## RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS

Other Exercises

- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.1
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.3
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables MCQS

Question 1.

In a parallelogram ABCD, write the sum of angles A and B.

Solution:

In ||gm ABCD,

∠A + ∠B = 180°

(Sum of consecutive angles of a ||gm)

Question 2.

In a parallelogram ABCD, if ∠D = 115°, then write the measure of ∠A.

Solution:

In ||gm ABCD,

∠D = 115°

But ∠A + ∠D = 180°

(Sum of consecutive angles of a ||gm)

⇒ ∠A + 115°= 180° ∠A = 180°- 115°

∴ ∠A = 65°

Question 3.

PQRS is a square such that PR and SQ intersect at O. State the measure of ∠POQ.

Solution:

In a square PQRS,

Diagonals PR and QS intersects each other at O.

∵ The diagonals of a square bisect each other at right angles.

∴ ∠POQ = 90°

Question 4.

If PQRS is a square then write the measure of ∠SRP.

Solution:

In square PQRS,

Join PR,

∵Diagonals of a square bisect are opposite angles

∴∠SRP = \(\frac { 1 }{ 2 }\)x ∠SRQ

= \(\frac { 1 }{ 2 }\) x 90° = 45°

Question 5.

If ABCD is a rhombus with ∠ABC = 56°, find the measure of ∠ACD.

Solution:

In rhombus ABCD,

Diagonals bisect each other at 0 at right angles.

∠ABC = 56°

But ∠ABC + ∠BCD = 180° (Sum of consecutive angles)

⇒ 56° + ∠BCD = 180°

⇒ ∠BCD = 180° – 56° = 124°

∵ Diagonals of a rhombus bisect the opposite angle

∴ ∠ACD = \(\frac { 1 }{ 2 }\) ∠BCD = \(\frac { 1 }{ 2 }\) x 124°

= 62°

Question 6.

The perimeter of a parallelogram is 22 cm. If the longer side measures 6.5 cm, what is the measure of the shorter side.

Solution:

Perimeter of a ||gm ABCD = 22cm

∴ Sum of two consecutive sides = \(\frac { 22 }{ 2 }\)

= 11cm

i.e. AB + BC = 11 cm

AB = 6.5 cm and let BC = x cm

∴ 6.5 + x = 11 cm

x = 11 – 6.5 = 4.5

∴ Shorter side = 4.5 cm

Question 7.

If the angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Then find the measure of the smallest angle.

Solution:

Ratio in the angles of a quadrilateral = 3 : 5 : 9 : 13

Let first angle = 3x

Second angle = 5x

Third angle = 9x

and fourth angle = 13x

∵ The sum of angles of a quadrilateral = 360°

∴ 3x + 5x + 9x + 13x = 360°

⇒ 30x = 360° ⇒ x = \(\frac { { 360 }^{ \circ } }{ 30 }\) = 12

∴ Smallest angle = 3x = 3 x 12° = 36°

Question 8.

In parallelogram ABCD if ∠A = (3x – 20°), ∠B = (y + 15)°, ∠C = (x + 40°), then find the value of x and y.

Solution:

In a ||gm ABCD,

∠A = (3x – 20°), ∠B = y + 15°,

∠C = x + 40°

Now, ∠A = ∠C (Opposite angles of a ||gm)

⇒ 3x – 20 = x + 40°

⇒ 3x – x = 40° + 20° ⇒ 2x = 60°

⇒ x = \(\frac { { 60 }^{ \circ } }{ 2 }\) = 30°

and ∠A + ∠B = 180° (Sum of the consecutive angles)

⇒ 3x-20° + y + 15° = 180°

⇒ 3x + y – 5° = 180°

⇒ 3 x 30° +y- 5° = 180°

⇒ 90° – 5° + y = 180

y = 180° – 90° + 5 = 95°

∴ x = 30°, y = 95°

Question 9.

If measures opposite angles of a parallelogram are (60 – x)° and (3x – 4)°, then find the measures of angles of the parallelogram.

