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

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 MCQS

**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

**Mark the correct alternative in each of the following:**

**Question 1.**

**The opposite sides of a quadrilateral have**

**(a) no common point**

**(b) one common point**

**(c) two common points**

**(d) infinitely many common points**

**Solution:**

The opposite sides of a quadrilateral have no common point. **(a)**

**Question 2.**

**The consecutive sides of a quadrilateral have**

**(a) no common point**

**(b) one common point**

**(c) two common points**

**(d) infinitely many common points**

**Solution:**

The consecutive sides of a quadrilateral have one common point. **(b)**

**Question 3.**

**PQRS is a quadrilateral, PR and QS intersect each other at O. In which of the following cases, PQRS is a parallelogram?**

**(a) ∠P = 100°, ∠Q = 80°, ∠R = 100°**

**(b) ∠P = 85°, ∠Q = 85°, ∠R = 95°**

**(c) PQ = 7 cm, QR = 7 cm, RS = 8 cm, SP = 8 cm**

**(d) OP = 6.5 cm, OQ = 6.5 cm, OR = 5.2 cm, OS = 5.2 cm**

**Solution:**

PQRS is a quadrilateral, PR and QS intersect each other at O. PQRS is a parallelogram if ∠P = 100°, ∠Q = 80°, ∠R = 100°** (a)**

**Question 4.**

**Which of the following quadrilateral is not a rhombus?**

**(a) All four sides are equal**

**(b) Diagonals bisect each other**

**(c) Diagonals bisect opposite angles**

**(d) One angle between the diagonals is 60°**

**Solution:**

A quadrilateral is not a rhombus if one angle between the diagonals is 60°. **(d)**

**Question 5.**

**Diagonals necessarily bisect opposite angles in a**

**(a) rectangle**

**(b) parallelogram**

**(c) isosceles trapezium**

**(d) square**

**Solution:**

Diagonals necessarily bisect opposite angles in a square.** (d)**

**Question 6.**

**The two diagonals are equal in a**

**(a) parallelogram**

**(b) rhombus**

**(c) rectangle**

**(d) trapezium**

**Solution:**

The two diagonals are equal in a rectangle. **(c)**

**Question 7.**

**We get a rhombus by joining the mid-points of the sides of a**

**(a) parallelogram**

**(b) rhombus**

**(c) rectangle**

**(d) triangle**

**Solution:**

We get a rhombus by joining the mid points of the sides of a rectangle.** (c)**

**Question 8.**

**The bisectors of any two adjacent angles of a parallelogram intersect at**

**(a) 30°**

**(b) 45°**

**(c) 60°**

**(d) 90°**

**Solution:**

The bisectors of any two adjacent angles of a parallelogram intersect at 90°.** (d)**

**Question 9.**

**The bisectors of the angle of a parallelogram enclose a**

**(a) parallelogram**

**(b) rhombus**

**(c) rectangle**

**(d) square**

**Solution:**

The bisectors of the angles of a parallelogram enclose a rectangle. **(c)**

**Question 10.**

**The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a**

**(a) parallelogram**

**(b) rectangle**

**(c) square**

**(d) rhombus**

**Solution:**

The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a parallelogram. **(a)**

**Question 11.**

**The figure formed by joining the mid-points of the adjacent sides of a rectangle is a**

**(a) square**

**(b) rhombus**

**(c) trapezium**

**(d) none of these**

**Solution:**

The figure formed by joining the mid-points of the adjacent sides of a rectangle is a rhombus. **(b)**

**Question 12.**

**The figure formed by joining the mid-points of the adjacent sides of a rhombus is a**

**(a) square**

**(b) rectangle**

**(c) trapezium**

**(d) none of these**

**Solution:**

The figure formed by the joining the mid-points of the adjacent sides of a rhombus is a rectangle. **(b)**

**Question 13.**

**The figure formed by joining the mid-points of the adjacent sides of a square is a**

**(a) rhombus**

**(b) square**

**(c) rectangle**

**(d) parallelogram**

**Solution:**

Tire figure formed by joining the mid-points of the adjacent sides of a square is a square. **(b)**

