# NCERT Solutions For Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3

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## NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3

NCERT TEXTBOOK EXERCISES

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

(ii) âˆ DCB = â€¦â€¦â€¦â€¦â€¦.
(iii) OC = â€¦â€¦â€¦â€¦â€¦â€¦.
(iv) mâˆ DAB + mâˆ CDA = â€¦â€¦â€¦â€¦..

Solution.

Opposite sides of a parallelogram are equal

(ii) âˆ DCB = âˆ DAB
Opposite angles of a parallelogram are equal

(iii) OC = OA
âˆµ Diagonals of a parallelogram bisect each other

(iv) mâˆ DAB + mâˆ CDA = 180Â°
âˆµ Adjacent angles of a parallelogram are supplementary.

Question 2. Consider the following paralleloÂ¬grams. Find the values of the unknowns x, y, z.

Solution.

(i) y = 100Â°
Opposite angles of a parallelogram are equal
x + 100Â° = 180Â°
Adjacent angles in a parallelogram are supplementary
â‡’ x = 180Â° â€“ 100Â°
â‡’ x = 80Â°
â‡’ z â€“ x = 80Â°
Opposite angles of a parallelogram are of equal measure

(ii) x + 50Â° = 180Â°
Adjacent angles in a parallelogram are supplementary
â‡’ x = 180Â° â€“ 50Â° = 130Â°
â‡’ y = x = 130Â°
The opposite angles of a parallelogram are of equal measure
180Â° â€“ z = 50Â°
Opposite angles of a parallelogram are of equal measure
â‡’ z = 180Â° â€“ 50Â° = 130Â°

(iii) x = 90Â°
Vertically opposite angles are equal
x + y + 30Â° = 180Â°
By angle sum property of a triangle
â‡’ 90Â° + y + 30Â° = 180Â°
â‡’ 120Â° + y = 180Â°
â‡’ y = 180Â° â€“ 120Â° = 60Â° z + 30Â° + 90Â° â€“ 180Â°
By angle sum property of a triangle
z = 60Â°

(iv) y = 80Â°
Opposite angles of a parallelogram are of equal measure
x + 80Â° = 180Â°
Adjacent angles in a parallelogram are supplementary
â‡’ x = 180Â° â€“ 80Â°
â‡’ x = 100Â°
â‡’ 180Â°-2+ 80Â°= 180Â°
Linear pair property and adjacent angles in a parallelogram are supplementary.
z = 80Â°

(v) y = 112Â°
Opposite angles of a parallelogram are equal
x + y + 40Â° = 180Â°
By angle sum property of a triangle
â‡’ x + 112Â° + 40Â° = 180Â°
â‡’ x + 152Â° = 180Â°
â‡’ x = 180Â°- 152Â°
â‡’ x = 28Â°
z = x = 28Â°.
Alternate interior angles

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) Can be, but need not be
(ii) No: in a parallelogram, opposite sides are equal; but here, AD â‰  BC.
(iii) No: in a parallelogram, opposite angles are of equal measure; but here âˆ A â‰  âˆ C.

Question 4. Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.

Solution.

A kite, for example

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 the two adjacent angles be 3xÂ° and 2xÂ°.
Then,
3xÂ° + 2xÂ° = 180Â°
âˆ´ Sum of the two adjacent angles of a parallelogram is 180Â°
â‡’ 5xÂ° = 180Â°
â‡’ { x }^{ \circ }=\frac { { 180 }^{ \circ } }{ 5 }
â‡’ xÂ° = 36Â°
â‡’ 3xÂ° = 3 x 36Â° = 108Â°
and
2xÂ° = 2 x 36Â° = 72Â°.
Since, the opposite angles of a parallelogram are of equal measure, therefore the measures of the angles of the parallelogram are 72Â°, 108Â°, 72Â°, and 108Â°.

Question 6. Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.

Solution.

