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2 years ago

Pls help with steps also

Mathematics

1 answer:

VMariaS [17]2 years ago

6 0

Part (i)

We're told that $\angle A = 60^{\circ}$ which means angle DAB is 60 degrees. This angle is cut into two equal halves of 30 degrees each (for angles DAP and PAB) due to segment AP being a bisector of angle DAB.

Since ABCD is a parallelogram, the adjacent angles A and B are supplementary. A = 60 leads to...

A+B = 180

B = 180 - A

B = 180-60

B = 120

Angle ABC is 120 degrees which splits in half to get 60. The angles PBA and PBC are 60 degrees each.

Focus on triangle PAB. We found A = 30 and B = 60 earlier. That must mean:

P+A+B = 180

P+30+60 = 180

P+90 = 180

P = 180-90

P = 90

Triangle PAB is a right triangle with angle P being the 90 degree angle. This is another way of saying angle APB is 90 degrees.

=====================================================

Part (ii)

From the previous part, we know that angles DAP and PAB are 30 degrees each. The alternate interior angles DPA and PAB are equal because we have a parallelogram, so angle DPA is also 30 degrees.

For triangle DPA, the base angles A and P are congruent (30 degrees each). This leads immediately to the fact triangle DPA is isosceles. The congruent sides are opposite the congruent base angles.

Therefore, AD = DP

-----------------

Also from part (i), we found that angles PBA and PBC were 60 degrees each.

Since we have a parallelogram, the alternate interior angles PBA and BPC are congruent, meaning that angle BPC is also 60.

For triangle PBC, we have the interior angles of:

P = 60
B = 60
C = x = unknown for now

P+B+C = 180

60+60+x = 180

120+x = 180

x = 180-120

x = 60

Each interior angle of triangle PBC is 60 degrees. Triangle PBC is equilateral. By definition, equilateral triangles have all sides the same length.

Therefore, PB = PC = BC.

=====================================================

Part (iii)

ABCD is a parallelogram with congruent opposite sides.

Let AD = BC = x

From part (ii), we found that AD and DP were the same length. So DP = x as well.

Also from part (ii), we found that BC = PC, so PC = x

We can then say:

DC = DP + PC

DC = x + x

DC = 2x

DC = 2*AD

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$\bold{\huge{\underline{ Solution }}}$

We have, 

Line segment AB
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Also, 

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According to the question,

The coordinates of midpoint and one of the end point of line segment AB are ( -8,8) and (-2,-20) .

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$\sf{ 16 = - 20 + x2 }$

$\sf{ y2 = 16 + 20 }$

$\bold{ y2 = 36 }$

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