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Times both sides b 2
2V=(bh)H
remember associative property
(ab)c=a(bc)
and
ab=ba
so
(bh)H=h(bH)
2V=h(bH)
divide both sides by bH
2V=(bh)H
remember associative property
(ab)c=a(bc)
and
ab=ba
so
(bh)H=h(bH)
2V=h(bH)
divide both sides by bH
Answer:
The diagram for the question is missing, but I found an appropriate diagram fo the question:
Proof:
since OC = CD = 297mm Therefore, Δ OCD is an isoscless triangle
∠BCO = 45°
∠BOC = 45°
∠PCO = 45°
∠POC = 45°
∠DOP = 22.5°
∠PDO = 67.5°
∠ADO = 22.5°
∠AOD = 67.5°
Step-by-step explanation:
Given:
AB = CD = 297 mm
AD = BC = 210 mm
BCPO is a square
∴ BC = OP = CP = OB = 210mm
Solving for OC
OCB is a right anlgled triangle
using Pythagoras theorem
(Hypotenuse)² = Sum of square of the other two sides
(OC)² = (OB)² + (BC)²
(OC)² = 210² + 210²
(OC)² = 44100 + 44100
OC = √(88200
OC = 296.98 = 297
OC = 297mm
An isosceless tringle is a triangle that has two equal sides
Therefore for △OCD
CD = OC = 297mm; Hence, △OCD is an isosceless triangle.
The marked angles are not given in the diagram, but I am assuming it is all the angles other than the 90° angles
Since BC = OB = 210mm
∠BCO = ∠BOC
since sum of angles in a triangle = 180°
∠BCO + ∠BOC + 90 = 180
(∠BCO + ∠BOC) = 180 - 90
(∠BCO + ∠BOC) = 90°
since ∠BCO = ∠BOC
∴ ∠BCO = ∠BOC = 90/2 = 45
∴ ∠BCO = 45°
∠BOC = 45°
∠PCO = 45°
∠POC = 45°
For ΔOPD
Note that DP = 297 - 210 = 87mm
∠PDO + ∠DOP + 90 = 180
∠PDO + 22.5 + 90 = 180
∠PDO = 180 - 90 - 22.5
∠PDO = 67.5°
∠ADO = 22.5° (alternate to ∠DOP)
∠AOD = 67.5° (Alternate to ∠PDO)
Hey!
Explanation:
↓↓↓↓↓↓↓↓↓↓
First, remove parenthesis.
Secondly, multiple by the numbers.
Third, apply exponents rule.
Finally, refine.
<u><em>Answer⇒⇒⇒⇒28</em></u>
Hope this helps!
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-Charlie