Solution:

Opposite angles of a ||gm ABCD are (60 – x)° and (3x – 4°)

But opposite angles of a ||gm are equal, the

60° – x° = 3x – 4° ⇒ 60° + 4° = 3x + x

⇒ 4x = 64° ⇒ x = \(\frac { { 64 }^{ \circ } }{{ 4 }^{ \circ } }\) = 16°

∴ ∠A = 60° – x = 60° – 16° = 44°

But ∠A + ∠B = 180° (sum of consecutive angle)

⇒ 44° + ∠B = 180°

⇒ ∠B = 180° – 44°

⇒ ∠B = 136°

But ∠A = ∠C and ∠B = ∠D (Opposite angles)

∴ Angles are 44°, 136°, 44°, 136°

Question 10.

In a parallelogram ABCD, the bisectors of ∠A also bisect BC at x, find AB : AD.

Solution:

In ||gm ABCD,

Bisectors of ∠A meets BC at X and BX = XC

Draw XY ||gm AB meeting AD at Y

Question 11.

In the figure, PQRS in an isosceles trapezium find x and y.

Solution:

∵ PQRS is an isosceles trapezium in which

SP = RQ and SR || PQ

∴ ∠P + ∠S = 180° (Sum of co-interior angles)

3x + 2x = 180° ⇒ 5x = 180°

⇒ x = \(\frac { { 180 }^{ \circ } }{ 5 }\) = 36°

But ∠P = ∠Qm (Base angles of isosceles trapezium)

y = 2x = 2 x 36° = 12°

∴ y = 12°

Hence x = 36°, y = 12°

Question 12.

In the figure ABCD is a trapezium. Find the values of x and y.

Solution:

In trapezium ABCD,

AB || CD

∴ ∠A + ∠D = 180° (Sum of cointerior angles)

x + 20° + 2x + 10° = 180°

3x + 30° = 180°

⇒ 3x= 180° – 30°

3x = 150°

x = \(\frac { { 150 }^{ \circ } }{ 3 }\) = 50°

Similarly, ∠B + ∠C = 180°

⇒ y + 92° = 180°

⇒ y = 180° – 92° = 88°

∴ x = 50°, y = 88°

Question 13.

In the figure, ABCD and AEFG are two parallelograms. If ∠C = 58°, find ∠F.

Solution:

In the figure, ABCD and AEFG are two parallelograms ∠C = 58°

∵ DC || GF and CB || FE (Sides of ||gms)

∴ ∠C = ∠F

But ∠C = 58°

∴ ∠F = 58°

Question 14.

Complete each of the following statements by means of one of those given in brackets against each:

(i) If one pair of opposite sides are equal and parallel, then the figure is ……… (parallelogram, rectangle, trapezium)

(ii) If in a quadrilateral only one pair of opposite sides are parallel, the quadrilateral is …….. (square, rectangle, trapezium)

(iii) A line drawn from the mid-point of one side of a triangle ………. another side intersects the third side at its mid-point, (perpendicular to, parallel to, to meet)

(iv) If one angle of a parallelogram is a right angle, then it is necessarily a …….. (rectangle, square, rhombus)

(v) Consecutive angle of a parallelogram are ……… (supplementary, complementary)

(vi) If both pairs of opposite sides of a quadrilateral are equal, then it is necessarily a ……… (rectangle, parallelogram, rhombus)

(vii) If opposite angles of a quadrilateral are equal, then it is necessarily a ………. (parallelogram, rhombus, rectangle)

(viii)If consecutive sides of a parallelogram are equal, then it is necessarily a …….. (kite, rhombus, square)

Solution:

(i) If one pair of opposite sides are equal and parallel, then the figure is parallelogram.

(ii) If in a quadrilateral only one pair of opposite sides are parallel, the quadrilateral is trapezium.

(iii) A line drawn from the mid-point of one side of a triangle parallel to another side intersects the third side at its mid-point,

(iv) If one angle of a parallelogram is a right angle, then it is necessarily a rectangle.

(v) Consecutive angle of a parallelogram are supplementary.

(vi) If both pairs of opposite sides of a quadrilateral are equal, then it is necessarily a parallelogram.

