties
In rectangle $ABCD$, $AD \parallel BC$, $AB \parallel CD$, and diagonals $AC = BD$, $OA = OC$, $OB = OD$, so $OA = OD$, $\angle ADC = 90^\circ$.
## Step2: Find $\angle ADB$
Given $\angle BDC = 55^\circ$, $\angle ADC = 90^\circ$, so $\angle ADB=\angle ADC - \angle BDC = 90^\circ - 55^\circ = 35^\circ$.
## Step3: Find $\angle CAD$
Since $OA = OD$, $\triangle AOD$ is isosceles, so $\angle CAD=\angle ADB = 35^\circ$.
# Answer:
$35^\circ$

### Problem 4 (2)
# Explanation:
## Step1: Recall rectangle diagonal property
In rectangle $ABCD$, diagonals bisect each other, so $OB = OC$, $OA = OD$, and $BD = AC = 2OC$.
## Step2: Calculate $BD$
Given $OC = 5\space\text{cm}$, then $BD = 2\times OC = 2\times5 = 10\space\text{cm}$.
# Answer:
$10\space\text{cm}$

### Problem 4 (3)
# Explanation:
## Step1: Recall square properties
In square $ABCD$, $\angle BAD = 90^\circ$, and diagonal $AC$ bisects $\angle BAD$.
## Step2: Find $\angle BAC$
So $\angle BAC=\frac{1}{2}\angle BAD=\frac{1}{2}\times90^\circ = 45^\circ$.
# Answer:
$45^\circ$

### Problem 5 (Proof)
# Explanation:
## Step1: Identify triangles
In $\triangle ABE$ and $\triangle ADF$,
## Step2: List given equal sides
Given $BE = DF$ (①),
## Step3: Use rhombus side property
Since $ABCD$ is a rhombus, $AB = AD$ (③),
## Step4: Use rhombus angle property
$\angle B=\angle D$ (②),
## Step5: Prove triangle congruence
By SAS (Side - Angle - Side) congruence criterion, $\triangle ABE\cong\triangle ADF$.
## Step6: Conclude equal sides
Therefore, $AE = AF$.

Filling the blanks:
In $\triangle ABE$ and $\triangle\boldsymbol{ADF}$,
given: $\boldsymbol{BE}=\boldsymbol{DF}$ $\cdots\cdots$①
Since quadrilateral $ABCD$ is a rhombus,
$\angle B=\angle D$ $\cdots\cdots$②
$\boldsymbol{AB}=\boldsymbol{AD}$ $\cdots\cdots$③
From ①, ②, and ③, since the two sides and the included angle are equal respectively,
$\triangle\boldsymbol{ABE}\cong\triangle\boldsymbol{ADF}$
Therefore, $\boldsymbol{AE}=\boldsymbol{AF}$

### Problem 6 (1)
# Explanation:
## Step1: Recall triangle area formula
The area of a triangle is $A=\frac{1}{2}\times base\times height$. For $\triangle ABC$, base $BC = 8\space\text{cm}$, height is the distance between $AD$ and $BC$ which is $10\space\text{cm}$.
## Step2: Calculate area
$A_{\triangle ABC}=\frac{1}{2}\times BC\times h=\frac{1}{2}\times8\times10 = 40\space\text{cm}^2$.
# Answer:
$40\space\text{cm}^2$

### Problem 6 (2)
# Explanation:
## Step1: Recall triangle area formula
For $\triangle ABD$, base $AD = 5\space\text{cm}$, height is the distance between $AD$ and $BC$ (which is $10\space\text{cm}$ as the trapezoid has $AD\parallel BC$).
## Step2: Calculate area
$A_{\triangle ABD}=\frac{1}{2}\times AD\times h=\frac{1}{2}\times5\times10 = 25\space\text{cm}^2$.
# Answer:
$25\space\text{cm}^2$

### Problem 6 (3)
# Explanation:
## Step1: Recall trapezoid area formula
The area of a trapezoid is $A=\frac{1}{2}\times(AD + BC)\times h$, where $AD = 5\space\text{cm}$, $BC = 8\space\text{cm}$, $h = 10\space\text{cm}$.
## Step2: Calculate area
$A=\frac{1}{2}\times(5 + 8)\times10=\frac{1}{2}\times13\times10 = 65\space\text{cm}^2$.
# Answer:
$65\space\text{cm}^2$

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QUESTION IMAGE

Question

Explanation:

Problem 4 (1)

Step1: Recall rectangle properties

Step2: Find \(\angle ADB\)

Step3: Find \(\angle CAD\)

Step1: Recall rectangle diagonal property

Step2: Calculate \(BD\)

Step1: Recall square properties

Step2: Find \(\angle BAC\)

Answer:

Problem 4 (2)