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### Access answers to Maths RD Sharma Solutions For Class 8 Exercise 17.1 Chapter 17 Understanding Shapes- III (Special Types of Quadrilaterals)

**1. Given below is a parallelogram ABCD. Complete each statement along with the definition or property used.(i) AD =**

**(ii) ∠DCB =(iii) OC =**

**(iv) ∠DAB + ∠CDA =**

**Solution:**

(i) AD = BC. Because, diagonals bisect each other in a parallelogram.

(ii) ∠DCB = ∠BAD. Because, alternate interior angles are equal.

(iii) OC = OA. Because, diagonals bisect each other in a parallelogram.

(iv) ∠DAB+ ∠CDA = 180°. Because sum of adjacent angles in a parallelogram is 180°.

**2. The following figures are parallelograms. Find the degree values of the unknowns x, y, z.**

**Solution:**

**(i)** ∠ABC = ∠y = 100^{o} (opposite angles are equal in a parallelogram)

∠x + ∠y = 180^{o} (sum of adjacent angles is = 180^{o }in a parallelogram)

∠x + 100^{o} = 180^{o}

∠x = 180^{o} – 100^{o}

= 80^{o}

∴ ∠x = 80^{o} ∠y = 100^{o}∠z = 80^{o} (opposite angles are equal in a parallelogram)

**(ii)** ∠RSP + ∠y = 180^{o} (sum of adjacent angles is = 180^{o }in a parallelogram)

∠y + 50^{o} = 180^{o}

∠y = 180^{o} – 50^{o}

= 130^{o}

∴ ∠x = ∠y = 130^{o} (opposite angles are equal in a parallelogram)

∠RSP = ∠RQP = 50^{o} (opposite angles are equal in a parallelogram)

∠RQP + ∠z = 180^{o} (linear pair)

50^{o} + ∠z = 180^{o}

∠z = 180^{o} – 50^{o}

= 130^{o}

∴ ∠x = 130^{o} ∠y = 130^{o} ∠z = 130^{o}

**(iii)** In ΔPMN

∠NPM + ∠NMP + ∠MNP = 180° [Sum of all the angles of a triangle is 180°]

30° + 90° + ∠z = 180°

∠z = 180°-120°

= 60°

∠y = ∠z = 60° [opposite angles are equal in a parallelogram]

∠z = 180°-120° [sum of the adjacent angles is equal to 180° in a parallelogram]

∠z = 60°

∠z + ∠LMN = 180° [sum of the adjacent angles is equal to 180° in a parallelogram]

60° + 90°+ ∠x = 180°

∠x = 180°-150°

∠x = 30°

∴ ∠x = 30^{o} ∠y = 60^{o} ∠z = 60^{o}

**(iv)** ∠x = 90° [vertically opposite angles are equal]

In ΔDOC

∠x + ∠y + 30° = 180° [Sum of all the angles of a triangle is 180°]

90° + 30° + ∠y = 180°

∠y = 180°-120°

∠y = 60°

∠y = ∠z = 60° [alternate interior angles are equal]

∴ ∠x = 90^{o} ∠y = 60^{o} ∠z = 60^{o}

**(v)** ∠x + ∠POR = 180° [sum of the adjacent angles is equal to 180° in a parallelogram]

∠x + 80° = 180°

∠x = 180°-80°

∠x = 100°

∠y = 80° [opposite angles are equal in a parallelogram]

∠SRQ =∠x = 100°

∠SRQ + ∠z = 180° [Linear pair]

100° + ∠z = 180°

∠z = 180°-100°

∠z = 80°

∴ ∠x = 100^{o} ∠y = 80^{o} ∠z = 80^{o}

**(vi)** ∠y = 112° [In a parallelogram opposite angles are equal]

∠y + ∠VUT = 180° [In a parallelogram sum of the adjacent angles is equal to 180°]

∠z + 40° + 112° = 180°

∠z = 180°-152°

∠z = 28°

∠z =∠x = 28° [alternate interior angles are equal]

∴ ∠x = 28^{o} ∠y = 112^{o} ∠z = 28^{o}

**3. Can the following figures be parallelograms? Justify your answer.**

**Solution:**

**(i)** No, opposite angles are equal in a parallelogram.

**(ii)** Yes, opposite sides are equal and parallel in a parallelogram.

**(iii)** No, diagonals bisect each other in a parallelogram.

**4. In the adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the geometrical truths you use to find them.**

**Solution:**

We know that

∠POH + 70° = 180° [Linear pair]

