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Nataly_w [17]
3 years ago
14

Please help me! I am trying to complete this question!​

Mathematics
2 answers:
Iteru [2.4K]3 years ago
7 0

Answer:

131.25 in ^2

Step-by-step explanation:

Find the area of the rectangle on the left

A = l*w

A = 11.75 * 6 = 70.5 in ^2

Now find the area of the rectangle on the right

The length is 9 and  the height is 11.75 - 5 = 6.75

A = 9 * 6.75

A = 60.75 in ^2

Add the areas together

70.5+60.75 = 131.25 in ^2

Natasha2012 [34]3 years ago
6 0

Answer:131.25

Step-by-step explanation:

11 3/4 x 6 = 70.5 (area of first rectangle)

11 3/4 - 5 = 6.75 (to find the height of the other figure)

6.75 x 9 = 60.75 (area of 2 figure)

60.75 + 70.5 = 131.25

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If ABCD is an A4 sheet and BCPO is the square, prove that △OCD is an isosceles triangle. And find the angles marked as 1 to 8 wi
Dmitry [639]

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

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∠BOC = 45°

∠PCO = 45°

∠POC = 45°

∠DOP = 22.5°

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∠ADO = 22.5°

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Step-by-step explanation:

Given:

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∴ 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

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OC = 296.98 = 297

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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°

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For ΔOPD

Tan\ \theta = \frac{opposite}{adjacent}\\ Tan\ (\angle DOP) = \frac{87}{210} \\(\angle DOP) = Tan^-1(0.414)\\(\angle DOP) = 22.5 ^{\circ}

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)

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