The document NCERT Solutions(Part- 3)- Understanding Quadrilaterals Class 8 Notes | EduRev is a part of the Class 8 Course Mathematics (Maths) Class 8.

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**EXERCISE 3.3**

**Question 1.** Given a parallelogram ABCD. Complete each statement along with the definition or property used.

(i) AD = _____

(ii) âˆ DCB = _____

(iii) OC = _____

(iv) mâˆ DAB + mâˆ CDA = _____

**Solution:**

(i) AD = BC [âˆµ Opposite sides are equal]

(ii) âˆ DCB = âˆ DAB [âˆµ Opposite angles are equal]

(iii) OC = OA [âˆµ Diagonals bisect each other]

(iv)mâˆ DAB + mâˆ CDA = 180Â° [âˆµ Adjacent angles are supplementary]**Question 2.** Consider the following parallelograms. Find the values of the unknowns x, y, z.**Solution:** **(i)** âˆ y = 100Â° [âˆµ Opposite angles of a parallelogram are equal.]

âˆµ Sum of interior angles of a parallelogram = 360Â°

âˆ´ x + y + z + âˆ B = 360Â°

or x + 100Â° + z + 100Â° = 360Â°

or x + z = 360Â° â€“ 100Â° â€“ 100Â° = 160Â°

But x = z

âˆ´ x = z = 160Â°/2 = 80Â°

Thus, **(ii)** âˆµ Opposite angles are equal.

âˆ´ mâˆ 1 = 50

Now, âˆ 1 + z = 180Â° [Linear pair]

or z = 180Â° â€“ âˆ 1 = 180Â° â€“ 50Â° = 130Â°

x + y + 50Â° + 50Â° = 360Â°

or x + y = 360Â° â€“ 50Â° â€“ 50Â° = 260Â°

But x = y

âˆ´ x = y = 260Â°/2Â° = 130Â°

**(iii) **âˆµ Vertically opposite angles are equal,

âˆ´ x = 90Â°

âˆµ Sum of the angles of a triangle = 180Â° âˆ´90Â° + 30Â° + y = 180Â°

or y = 180Â° â€“ 30Â° â€“ 90Â° = 60Â°

In the figure, ABCD is a parallelogram.

âˆ´ AD || BC and BD is a transversal.

âˆ´ y = z [Alternate angles]

But y = 60Â°

âˆ´ z = 60Â°

Thus, x = 90Â°, y = 60Â° and z = 60Â°.**(iv)** ABCD is a parallelogram.

âˆ´ Opposite angles are equal.

âˆ´ y = 80Â°

AB || CD and BC is a transversal.

âˆ´ x + 80Â° = 180Â° [Interior opposite angles]

or x = 180Â° â€“ 80Â° = 100Â°

Again BC || AD and CD is a transversal,

âˆ´ z = 80Â° [Corresponding angles]

Thus, x = 100Â°, y = 80Â° and z = 80Â°**(v)** âˆµ In a parallelogram, opposite angles are equal.

âˆ´ y = 112Â°

In Î” ACD,

x + y + 40Â° = 180Â°

x + 112Â° + 40Â° = 180Â°

âˆ´ x = 180Â° â€“ 112Â° â€“ 40Â° = 28Â°

âˆµ AD || BC and AC is a transversal.

âˆ´ x = z [âˆµ Alternate angles are equal]

and z = 28Â°

Thus, x = 28Â°, y = 112Â° and z = 28Â°**Question 3.** Can a quadrilateral ABCD be a parallelogram if

(i) âˆ D + âˆ B = 180Â°?

(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?

(iii) âˆ A = 70Â° and âˆ C = 65Â°?**Solution: **(i) In a quadrilateral ABCD,

âˆ A + âˆ B = Sum of adjacent angles = 180Â°**âˆ´ **The quadrilateral may be a parallelogram but not always.

(ii) In a quadrilateral ABDC,

AB = DC = 8 cm

AD = 4 cm

BC = 4.4 cm

âˆµ Opposite sides AB and BC are not equal.**âˆ´** It cannot be a parallelogram.

(iii) In a quadrilateral ABCD,

âˆ A = 70Â° and âˆ C = 65Â°

âˆµ Opposite angles âˆ A â‰ âˆ C**âˆ´ **It cannot be a parallelogram.**Question 4.** Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.**Solution: **

In the adjoining figure, ABCD is not a parallelogram such that opposite angles â€“B and â€“D are equal. It is a kite.

**Question 5. **The measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measure of each of the angles of the parallelogram.**Solution: **Let ABCD be a parallelogram in which adjacent angles âˆ A and âˆ B are 3x and 2x respectively, since adjacent angles are supplementary.**âˆ´ **âˆ A + âˆ B = 180Â°**âˆ´ **3x + 2x = 180Â°

or 5x = 180Â°

or x = 180Â°/5Â° = 36Â°**âˆ´ **âˆ A = 3 x 36Â° = 108Â°

and âˆ B = 2 x 36Â° = 72Â°

âˆµ Opposite angles are equal.**âˆ´ ** âˆ D = âˆ B = 72Â°

and âˆ C= âˆ A = 108Â°**âˆ´ ** âˆ A = 108Â°, âˆ B = 72Â°, âˆ C = 108Â° and âˆ D = 72Â°**Question 6. **Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.**Solution:** Let ABCD be a parallelogram such that adjacent angles âˆ A = âˆ B.