∠POH = 180°-70°

∠POH = 110°

∠POH = ∠x = 110° [opposite angles are equal in a parallelogram]

∠x + ∠z + 40° = 180° [sum of the adjacent angles is equal to 180° in a parallelogram]

110° + ∠z + 40° = 180°

∠z = 180° – 150°

∠z = 30°

∠z +∠y = 70°

∠y + 30° = 70°

∠y = 70°- 30°

∠y = 40°

**5.** **In the following figures GUNS and RUNS are parallelograms. Find x and y.**

**Solution:**

**(i)** 3y – 1 = 26 [opposite sides are of equal length in a parallelogram]

3y = 26 + 1

y = 27/3

y = 9

3x = 18 [opposite sides are of equal length in a parallelogram]

x = 18/3

x = 6

∴ x = 6 and y = 9

**(ii)** y – 7 = 20 [diagonals bisect each other in a parallelogram]

y = 20 + 7

y = 27

x – y = 16 [diagonals bisect each other in a parallelogram]

x -27 = 16

x = 16 + 27

= 43

∴ x = 43 and y = 27

**6. In the following figure RISK and CLUE are parallelograms. Find the measure of x.**

**Solution:**

In parallelogram RISK

∠RKS + ∠KSI = 180° [sum of the adjacent angles is equal to 180° in a parallelogram]

120° + ∠KSI = 180°

∠KSI = 180° – 120°

∠KSI = 60°

In parallelogram CLUE

∠CEU = ∠CLU = 70° [opposite angles are equal in a parallelogram]

In ΔEOS

70° + ∠x + 60° = 180° [Sum of angles of a triangles is 180°]

∠x = 180° – 130°

∠x = 50°

∴ x = 50°

**7. Two opposite angles of a parallelogram are (3 x – 2)^{o} and (50 – x)^{o}. Find the measure of each angle of the parallelogram.**

**Solution:**

We know that opposite angles of a parallelogram are equal.

So, (3x – 2)° = (50 – x)°

3x^{o} – 2° = 50° – x^{o}

3x° + x^{o} = 50° + 2°

4x^{o} = 52°

x^{o} = 52^{o}/4

= 13^{o}

Measure of opposite angles are,

(3x – 2)^{o} = 3×13 – 2 = 37°

(50 – x)^{o} = 50 – 13 = 37°

We know that Sum of adjacent angles = 180°

Other two angles are 180° – 37° = 143°

∴ Measure of each angle is 37^{o}, 143^{o}, 37^{o}, 143^{o}

**8. If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.**

**Solution:**

Let us consider one of the adjacent angle as x°

Other adjacent angle is = 2x^{o}/3

We know that sum of adjacent angles = 180°

So,

x^{o} + 2x^{o}/3 = 180^{o}

(3x^{o} + 2x^{o})/3 = 180^{o}

5x^{o}/3 = 180^{o}

x^{o} = 180^{o}×3/5

= 108^{o}

Other angle is = 180^{o} – 108^{o} = 72^{o}

∴ Angles of a parallelogram are 72^{o}, 72^{o}, 108^{o}, 108^{o}

**9. The measure of one angle of a parallelogram is 70°. What are the measures of the remaining angles?**

**Solution:**

Let us consider one of the adjacent angle as x°

Other adjacent angle = 70°

We know that sum of adjacent angles = 180°

So,

x^{o} + 70^{o} = 180^{o}

x^{o} = 180^{o} – 70^{o}

= 110^{o}

∴ Measures of the remaining angles are 70°, 70°, 110° and 110°

**10. Two adjacent angles of a parallelogram are as 1 : 2. Find the measures of all the angles of the parallelogram.**

**Solution:**

Let us consider one of the adjacent angle as x°

Other adjacent angle = 2x°

We know that sum of adjacent angles = 180°

So,

x^{o} + 2x^{o} = 180^{o}

3x^{o} = 180^{o}

x^{o} = 180^{o}/3

= 60^{o}

So other angle is 2x = 2×60 = 120^{o}

∴ Measures of the remaining angles are 60°, 60°, 120° and 120°

**11. In a parallelogram ABCD, ∠D= 135°, determine the measure of ∠A and ∠B.**

**Solution:**

Given, one of the adjacent angle ∠D = 135°

Let other adjacent angle ∠A be = x°

We know that sum of adjacent angles = 180°

x^{o} + 135^{o} = 180^{o}

x^{o} = 180^{o} – 135^{o}

= 45^{o}

∠A = x^{o} = 45^{o}

We know that measure of opposite angles are equal in a parallelogram.