**Question 14.**

**The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a**

**(a) rectangle**

**(b) parallelogram**

**(b) rhombus**

**(d) square**

**Solution:**

The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a parallelogram. **(b)**

**Question 15.**

**If one angle of a parallelogram is 24° less than twice the smallest angle, then the measure of the largest angle of the parallelogram is**

**(a) 176°**

**(b) 68°**

**(c) 112°**

**(d) 102°**

**Solution:**

Let the smallest angle be x

The largest angle = 2x – 24°

But sum of two adjacent angles = 180°

**Question 16.**

**In a parallelogram ABCD, If ∠DAB = 75° and ∠DBC = 60°, then ∠BDC =**

**(a) 75°**

**(b) 60°**

**(c) 45°**

**(d) 55°**

**Solution:**

In ||gm ABC,

∠A = 75°, ∠DBC = 60°

But ∠A + ∠B = 180° (Sum of two consecutive angles)

⇒ 75° + ∠B = 180°

⇒ ∠B = 180°- 75“= 105°

But ∠DBC = 60°

∴ ∠DBA = 105°-60° = 45°

But ∠BDC = ∠DBA (Alternate angles)

∴ ∠BDC = 45° **(c)**

**Question 17.**

**ABCD is a parallelogram and E and F are the centroids of triangles ABD and BCD respectively, then EF =**

**(a) AE**

**(b) BE**

**(c) CE**

**(d) DE**

**Solution:**

In ||gm ABCD, BD is joined forming two triangles ABD and BCD

E and F are the centroid of ∆ABD and ∆BCD

Now E and F trisect AC

i.e. AE = EF = FC

∴ EF = AE **(a)**

**Question 18.**

**ABCD is a parallelogram, M is the mid¬point of BD and BM bisects ∠B. Then, ∠AMB =**

**(a) 45°**

**(b) 60°**

**(c) 90°**

**(d) 75°**

**Solution:**

In ||gm ABCD, M is mid-point of BD and

BM bisects ∠B

AM is joined

∴AM bisects ∠A

But ∠A + ∠B = 180° (Sum of two consecutive angles)

∴ ∠AMB = 90° **(c)**

**Question 19.**

**If an angle of a parallelogram is two-third of its adjacent angle, the smallest angle of the parallelogram is**

**(a) 108°**

**(b) 54°**

**(c) 12°**

**(d) 81°**

**Solution:**

Let adjacent angle of a ||gm = x

Then second angle = \(\frac { 2 }{ 3 }\) x

∴ x+ \(\frac { 2 }{ 3 }\) x= 180°

(Sum of two adjacent angles of a ||gm is 180°)

**Question 20.**

**If the degree measures of the angles of quadrilateral are Ax, lx, 9x and 10JC, what is the sum of the measures of the smallest angle and largest angle?**

**(a) 140°**

**(b) 150°**

**(c) 168°**

**(d) 180°**

**Solution:**

Sum of the angles of a quadrilateral = 360°

∴ 4x + 1x + 9x + 10x = 360°

⇒ 30x = 360°

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

Now sum of smallest and largest angle = 4 x 12° + 10 x 12°

= 48° + 120° = 168° **(c)**

**Question 21.**

**If the diagonals of a rhombus are 18 cm and 24 cm respectively, then its side is equal to**

**(a) 16 cm**

**(b) 15 cm**

**(c) 20 cm**

**(d) 17 cm**

**Solution:**

Diagonals of a rhombus are 18 cm and 24 cm But diagonals of a rhombus bisect each other at right angles

