Given kite ABCD with AD=AB, AO=3, and BD=8. Find AD.
This is to find the missing side with Vertex angles of Kite ∠A and ∠C, non-vertex angles are ∠D & ∠B, Main Diagonal bisects ∠A and ∠C and cross diagonal bisects ∠D and ∠B. Please help me find side AD
Answers
Answer:
AD = 5
Step-by-step explanation:
Given:
A kite ABCD with following dimensions:
AD = AB
AO = 3 and
BD = 8
To find:
AD = ?
Solution:
Kindly refer to the attached image for the figure of given dimensions.
In a kite, the following properties hold true:
1. The diagonals of a kite are perpendicular to each other.
2. The longer diagonal(along the vertex angles) bisects the shorter one.
Using above properties:
Looking at the [tex]\triangle AOD:[/tex]
[tex]\angle AOD=90^\circ[/tex]
AO = 3
OD = 4
We know that, Pythagoras theorem hold true in every right angled triangle.
According to Pythagorean theorem:
[tex]\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}[/tex]
[tex]AD^{2} = 4^{2} + 3^{2}\\\Rightarrow AD = \bold{5}[/tex]
The value of AD should be 5.
Calculation of the value of AD:Since kite ABCD with AD=AB, AO=3, and BD=8
Here angle AOD = 90 degrees
And, AO = 3 and OD = 4
So, here we use the Pythagorean theorem
[tex]= \sqrt 3^2 + 4^2\\\\= \sqrt{9 + 16} \\\\ = \sqrt{25}[/tex]
= 5
Hence, The value of AD should be 5.
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Your Assignment
The most massive planet in the solar system, Jupiter, has a mass of about 1,898,000,000,000,000,000,000,000,000 kg.
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Write your answer in the space below.
2. Estimate Mercury's mass using a one-digit whole number times a power of 10. Be sure to include units.
Write your answer in the space below.
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Answers
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2. Mercury, has a mass of about 330,000,000,000,000,000,000,000 kg.
The scientific notation is:- 3.30 x 10²³ kg.
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Answer:
The expression shows the number of dollars he will have saved by the end of the year is given by 35(7)+40(5)=\$44535(7)+40(5)=$445
Step-by-step explanation:
Since we have given that
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35(7)+40(5)=\$44535(7)+40(5)=$445
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Answers
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Will give Brainliest
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5. Multiply by 100 to convert annual rate into a percentage.
Don't know if you still need this, but thought I could help!