(vii) If opposite angles of a quadrilateral are equal, then it is necessarily a parallelogram.

(viii) If consecutive sides of a parallelogram are equal, then it is necessarily a rhombus.

Question 15.

In a quadrilateral ABCD, bisectors of A and B intersect at O such that ∠AOB = 75°, then write the value of ∠C + ∠D.

Solution:

In quadrilateral ABCD,

Bisectors of ∠A and ∠B meet at O and ∠AOB = 75°

In AOB, ∠AOB = 75°

∴ ∠OAB + ∠OBA = 180° – 75° = 105°

But OA and OB are the bisectors of ∠A and ∠B.

∴ ∠A + ∠B = 2 x 105° = 210°

But ∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quad.)

∴ 210° + ∠C + ∠D = 360°

⇒ ∠C + ∠D = 360° – 210° = 150°

Hence ∠C + ∠D = 150°

Question 16.

The diagonals of a rectangle ABCD meet at O. If ∠BOC = 44° find ∠OAD.

Solution:

In rectangle ABCD,

Diagonals AC and BD intersect each other at O and ∠BOC = 44°

But ∠AOD = ∠BOC (Vertically opposite angles)

∴ ∠AOD = 44°

In ∆AOD,

∠AOD + ∠OAD + ∠ODA = 180° (Sum of angles of a triangle)

⇒ 44° + ∠OAD + ∠OAD = 180° [∵ OA = OD, ∠OAD = ∠ODA]

⇒ 2∠OAD = 180° – 44° = 136°

∴ ∠OAD = \(\frac { { 136 }^{ \circ } }{ 2 }\) = 68°

Question 17.

If ABCD is a rectangle with ∠BAC = 32°, find the measure if ∠DBC.

Solution:

In rectangle ABCD,

Diagonals bisect each other at O

∠BAC = 32°

∵ OA = OB

∴ ∠OBA Or ∠DBA = ∠BAC = 32°

But ∠ABC = 90° (Angle of a rectangles)

∴ ∠DBC = ∠ABC – ∠DBA

= 90° – 32° = 58°

Question 18.

If the bisectors of two adjacent angles A and B of a quadrilateral ABCD intersect at a point O. Such that ∠C + ∠D = k(∠AOB), then find the value of k.

Solution:

In quadrilateral ABCD,

Bisectors of ∠A and ∠B meet at O

Such that ∠C + ∠D = k (∠AOB)

Question 19.

In the figure, PQRS is a rhombus in which the diagonal PR is produced to T. If ∠SRT = 152°, find x, y and z.

Solution:

In rhombus PQRS,

Diagonal PR and SQ bisect each other at right angles and PR is produced to T such that ∠SRT = 152°

But ∠SRT + ∠SRP = 180° (Linear pair)

⇒ 152° +∠SRP = 180°

⇒ ∠SRP =180°- 152° = 28°

But ∠SPR = ∠SRP (∵ PR bisects ∠P and ∠R)

⇒ z = 28°

y = 90° (∵ Diagonals bisect each other at right angles)

∠RPQ = z = 28°

∴ In ∆POQ,

z + x = 90° ⇒ 28° + x = 90°

⇒ x = 90° – 28° = 62°

∴ x = 62°, y = 90°, z = 28°

Question 20.

In the figure, ABCD is a rectangle in which diagonal AC is produced to E. If ∠ECD = 146°, find ∠AOB.

Solution:

In rectangle ABCD,

Diagonals AC and BD bisect each other at O

AC is produced to E and ∠DCE = 146°

∠DCE + ∠DCA = 180° (Linear pair)

⇒ 146°+ ∠DCA= 180°

⇒ ∠DCA = 180°- 146°

⇒ ∠DCA = 34°

∴ ∠CAB = ∠DCA (Alternate angles)

= 34°

Now in ∆AOB,

∠AOB = 180° – (∠DAB + ∠OBA)

= 180° – (34° + 34°)

= 1803 – 68° = 112°

Hope given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS are helpful to complete your math homework.

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