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) ∠DCB = _____

(iii) OC = _____

(iv) m∠DAB + m∠CDA = _____

**Solution:**

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

(ii) ∠DCB = ∠DAB [∵ Opposite angles are equal]

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

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

∵ Sum of interior angles of a parallelogram = 360°

∴ x + y + z + ∠B = 360°

or x + 100° + z + 100° = 360°

or x + z = 360° – 100° – 100° = 160°

But x = z

∴ x = z = 160°/2 = 80°

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

∴ m∠1 = 50

Now, ∠1 + z = 180° [Linear pair]

or z = 180° – ∠1 = 180° – 50° = 130°

x + y + 50° + 50° = 360°

or x + y = 360° – 50° – 50° = 260°

But x = y

∴ x = y = 260°/2° = 130°

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

∴ x = 90°

∵ Sum of the angles of a triangle = 180° ∴90° + 30° + y = 180°

or y = 180° – 30° – 90° = 60°

In the figure, ABCD is a parallelogram.

∴ AD || BC and BD is a transversal.

∴ y = z [Alternate angles]

But y = 60°

∴ z = 60°

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

∴ Opposite angles are equal.

∴ y = 80°

AB || CD and BC is a transversal.

∴ x + 80° = 180° [Interior opposite angles]

or x = 180° – 80° = 100°

Again BC || AD and CD is a transversal,

∴ z = 80° [Corresponding angles]

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

∴ y = 112°

In Δ ACD,

x + y + 40° = 180°

x + 112° + 40° = 180°

∴ x = 180° – 112° – 40° = 28°

∵ AD || BC and AC is a transversal.

∴ x = z [∵ Alternate angles are equal]

and z = 28°

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

(i) ∠D + ∠B = 180°?

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

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

∠A + ∠B = Sum of adjacent angles = 180°**∴ **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

∵ Opposite sides AB and BC are not equal.**∴** It cannot be a parallelogram.

(iii) In a quadrilateral ABCD,

∠A = 70° and ∠C = 65°

∵ Opposite angles ∠A ≠ ∠C**∴ **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 –B and –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 ∠A and ∠B are 3x and 2x respectively, since adjacent angles are supplementary.**∴ **∠A + ∠B = 180°**∴ **3x + 2x = 180°

or 5x = 180°

or x = 180°/5° = 36°**∴ **∠A = 3 x 36° = 108°

and ∠B = 2 x 36° = 72°

∵ Opposite angles are equal.**∴ ** ∠D = ∠B = 72°

and ∠C= ∠A = 108°**∴ ** ∠A = 108°, ∠B = 72°, ∠C = 108° and ∠D = 72°**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 ∠A = ∠B.

Since ∠A + ∠B = 180°**∴ **A = ∠B = 180°/2 = 90°

Since, opposite angles of a parallelogram are equal.**∴ ** ∠A= ∠C = 90°

and ∠B= ∠D = 90°

Thus, ∠A = 90°, ∠B = 90°,∠C = 90° and ∠D = 90°.

**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 = 70° …(1)

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

∴ In Δ ∠HOP, ∠HOP = 180° -(y + z)

= 180° – 70°

= 110°

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

∴ x = 110° EH || OP and PH is a transversal.

∴ y = 40° [Alternate angles are equal]

From (1), 40° + z = 70°

∴ z = 70° – 40° = 30°

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

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

∴ Its opposite sides are equal.

∴ GS = NU and SN = GU

or 3x = 18 and 26 = 3y – 1

Now 3x = 18

⇒ x = 18/3 = 6

3y – 1 = 26

⇒ 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.

∴ x + y = 16 and 7 + y = 20

i.e. y = 20 – 7

or y = 13

∴ From x + y = 16, we have

x + 13 = 16

or x = 16 – 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.

∴ ∠R + ∠K = 180° [∵ Adjacent angles of a parallelogram are supplementary]

or ∠R + 120° = 180°

⇒ ∠R = 180° – 120° = 60°

But ∠R and ∠S are opposite angles.

∴ ∠S = 60°

CLUE is also a parallelogram.

∴ Its opposite angles are equal.

∴ ∠E= ∠L = 70°

Now, in Δ ESO, we have

∠E + ∠S + x = 180°

∴ 70° + 60° + x = 180°

or x = 180° – 60° – 70°

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

i.e. ∠M and ∠L are supplementary.

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

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

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

∴ Interior opposite angles along BC are supplementary.

∴ m∠B + m∠C = 180°

or m∠C = 180° – m∠B

∴ m∠C = 180° – 120° [∵ ∠B = 120°]

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

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

∴ m∠P + m∠Q = 180° [Interior opposite angles are supplementary]

or m∠P + 130° = 180° or m∠P = 180° – 130° = 50°

Again SP || RQ and RS is a transversal.

∴ m∠S + m–R = 180°

or m–S + 90° = 180°

∴ m∠S = 180° – 90° = 90°

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

∴ m∠P + m–Q + m∠R + m∠S = 360°

or m∠P + 130° + 90° + 90° = 360°

or m∠P = 360° – 130° – 90° – 90° = 50°

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