So, ∠A = ∠C = 45^{o}

And ∠D = ∠B = 135^{o}

**12. ABCD is a parallelogram in which ∠A = 70°. Compute ∠B, ∠C and ∠D.**

**Solution:**

Given, one of the adjacent angle ∠A = 70^{o}

Other adjacent angle ∠B be = x^{o}

We know that sum of adjacent angles = 180°

x^{o} + 70^{o} = 180^{o}

x^{o} = 180^{o} – 70^{o}

= 110^{o}

∠B = x^{o} = 110^{o}

We know that measure of opposite angles are equal in a parallelogram.

So, ∠A = ∠C = 70^{o}

And ∠D = ∠B = 110^{o}

**13. The sum of two opposite angles of a parallelogram is 130°. Find all the angles of the parallelogram.**

**Solution:**

Consider ABCD as a parallelogram

∠A + ∠C = 130^{0}

Here ∠A and ∠C are opposite angles

So ∠C = 130/2 = 65^{0}

We know that sum of adjacent angles is 180^{0}

∠B + ∠D = 180^{0}

65^{0} + ∠D = 180^{0}

∠D = 180^{0} – 65^{0} = 115^{0}

∠D = ∠B = 115^{0} ( Opposite angles)

Therefore, ∠A = 65^{0}, ∠B = 115^{0}, ∠C = 65^{0} and ∠D = 115^{0}.

**14. All the angles of a quadrilateral are equal to each other. Find the measure of each. Is the quadrilateral a parallelogram? What special type of parallelogram is it?**

**Solution:**

Let us consider each angle of a parallelogram as x^{o}

We know that sum of angles = 360^{o}

x^{o} + x^{o} + x^{o} + x^{o} = 360^{o}

4 x^{o} = 360^{o}

x^{o} = 360^{o}/4

= 90^{o}

∴ Measure of each angle is 90^{o}

Yes, this quadrilateral is a parallelogram.

Since each angle of a parallelogram is equal to 90°, so it is a rectangle.

**15. Two adjacent sides of a parallelogram are 4 cm and 3 cm respectively. Find its perimeter.**

**Solution:**

We know that opposite sides of a parallelogram are parallel and equal.

So, Perimeter = Sum of all sides (there are 4 sides)

Perimeter = 4 + 3 + 4 + 3

= 14 cm

∴ Perimeter is 14cm.

**16. The perimeter of a parallelogram is 150 cm. One of its sides is greater than the other by 25 cm. Find the length of the sides of the parallelogram.**

**Solution:**

Given, Perimeter of the parallelogram = 150 cm

Let us consider one of the sides as = ‘x’ cm

Other side as = (x + 25) cm

We know that opposite sides of a parallelogram are parallel and equal.

So, Perimeter = Sum of all sides

x + x + 25 + x + x + 25 = 150

4x + 50 = 150

4x = 150 – 50

x = 100/4

= 25

∴ Sides of the parallelogram are (x) = 25 cm and (x+25) = 50 cm.

**17. The shorter side of a parallelogram is 4.8 cm and the longer side is half as much again as the shorter side. Find the perimeter of the parallelogram.**

**Solution:**

Given, Shorter side of the parallelogram = 4.8 cm

Longer side of the parallelogram = 4.8 + 4.8/2

= 4.8 + 2.4

= 7.2cm

We know that opposite sides of a parallelogram are parallel and equal.

So, Perimeter = Sum of all sides

Perimeter of the parallelogram = 4.8 + 7.2 + 4.8 + 7.2

= 24cm

∴ Perimeter of the parallelogram is 24 cm.

**18. Two adjacent angles of a parallelogram are (3 x-4)^{o} and (3x+10)^{o}. Find the angles of the parallelogram.**

**Solution:**

We know that adjacent angles of a parallelogram are equal.