**Question 22.**

**ABCD is a parallelogram in which diagonal AC bisects ∠BAD. If ∠BAC = 35°, then ∠ABC =**

**(a) 70°**

**(b) 110°**

**(c) 90°**

**(d) 120°**

**Solution:**

In ||gm ABCD, AC is its diagonal which bisect ∠BAD

∠BAD = 35°

∴ ∠BAD = 2 x 35° = 70°

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

⇒ 70° + ∠B = 180°⇒ ∠B = 180° – 70°

∴ ∠B = 110°

⇒ ABC = 110° **(b)**

**Question 23.**

**In a rhombus ABCD, if ∠ACB = 40°, then ∠ADB =**

**(a) 70°**

**(b) 45°**

**(c) 50°**

**(d) 60°**

**Solution:**

In rhombus ABCD, ∠ACB = 40°

∴ ∠BCD = 2 x ∠ACB

= 2 x 40° = 80°

But ∠BCD + ∠ADC = 180° (Sum of consecutive angles of ||gm)

⇒ 80° + ∠ADC = 180°

⇒ ∠ADC = 180° – 80° = 100°

∴ ∠ADB = \(\frac { 1 }{ 2 }\)∠ADC = \(\frac { 1 }{ 2 }\)x 100° = 50° **(c)**

**Question 24.**

**In ∆ABC, ∠A = 30°, ∠B = 40° and ∠C = 110°. The angles of the triangle formed by joining the mid-points of the sides of this triangle are**

**(a) 70°, 70°, 40°**

**(b) 60°, 40°, 80°**

**(c) 30°, 40°, 110°**

**(d) 60°, 70°, 50°**

**Solution:**

In ∆ABC,

∠A = 30°, ∠B = 40°, ∠C = 110°

D, E and F are mid-points of the sides of the triangle. By joining them in order,

DEF is a triangle formed

Now BDEF, CDFE and AFDE are ||gms

∴ ∠A = ∠D = 30°

∠B = ∠E = 40°

∠C = ∠F= 110°

∴ Angles are 30°, 40°, 110° **(c)**

**Question 25.**

**The diagonals of a parallelogram ABCD intersect at O. If ∠BOC = 90° and ∠BDC = 50°, then ∠OAB =**

**(a) 40°**

**(b) 50°**

**(c) 10°**

**(d) 90°**

**Solution:**

In ||gm ABCD, diagonals AC and BD intersect each other at O

BOC = 90°, ∠BDC = 50°

∵ ∠BOC = 90°

∴ Diagonals of ||gm bisect each other at 90°

∴∠COD = 90°

In ∆COD,

∠OCD = 90° – 50° = 40°

But ∠OAB = ∠OCD (Alternate angles)

∴∠OAB = 40°** (a)**

**Question 26.**

**ABCD is a trapezium in which AB || DC. M and N are the mid-points of AD and BC respectively. If AB = 12 cm, MN = 14 cm, then CD =**

**(a) 10 cm**

**(b) 12 cm**

**(c) 14 cm**

**(d) 16 cm**

**Solution:**

In trapezium AB || DC

M and N are mid-points of sides AD and BC and MN are joined

AB = 12 cm, MN = 14 cm

∵ MN = \(\frac { 1 }{ 2 }\)(AB + CD)

⇒ 2MN = AB + CD

⇒ 2 x 14 = 12 + CD

CD = 2 x 14 – 12 = 28 – 12 = 16 cm** (d)**

**Question 27.**

**Diagonals of a quadrilateral ABCD bisect each other. If ∠A = 45°, then ∠B =**

**(a) 115°**

**(b) 120°**

**(c) 125°**

**(d) 135°**

**Solution:**

Diagonals AC and BD of quadrilateral ABCD bisect each other at O

∴ AO = OC, BO = OD

∴ ABCD is a ||gm ∠A = 45°

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

∴ ∠B = 180° – ∠A = 180° – 45°

= 135° **(d)**

**Question 28.**

**P is the mid-point of side BC of a paralleogram ABCD such that ∠BAP = ∠DAP. If AD = 10 cm, then CD =**

**(a) 5 cm**

**(b) 6 cm**

**(c) 8 cm**

**(d) 10 cm**

**Solution:**

In ||gm ABCD, P is mid-point of BC

AD = 10cm

∠BAP = ∠DAP

Through P, draw PQ || AB

∴ ABPQ is rhombus

∴ AB = BP = AQ

= \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) x 10 = 5 cm

But CD = AB (Opposite sides of ||gm)