Let the two adjacent angles of a parallelogram be xÂ° each.
Then,
xÂ° + xÂ° = 180Â°
âˆ´ Sum of the two adjacent angles of a parallelogram is 180Â°.
â‡’ 2xÂ° = 180Â°
â‡’ ${ x }^{ \circ }=\frac { { 180 }^{ \circ } }{ 2 }$
â‡’ xÂ° = 90Â°.
Since the opposite angles of a parallelogram are of equal measure, therefore the measure of each of the angles of the parallelogram is 90Â°, i.e., each angle of the parallelogram is a right angle.

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.

x = 180Â° â€“ 70Â° = 110Â°
Linear pair property and the opposite angles of a parallelogram are of equal measure.
âˆµ HOPE is a || gm
âˆ´ HE || OP
and HP is a transversal
âˆ´ y = 40Â°
alternate interior angles
40Â° + z + x = 180Â°
The adjacent angles in a parallelogram are supplementary

â‡’ 40Â° + z + 110Â° = 180Â°
â‡’ z + 150Â° = 180Â°
â‡’ z = 180Â° â€“ 150Â°
â‡’ z = 30Â°.

Question 8. The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm)

(I)

(ii)

Solution.

(i)
For Figure GUNS
Since the opposite sides of a parallelogram are of equal length, therefore,
â‡’ 3x = 18
â‡’ x=183=6
and, 3y â€“ 1 = 26
â‡’ 3y = 26 + 1
â‡’ 3y = 27
y=273=9
Hence, x = 6; y = 9.

(ii)
For Figure RUNS
Since the diagonals of a parallelogram bisect each other, therefore,
â‡’ x + y = 16 â€¦(1)
and, y + 7 = 20 â€¦(2)
From (2),
â‡’ y â€“ 20 â€“ 7 = 13
Putting y = 13 in (1), we get
â‡’ x + 13 = 16 â‡’ x = 16 â€“ 13 = 3.
Hence, x = 3; y = 13.

Question 9. In the below figure both RISK and CLUE are parallelograms. Find the value of x.

Solution.

Question 10. Explain how this figure is a trapezium. Which of its two sides is parallel?

Solution.

âˆµ âˆ KLM + âˆ NML = 80Â° + 100Â° = 180Â°
âˆ´ KL || NM
âˆµ The sum of consecutive interior angles is 180Â°
âˆ´ Figure KLMN is a trapezium.
Its two sides $\overline { KL }$ and $\overline { NM }$ are parallel.

Question 11. Find mâˆ C in the figure, if $\overline { AB }$ || $\overline { DC }$.

Solution.

âˆµ $\overline { AB }$ || $\overline { DC }$
âˆ´ mâˆ C + mâˆ B = 180Â°
âˆµ The sum of consecutive interior angles is 180Â°
mâˆ C+ 120Â° = 180Â°
â‡’ mâˆ C = 180Â° â€“ 120Â° = 60Â°.

Question 12. Find the measure of âˆ P and âˆ S, if $\overline { SP }$ || $\overline { RQ }$ in the figure. (If you find mZ R, is there more than one method to find mâˆ P ?)

Solution.

âˆµ $\overline { SP }$ || $\overline { RQ }$
âˆ´ mâˆ P+mâˆ Q = 180Â°
âˆµ The sum of consecutive interior angles is 180Â°
â‡’ mâˆ P + 130Â° = 180Â°
â‡’ mâˆ P = 180Â° â€“ 130Â°
â‡’ mâˆ P = 50Â°
Again, mâˆ R + mâˆ S = 180Â°
âˆµ The sum of consecutive interior angles is 180Â°
â‡’ 90Â° + m Z S = 180Â°
â‡’ mâˆ S = 180Â° â€“ 90Â° = 90Â°
Yes; there is one more method of finding mâˆ P if mâˆ R is given and that is by using the angle sum property of a quadrilateral.
We have,
mâˆ P + mâˆ Q + mâˆ R + mâˆ S = 360Â°
â‡’ mâˆ P + 130Â° + 90Â° + 90Â° = 360Â°
â‡’ mâˆ P + 310Â° = 360Â°
â‡’ mâˆ P = 360Â° â€“ 310Â° = 50Â°.

## NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.3 PDF

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