So, (3x – 4)° + (3x + 10)° = 180^{o}

3x° + 3x^{o} – 4 + 10 = 180°

6x = 180° – 6^{o}

x = 174^{o}/6

= 29^{o}

Measure of adjacent angles are,

(3x – 4)^{o} = 3×29 – 4 = 83°

(3x + 10)^{o} = 3×29 + 10 = 97°

We know that Sum of adjacent angles = 180°

∴ Measure of each angle is 83^{o}, 97^{o}, 83^{o}, 97^{o}

**19. In a parallelogram ABCD, the diagonals bisect each other at O. If ∠ABC =30°, ∠BDC= 10° and ∠CAB =70°. Find:∠DAB, ∠ADC, ∠BCD, ∠AOD, ∠DOC, ∠BOC, ∠AOB, ∠ACD, ∠CAB, ∠ADB, ∠ACB, ∠DBC, and ∠DBA.**

**Solution:**

Firstly let us draw a parallelogram

Given, ∠ABC = 30^{o},

∠ABC = ∠ADC = 30° [We know that measure of opposite angles are equal in a parallelogram]

∠BDC = 10°

∠CAB =70°

∠BDA = ∠ADB = ∠ADC – ∠BDC = 30° – 10° = 20°

∠DAB = 180° – 30° = 150°

∠ADB = ∠DBC = 20^{o} (alternate angles)

∠BCD = ∠DAB = 150° [we know, opposite angles are equal in a parallelogram]

∠DBA = ∠BDC = 10° [we know, Alternate interior angles are equal]

In ΔABC

∠CAB + ∠ABC + ∠BCA = 180° [since, sum of all angles of a triangle is 180°]

70^{o} + 30^{o} + ∠BCA = 180°

∠BCA = 180^{o} – 100^{o}

= 80^{o}

∠DAB = ∠DAC + ∠CAB = 70^{o} + 80^{o} = 150^{o}

∠BCD = 150^{o} (opposite angle of the parallelogram)

∠DCA = ∠CAB = 70^{o}

In ΔDOC

∠BDC + ∠ACD + ∠DOC = 180° [since, sum of all angles of a triangle is 180°]

10° + 70° + ∠DOC = 180°

∠DOC = 180°- 80°

∠DOC = 100°

So, ∠DOC = ∠AOB = 100° [Vertically opposite angles are equal]

∠DOC + ∠AOD = 180° [Linear pair]

100° + ∠AOD = 180°

∠AOD = 180°- 100°

∠AOD = 80°

So, ∠AOD = ∠BOC = 80° [Vertically opposite angles are equal]

∠CAB = 70^{o}

∠ABC + ∠BCD = 180° [In a parallelogram sum of adjacent angles is 180°]

30° + ∠ACB + ∠ACD = 180°

30° + ∠ACB + 70° = 180°

∠ACB = 180° – 100°

∠ACB = 80°

∴ ∠DAB = 150^{o}, ∠ADC = 30, ∠BCD = 150^{o}, ∠AOD = 80^{o}, ∠DOC = 100^{o}, ∠BOC = 80^{o}, ∠AOB = 100^{o}, ∠ACD = 70^{o}, ∠CAB = 70^{o}, ∠ADB = 20^{o}, ∠ACB = 80^{o}, ∠DBC = 20^{o}, and ∠DBA = 10^{o}.

**20. Find the angles marked with a question mark shown in Figure.**

**Solution:**

In ΔBEC

∠BEC + ∠ECB +∠CBE = 180° [Sum of angles of a triangle is 180°]

90° + 40° + ∠CBE = 180°

∠CBE = 180°-130°

∠CBE = 50°

∠CBE = ∠ADC = 50° (Opposite angles of a parallelogram are equal)

∠B = ∠D = 50° [Opposite angles of a parallelogram are equal]

∠A + ∠B = 180° [Sum of adjacent angles of a triangle is 180°]

∠A + 50° = 180°

∠A = 180°-50°

So, ∠A = 130°

In ΔDFC

∠DFC + ∠FCD +∠CDF = 180° [Sum of angles of a triangle is 180°]