∴ CD = 5 cm** (a)**

**Question 29.**

**In ∆ABC, E is the mid-point of median AD such that BE produced meets AC at E If AC = 10.5 cm, then AF =**

**(a) 3 cm**

**(b) 3.5 cm**

**(c) 2.5 cm**

**(d) 5 cm**

**Solution:**

In ∆ABC, E is the mid-point of median AD

Such that BE produced meets AC at F

AC = 10.5 cm

Draw DG || AF

In ∆ADG

E is mid-point of AD and EF || DG

∴ F is mid-point of AG

⇒ AF = FG …(i)

In ∆BCF

D is mid-point of BC and DG || BF

∴ G is mid-point of FC

∴ FG = GC …(i)

From (i) and (ii)

AF = FG = GC = \(\frac { 1 }{ 3 }\) AC

But AC = 10.5 cm

∴ AF = \(\frac { 1 }{ 3 }\) AC = \(\frac { 1 }{ 3 }\) x 10.5 = 3.5 cm** (b)**

**Question 30.**

**ABCD is a parallelogram and E is the mid-point of BC. DE and AB when produced meet at F. Then, AF =**

**Solution:**

In ||gm ABCD, E is mid-point of BC DE and AB are produced to meet at F

∵ E is mid point of BC

∴ BE = EC

In ∆BEF and ∆CDE

BE = EC

∠BEF = ∠CED (Vertically opposite angle)

and ∠EBF = ∠ECD (Alternate angles)

∴ ∆BEF ≅ ∆CDE (ASA criterian)

∴ DC = BF

But DC = AB

∴ AB = BF

AF = AB + BF = AB + AB

= 2AB** (b)**

**Question 31.**

**In a quadrilateral ABCD, ∠A + ∠C is 2 times ∠B + ∠D. If ∠A = 140° and ∠D = 60°, then ∠B =**

**(a) 60°**

**(b) 80°**

**(c) 120°**

**(d) None of these**

**Solution:**

In quadrilateral ABCD

⇒ ∠A + ∠C = 2(∠B + ∠D)

⇒ ∠A + ∠C = 2∠B + 2∠D

Adding 2∠A + 2∠C both sides

2∠A + 2∠C + ∠A + ∠C = 2∠A + 2∠B + 2∠C + 2∠D

⇒ 3∠A + 3∠C = 2(∠A + ∠B + ∠C + ∠D)

⇒ 3(∠A + ∠C) = 2 x 360° = 720°

∴ ∠A + ∠C = \(\frac { { 720 }^{ \circ } }{ 3 }\) = 240°

⇒ 40° + ∠C = 240° (∵ ∠A = 40°)

∠C = 240° – 40° = 200°

Now 2(∠B + ∠D) = ∠A + ∠C = 240°

∠B + ∠D = \(\frac { { 240 }^{ \circ } }{ 2 }\) = 120°

∴ ∠B = 60° = 120°

∴ ∠B = 60°** (a)**

**Question 32.**

**The diagonals AC and BD of a rectangle ABCD intersect each other at P. If ∠ABD = 50°, then ∠DPC =**

**(a) 70°**

**(b) 90°**

**(c) 80°**

**(d) 100°**

**Solution:**

In rectangle ABCD, diagonals AC and BD intersect each other at P

∠ABD = 50°

∴ ∠CAB = ∠ABD = 50° (∵ AP = BP)

Now in ∆APB

∠CAB + ∠ABD + ∠APB = 180° (Angles of a triangle)

⇒ ∠PAB + ∠PBA + ∠APB = 180°

⇒ 50° + 50° + ∠APB = 180°

⇒ ∠APB = 180° – 50° – 50° = 80°

But ∠DPC = ADB (Vertically opposite angles)

∴ ∠DPC = 80°** (c)**

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

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