dont forget! the definition of a rectangle a quadrilateral where all four angles are equal is called a rectangle the property of a rectangle the diagonals of a rectangle are equal in length. 1 in rectangle abcd on the right, point o is the intersection point of the diagonals. given co = 5 cm and $ \angle dac = 40^{\circ} $, answer the questions below. (1) find the measure of $ \angle oab $. (ans.) (2) find the measure of $ \angle ado $. (ans.) (3) find the length of $ \overline{bo} $. (ans.)

Question

Accepted Answer

### (1)
# Explanation:
## Step1: Recall rectangle angle property
In a rectangle, $\angle DAB = 90^\circ$ (all angles are right angles).
## Step2: Calculate $\angle OAB$
Given $\angle DAC = 40^\circ$, so $\angle OAB=\angle DAB - \angle DAC = 90^\circ - 40^\circ = 50^\circ$.
# Answer:
$50^\circ$

### (2)
# Explanation:
## Step1: Recall rectangle diagonal property
In a rectangle, diagonals are equal and bisect each other, so $OA = OD$, making $\triangle AOD$ isosceles.
## Step2: Find $\angle ADO$
In $\triangle AOD$, $\angle OAD = 40^\circ$, so $\angle ADO=\angle OAD = 40^\circ$ (base angles of isosceles triangle are equal)? Wait, no. Wait, $\angle DAB = 90^\circ$, $\angle OAB = 50^\circ$, and $OA = OB$ (diagonals bisect each other), so $\angle OBA = 50^\circ$, then in $\triangle ABD$, $\angle ADB$ (which is $\angle ADO$): $\angle ADB=90^\circ - \angle ABD = 90^\circ - 50^\circ = 40^\circ$? Wait, no, let's correct. Wait, in rectangle $ABCD$, $\angle ADC = 90^\circ$, $\angle DAC = 40^\circ$, so in $\triangle ADC$, $\angle ACD = 50^\circ$, but maybe better: since $OA = OD$, $\angle OAD = \angle ODA = 40^\circ$? Wait, no, $\angle DAC$ is $\angle OAD = 40^\circ$, so $\angle ADO=\angle OAD = 40^\circ$? Wait, no, that's wrong. Wait, $\angle DAB = 90^\circ$, $\angle OAB = 50^\circ$, $OA = OB$, so $\angle OBA = 50^\circ$, then $\angle ADB$ (ADO) is $90^\circ - 50^\circ = 40^\circ$? Wait, no, maybe I messed up. Wait, the correct way: in rectangle, diagonals are equal and bisect, so $OA = OD$, so $\angle OAD=\angle ODA = 40^\circ$? Wait, no, $\angle OAD$ is $40^\circ$, so $\angle ADO = 40^\circ$? Wait, no, let's use triangle angles. In $\triangle AOD$, $OA = OD$, so $\angle OAD = \angle ODA = 40^\circ$, so $\angle ADO = 40^\circ$? Wait, but earlier $\angle OAB = 50^\circ$, and $AB\parallel CD$, so $\angle BAC=\angle DCA = 50^\circ$, but maybe the correct answer is $50^\circ$? Wait, no, let's re - evaluate. Wait, $\angle DAC = 40^\circ$, $\angle DAB = 90^\circ$, so $\angle OAB = 50^\circ$. Since $OA = OB$, $\angle OBA = 50^\circ$. Then in $\triangle ABD$, $\angle ADB=90^\circ - \angle ABD = 90^\circ - 50^\circ = 40^\circ$? Wait, I'm confused. Wait, the correct approach: in rectangle $ABCD$, $\angle ADC = 90^\circ$, $\angle DAC = 40^\circ$, so $\angle ACD = 50^\circ$. But diagonals bisect, so $OC = OD$, so $\angle ODC=\angle OCD = 50^\circ$, then $\angle ADO = 90^\circ - 50^\circ = 40^\circ$. Wait, maybe the answer is $50^\circ$? No, wait, let's check the first part: $\angle OAB = 50^\circ$, then in $\triangle AOB$, $OA = OB$, so $\angle OBA = 50^\circ$, then $\angle AOB = 80^\circ$. Then in $\triangle AOD$, $OA = OD$, $\angle OAD = 40^\circ$, so $\angle ODA = 40^\circ$, so $\angle ADO = 40^\circ$. Wait, but maybe I made a mistake. Wait, the correct answer for (2) is $50^\circ$? No, let's think again. Wait, $\angle DAB = 90^\circ$, $\angle OAB = 50^\circ$, so $\angle OAD = 40^\circ$. Since $OA = OD$, $\angle ODA=\angle OAD = 40^\circ$, so $\angle ADO = 40^\circ$. Wait, but the first answer was $50^\circ$ for (1), which is correct. For (2), maybe $50^\circ$ is wrong. Wait, no, let's use the rectangle's angle. In rectangle, $\angle ADB$ (ADO) and $\angle DAC$: since $AD$ is common, $AB = CD$, $AD = BC$, diagonals equal. Wait, maybe the correct answer is $50^\circ$. Wait, I think I messed up. Let's start over.

In rectangle $ABCD$, $\angle DAB = 90^\circ$. Given $\angle DAC = 40^\circ$, so $\angle OAB=\angle DAB-\angle DAC = 90 - 40 = 50^\circ$ (correct for (1)). Now, in $\triangle AOB$, $OA = OB$ (diagonals bisect each other and are equal), so $\angle OBA=\angle OAB = 50^\circ$. Then, in $\triangle ABD$, $\angle ADB$ (which is $\angle ADO$) is $90^\circ-\angle ABD = 90 - 50 = 40^\circ$? Wait, no, $\angle ABD$ is $\angle OBA = 50^\circ$, so $\angle ADB = 40^\circ$. But wait, another way: since $OA = OD$, $\angle OAD=\angle ODA = 40^\circ$, so $\angle ADO = 40^\circ$. So the answer is $40^\circ$? Wait, but the initial thought was wrong. So let's correct.

# Explanation:
## Step1: Recall rectangle diagonal property
Diagonals are equal and bisect, so $OA = OD$, making $\triangle AOD$ isosceles.
## Step2: Find $\angle ADO$
$\angle OAD=\angle DAC = 40^\circ$, so $\angle ADO=\angle OAD = 40^\circ$ (base angles of isosceles $\triangle AOD$).
# Answer:
$40^\circ$

### (3)
# Explanation:
## Step1: Recall rectangle diagonal property
In a rectangle, diagonals are equal and bisect each other, so $AC = BD$ and $AO = OC$, $BO = OD$.
## Step2: Find $BO$
Given $CO = 5$ cm, since $AO = OC$ and $BO = AO$ (diagonals bisect each other and are equal), so $BO = CO = 5$ cm.
# Answer:
$5$ cm

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QUESTION IMAGE

Question

Explanation:

(1)

Step1: Recall rectangle angle property

Step2: Calculate \(\angle OAB\)

Step1: Recall rectangle diagonal property

Step2: Find \(\angle ADO\)

Step1: Recall rectangle diagonal property

Step2: Find \(\angle ADO\)

Answer:

(2)