90° + ∠FCD + 50° = 180°

∠FCD = 180°-140°

∠FCD = 40°

∠A = ∠C = 130° [Opposite angles of a parallelogram are equal]

∠C = ∠FCE +∠BCE + ∠FCD

∠FCD + 40° + 40° = 130°

∠FCD = 130° – 80°

∠FCD = 50°

∴ ∠EBC = 50^{o}, ∠ADC = 50^{o} and ∠FCD = 50^{o}

**21. The angle between the altitudes of a parallelogram, through the same vertex of an obtuse angle of the parallelogram is 60°. Find the angles of the parallelogram.**

**Solution:**

Let us consider a parallelogram, ABCD. Where, DP⊥AB and DQ ⊥BC.

Given ∠PDQ = 60°

In quadrilateral DPBQ

∠PDQ + ∠DPB + ∠B + ∠BQD = 360° [Sum of all the angles of a Quadrilateral is 360°]

60° + 90° + ∠B + 90° = 360°

∠B = 360° – 240°

∠B = 120°

∠B = ∠D = 120° [Opposite angles of parallelogram are equal]

∠B + ∠C = 180° [Sum of adjacent interior angles in a parallelogram is 180°]

120° + ∠C = 180°

∠C = 180° – 120° = 60°

∠A = ∠C = 60° (Opposite angles of parallelogram are equal)

∴ Angles of a parallelogram are 60^{o}, 120^{o}, 60^{o}, 120^{o}

**22. In Figure, ABCD and AEFG are parallelograms. If ∠C =55°, what is the measure of ∠F?**

**Solution:**

In parallelogram ABCD

∠C = ∠A = 55° [In a parallelogram opposite angles are equal in a parallelogram]

In parallelogram AEFG

∠A = ∠F = 55° [In a parallelogram opposite angles are equal in a parallelogram]

∴ Measure of ∠F = 55°

**23. In Figure, BDEF and DCEF are each a parallelogram. Is it true that BD = DC? Why or why not?**

**Solution:**

In parallelogram BDEF

BD = EF [In a parallelogram opposite sides are equal]

In parallelogram DCEF

DC = EF [In a parallelogram opposite sides are equal]

Since, BD = EF = DC

So, BD = DC

**24. In Figure, suppose it is known that DE = DF. Then, is ΔABC isosceles? Why or why not?**

**Solution:**

In parallelogram BDEF

BD = EF and BF = DE [opposite sides are equal in a parallelogram]

In parallelogram DCEF

DC = EF and DF = CE [opposite sides are equal in a parallelogram]

In parallelogram AFDE

AF = DE and DF = AE [opposite sides are equal in a parallelogram]

So, DE = AF = BF

Similarly: DF = CE = AE

Given, DE = DF

Since, DF = DF

AF + BF = CE + AE

AB = AC

∴ ΔABC is an isosceles triangle.

**25. Diagonals of parallelogram ABCD intersect at O as shown in Figure. XY contains O, and X, Y are points on opposite sides of the parallelogram. Give reasons for each of the following:**

**(i) OB = OD**

**(ii) ∠OBY = ∠ODX**

**(iii) ∠BOY = ∠DOX**

**(iv) ΔBOY = ΔDOX**

**Now, state if XY is bisected at O.**

**Solution:**

(i) OB = OD

OB = OD. Since diagonals bisect each other in a parallelogram.

(ii) ∠OBY =∠ODX

∠OBY =∠ODX. Since alternate interior angles are equal in a parallelogram.

(iii) ∠BOY= ∠DOX

∠BOY= ∠DOX. Since vertical opposite angles are equal in a parallelogram.

(iv) ΔBOY ≅ ΔDOX

ΔBOY and ΔDOX. Since OB = OD, where diagonals bisect each other in a parallelogram.

∠OBY =∠ODX [Alternate interior angles are equal]

∠BOY= ∠DOX [Vertically opposite angles are equal]

ΔBOY ≅ΔDOX [by ASA congruence rule]

OX = OY [Corresponding parts of congruent triangles]

∴ XY is bisected at O.