Since âˆ A + âˆ B = 180Â°**âˆ´ **A = âˆ B = 180Â°/2 = 90Â°

Since, opposite angles of a parallelogram are equal.**âˆ´ ** âˆ A= âˆ C = 90Â°

and âˆ B= âˆ D = 90Â°

Thus, âˆ A = 90Â°, âˆ B = 90Â°,âˆ C = 90Â° and âˆ D = 90Â°.

**Question 7. The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.****Solution:**

y + z = 70Â° â€¦(1)

In a triangle, exterior angles is equal to the sum of interior angles opposite.

âˆ´ In Î” âˆ HOP, âˆ HOP = 180Â° -(y + z)

= 180Â° â€“ 70Â°

= 110Â°

Now x = âˆ HOP [Opposite angles of a parallelogram are equal]

âˆ´ x = 110Â° EH || OP and PH is a transversal.

âˆ´ y = 40Â° [Alternate angles are equal]

From (1), 40Â° + z = 70Â°

âˆ´ z = 70Â° â€“ 40Â° = 30Â°

Thus, x = 110Â°, y = 40Â° and z = 30Â°**Question 8.** The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm.)

**Solution:** (i) âˆµ GUNS is a parallelogram.

âˆ´ Its opposite sides are equal.

âˆ´ GS = NU and SN = GU

or 3x = 18 and 26 = 3y â€“ 1

Now 3x = 18

â‡’ x = 18/3 = 6

3y â€“ 1 = 26

â‡’ y = 26 + 1/3 = 27/3 = 9

Thus, x = 6 cm and y = 9 cm

(ii) RUNS is a parallelogram and thus its diagonals bisect each other.

âˆ´ x + y = 16 and 7 + y = 20

i.e. y = 20 â€“ 7

or y = 13

âˆ´ From x + y = 16, we have

x + 13 = 16

or x = 16 â€“ 13 = 3

Thus, x = 3 cm and y = 13 cm.**Question 9.** In the following figure, both RISK and CLUE are parallelogram. Find the value of x.**Solution: **

RISK is a parallelogram.

âˆ´ âˆ R + âˆ K = 180Â° [âˆµ Adjacent angles of a parallelogram are supplementary]

or âˆ R + 120Â° = 180Â°

â‡’ âˆ R = 180Â° â€“ 120Â° = 60Â°

But âˆ R and âˆ S are opposite angles.

âˆ´ âˆ S = 60Â°

CLUE is also a parallelogram.

âˆ´ Its opposite angles are equal.

âˆ´ âˆ E= âˆ L = 70Â°

Now, in Î” ESO, we have

âˆ E + âˆ S + x = 180Â°

âˆ´ 70Â° + 60Â° + x = 180Â°

or x = 180Â° â€“ 60Â° â€“ 70Â°

â‡’ x = 50**Question 10.** Explain how this figure is a trapezium. Which of its two sides are parallel?Â°**Solution: **Since, 100Â° + 80Â° = 180Â°

i.e. âˆ M and âˆ L are supplementary.

[âˆµ If interior opposite angles along the transversal are supplementary]

**Question 11**. Find mâˆ C in the adjoining figure if .**Solution:**

âˆµ ABCD is a trapezium in which and BC is a transversal.

âˆ´ Interior opposite angles along BC are supplementary.

âˆ´ mâˆ B + mâˆ C = 180Â°

or mâˆ C = 180Â° â€“ mâˆ B

âˆ´ mâˆ C = 180Â° â€“ 120Â° [âˆµ âˆ B = 120Â°]

or mâˆ C = 60Â°**Question 12.** Find the measure of âˆ P and âˆ S if in figure. (If you find mâˆ R, is there more than one method to find mâˆ P?)**Solution: **

PQRS is a trapezium such that and PQ is a transversal.

âˆ´ mâˆ P + mâˆ Q = 180Â° [Interior opposite angles are supplementary]

or mâˆ P + 130Â° = 180Â° or mâˆ P = 180Â° â€“ 130Â° = 50Â°

Again SP || RQ and RS is a transversal.

âˆ´ mâˆ S + mâ€“R = 180Â°

or mâ€“S + 90Â° = 180Â°

âˆ´ mâˆ S = 180Â° â€“ 90Â° = 90Â°

Yes, using the angle sum property of a quadrilateral, we can find mâˆ P when mâˆ R is known.

âˆ´ mâˆ P + mâ€“Q + mâˆ R + mâˆ S = 360Â°

or mâˆ P + 130Â° + 90Â° + 90Â° = 360Â°

or mâˆ P = 360Â° â€“ 130Â° â€“ 90Â° â€“ 90Â° = 50Â°

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