could i please ask for help here :(
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Answer:
1. In a parallelogram, opposite angles are congruent (equal) and any two consecutive angles are supplementary (add up to 180 degrees).
2. Since opposite angles in a parallelogram are congruent, angle Q and angle R are also equal to 40 degrees each.
3.
(a) Properties of a rectangle that are not true for all parallelograms include: all angles are right angles and opposite sides are congruent.
(b) Properties of a rhombus that are not true for all parallelograms include: all sides are congruent and opposite angles are congruent.
(c) Properties of a square that are not true for all parallelograms include: all angles are right angles, all sides are congruent, and opposite sides are parallel.
4. A quadrilateral is a parallelogram if any of the following conditions are met:
- Opposite sides are parallel.
- Opposite sides are congruent.
- Opposite angles are congruent.
- Consecutive angles are supplementary (add up to 180 degrees).
- Diagonals bisect each other.
5.
a. PR is perpendicular to QS.
b. QR is congruent to PS and PQ is congruent to RS.
c. To find the area of the kite, we can use the formula A = (d1 x d2)/2, where d1 and d2 are the lengths of the diagonals. In this case, d1 is PR (which we know is 20cm) and d2 is QS (which we know is 12cm), so plugging in the values, we get:
A = (20cm x 12cm)/2
A = 120 square cm
6. In a rhombus, opposite angles are congruent, so we can set up an equation:
4x = 2x + 30
2x = 30
x = 15
So m/I = m angle I / m angle E = (4x)/(2x + 30) = 60/60 = 1.
7. Since D and E are midpoints of AB and AC respectively, we know that DE is half of BC. So BC = 2 x DE = 2 x 18cm = 36cm.
8. Since MATH is an isosceles trapezoid, we know that MA = TH. Let x be the length of segment KL. Then we can set up an equation:
MA + AK + KL + LH + HT = MA + TH
27cm + AK + x + LH + 43cm = 27cm + 27cm
AK + LH + x = 27cm
Since AK and LH are parallel to each other and to MT, we know that they are congruent, so we can set them equal to each other:
AK = LH
AK + AK + x = 27cm
2AK + x = 27cm
We also know that AK + KH = MA = 27cm, so we can substitute in:
AK + x + KH = 27cm
AK + x + AK = 27cm
2AK + x = 27cm
Substituting in AK = LH, we get:
2LH + x = 27cm
Since LH and MT are parallel, we know that they are congruent, so LH = MT = (HT + TH)/2 = (43cm - 27cm)/2 = 8cm. Substituting in:
2(8cm) + x = 27cm
16cm + x = 27cm
x = 11cm
Therefore, the length of segment KL is 11cm.
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