**26. In Fig. 17.31, ABCD is a parallelogram, CE bisects ∠C and AF bisects ∠A. In each of the following, if the statement is true, give a reason for the same:**

**(i) ∠A = ∠C**

**(ii) ∠FAB = ½ ∠A**

**(iii) ∠DCE = ½ ∠C**

**(iv) ∠CEB = ∠FAB**

**(v) CE ∥ AF**

**Solution:**

(i) ∠A = ∠C

True, Since ∠A =∠C = 55° [opposite angles are equal in a parallelogram]

(ii) ∠FAB = ½ ∠A

True, Since AF is the angle bisector of ∠A.

(iii) ∠DCE= ½ ∠C

True, Since CE is the angle bisector of angle ∠C.

(iv) ∠CEB= ∠FAB

True,

Since ∠DCE = ∠FAB (opposite angles are equal in a parallelogram).

∠CEB = ∠DCE (alternate angles)

½ ∠C = ½ ∠A [AF and CE are angle bisectors]

(v) CE || AF

True, since one pair of opposite angles are equal, therefore quad. AEFC is a parallelogram.

**27. Diagonals of a parallelogram ABCD intersect at O. AL and CM are drawn perpendiculars to BD such that L and M lie on BD. Is AL = CM? Why or why not?**

**Solution:**

Given, AL and CM are perpendiculars on diagonal BD.

In ΔAOL and ΔCOM

∠AOL = ∠COM (vertically opposite angle) ….. (i)

∠ALO = ∠CMO = 90^{o} (each right angle) ……. (ii)

By using angle sum property

∠AOL + ∠ALO + ∠LAO = 180^{o} ……… (iii)

∠COM + ∠CMO + ∠OCM = 180^{o} ……. (iv)

From (iii) and (iv)

∠AOL + ∠ALO + ∠LAO = ∠COM + ∠CMO + ∠OCM

∠LAO = ∠OCM (from (i) and (ii))

In ΔAOL and ΔCOM

∠ALO = ∠CMO (each right angle)

AO = OC (diagonals of a parallelogram bisect each other)

∠LAO = ∠OCM (proved)

So, ΔAOL is congruent to ΔCOM

∴ AL = CM (Corresponding parts of congruent triangles)

**28. Points E and F lie on diagonals AC of a parallelogram ABCD such that AE = CF. what type of quadrilateral is BFDE?**

**Solution:**

In parallelogram ABCD:

AO = OC…………….. (i) (Diagonals of a parallelogram bisect each other)

AE = CF…….. (ii) Given

On subtracting (ii) from (i)

AO – AE = OC – CF

EO = OF ….. (iii)

In ΔDOE and ΔBOF

EO = OF (proved)

DO = OB (Diagonals of a parallelogram bisect each other)

∠DOE = ∠BOF (vertically opposite angles are equal in a parallelogram)

By the rule of SAS congruence ΔDOE ≅ ΔBOF

So, DE = BF (Corresponding parts of congruent triangles)

In ΔBOE and ΔDOF

EO = OF (proved)

DO = OB (diagonals of a parallelogram bisect each other)

∠DOF = ∠BOE (vertically opposite angles are equal in a parallelogram)

By the rule of SAS congruence ΔDOE ≅ ΔBOF

∴ DF = BE (Corresponding parts of congruent triangles)

∴ BFDE is a parallelogram, since one pair of opposite sides are equal and parallel.

**29. In a parallelogram ABCD, AB = 10cm, AD = 6 cm. The bisector of ∠A meets DC in E, AE and BC produced meet at F. Find the length CF.**

**Solution:**

In a parallelogram ABCD

Given, AB = 10 cm, AD = 6 cm

⇒ CD = AB = 10 cm and AD = BC = 6 cm [In a parallelogram opposite sides are equal]

AE is the bisector of ∠DAE = ∠BAE = x

∠BAE = ∠AED = x (alternate angles are equal)

ΔADE is an isosceles triangle. Since opposite angles in ΔADE are equal.

AD = DE = 6cm (opposite sides are equal)

CD = DE + EC

EC = CD – DE

= 10 – 6

= 4cm

∠DEA = ∠CEF = x (vertically opposite angle are equal)

∠EAD = ∠EFC = x (alternate angles are equal)

ΔEFC is an isosceles triangle. Since opposite angles in ΔEFC are equal.

CF = CE = 4cm (opposite side are equal to angles)

∴ CF = 4cm.

**All Chapter RD Sharma Solutions For Class 8 Maths**

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