## RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS

Other Exercises

- RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.1
- RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2
- RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3
- RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS
- RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS

Mark the correct alternative in each of the following:

Question 1.

Two parallelograms are on the same base and between the same parallels. The ratio of their areas is

(a) 1 : 2

(b) 2 : 1

(c) 1 : 1

(d) 3 : 1

Solution:

Two parallelograms which are on the same base and between the same parallels are equal in area

∴ Ratio in their areas =1 : 1 (c)

Question 2.

A triangle and a parallelogram are on the same base and between the same parallels. The ratio of the areas of triangle and parallelogram is

(a) 1 : 1

(b) 1 : 2

(c) 2 : 1

(d) 1 : 3

Solution:

A triangle and a parallelogram which are on the same base and between the same parallels, then area of triangle is half the area of the parallelogram

∴ Their ratio =1:2 (c)

Question 3.

Let ABC be a triangle of area 24 sq. units and PQR be the triangle formed by the mid-points of sides of ∆ABC. Then the area of ∆PQR is

(a) 12 sq. units

(b) 6 sq. units

(c) 4 sq. units

(d) 3 sq. units

Solution:

Area of ∆ABC = 24 sq. units

Question 4.

The median of a triangle divides it into two

(a) congruent triangle

(b) isosceles triangles

(c) right triangles

(d) triangles of equal areas

Solution:

The median of a triangle divides it into two triangles equal in area (d)

Question 5.

In a ∆ABC, D, E, F are the mid-points of sides BC, CA and AB respectively. If

ar(∆ABC) = 16 cm2, then ar(trapezium FBCE) =

(a) 4 cm²

(b) 8 cm²

(c) 12 cm²

(d) 10 cm²

Solution:

In ∆ABC, D, E and F are the mid points of sides BC, CA and AB respectively

ar(∆ABC) = 16 cm²

Question 6.

ABCD is a parallelogram. P is any point on CD. If ar(∆DPA) = 15 cm² and ar(∆APC) = 20 cm², then ar(∆APB) =

(a) 15 cm²

(b) 20 cm²

(c) 35 cm²

(d) 30 cm²

Solution:

In ||gm ABCD, P is any point on CD

AP, AC and PB are joined

ar(∆DPA) =15 cm²

ar(∆APC) = 20 cm²

Adding, ar(∆ADC) = 15 + 20 = 35 cm²

∵ AC divides it into two triangles equal in area

∴ ar(∆ACB) = ar(∆ADC) = 35 cm²

∵ ∆APB and ∆ACB are on the same base

AB and between the same parallels

∴ ar(∆APB) = ar(∆ACB) = 35 cm²(c)

Question 7.

The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 16 cm and 12 cm is

(a) 28 cm²

(b) 48 cm²

(c) 96 cm²

(d) 24 cm²

Solution:

In rhombus ABCD,

P, Q, R and S are the mid points of sides AB, BC, CD and DA respectively and are joined in order to get a quad. PQRS

Question 8.

A, B, C, D are mid points of sides of parallelogram PQRS. If ar(PQRS) = 36 cm²,then ar(ABCD) =

(a) 24 cm²

(b) 18 cm²

(c) 30 cm²

(d) 36 cm²

Solution:

A, B, C and D are the mid points of a ||gm PQRS

Area of PQRS = 36 cm²

The area of ||gm formed by joining AB, BC, CD and DA

Question 9.

The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is

(a) a rhombus of area 24 cm²

(b) a rectangle of area 24 cm²

(c) a square of area 26 cm²

(d) a trapezium of area 14 cm²

Solution:

Let P, Q, R, S be the mid points of sides of a rectangle ABCD. Whose sides 8 cm and 6 cm

Their PQRS is a rhombus

Question 10.

If AD is median of ∆ABC and P is a point on AC such that ar(∆ADP) : ar(∆ABD) = 2:3, then ar(∆PDC) : ar(∆ABC) is

(a) 1 : 5

(b) 1 : 5

(c) 1 : 6

(d) 3 : 5

Solution:

AD is the median of ∆ABC,

P is a point on AC such that

ar(∆ADP) : ar(∆ABD) = 2:3

Let area of ∆ADP = 2×2

Then area of ∆ABD = 3×2

But area of AABD = \(\frac { 1 }{ 2 }\) area AABC

∴ Area ∆ABC = 2 x area of ∆ABD

= 2 x 3x² = 6x²

and area of ∆PDC = area ∆ADC – (ar∆ADP) = area ∆ABD – ar ∆ADP

= 3x² – 2x² = x²

∴ Ratio = x² : 6x²

= 1 : 6 (c)

Question 11.

Medians of AABC, intersect at G. If ar(∆ABC) = 27 cm2, then ar(∆BGC) =

(a) 6 cm2

(b) 9 cm2

(c) 12 cm2

(d) 18 cm2

Solution:

In ∆ABC, AD, BE and CF are the medians which intersect each other at G

Question 12.

In a ∆ABC if D and E are mid-points of BC and AD respectively such that ar(∆AEC) = 4 cm², then ar(∆BEC) =

(a) 4 cm²

(b) 6 cm²

(c) 8 cm²

(d) 12 cm²

Solution:

In ∆ABC, D and E are the mid points of BC and AD

Join BE and CE ar(∆AEC) = 4 cm²

In ∆ABC,

∵ AD is the median of BC

∴ ar(∆ABD) = ar(∆ACD)

Similarly in ∆EBC,

ED is the median

∴ ar(∆EBD) = ar(∆ECD)

and in ∆ADC, CE is the median

∴ ar(∆FDC) = ar(∆AEC)

= 4 cm

∴ar∆EBC = 2 x ar(∆EDC)

= 2 x 4 = 8 cm (c)

Question 13.

In the figure, ABCD is a parallelogram. If AB = 12 cm, AE = 7.5 cm, CF = 15 cm, then AD =

(a) 3 cm

(b) 6 cm

(c) 8 cm

(d) 10.5 cm

Solution:

In ||gm ABCD, AB = 12 cm AE = 7.5 cm

∴ Area of ||gm ABCD = base x height = AB x AE = 12 x 7.5 cm² = 90 cm²

Now area ||gm ABCD = 90 cm²

and altitude CF = 15 cm

∴ Base AD = \(\frac { Area }{ Altitude }\) = \(\frac { 90 }{ 15 }\) = 6 cm (b)

Question 14.

In the figure, PQRS is a parallelogram. If X and Y are mid-points of PQ and SR respectively and diagonal SQ is joined. The ratio ar(||gm XQRY) : ar(∆QSR) =

(a) 1 : 4

(b) 2 : 1

(c) 1 : 2

(d) 1 : 1

Solution:

In ||gm PQRS, X and Y are the mid points of PQ and SR respectively XY and SQ are joined.

∵ XY bisects PQ and SR

∴ PXYS and XQRY are also ||gms and ar(∆PXYS) = nr(∆XQRY)

∵ ||gm PQRS and AQSR are on the same base and between the same parallel lines

Question 15.

Diagonal AC and BD of trapezium ABCD, in which AB || DC, intersect each other at O. The triangle which is equal in area of ∆AOD is

(a) ∆AOB

(b) ∆BOC

(c) ∆DOC

(d) ∆ADC

Solution:

In trapezium ABCD, diagonals AC and BD intersect each other at O. AB || DC

∆ABC and ∆ABD are on the same base and between the same parallels

∴ ar(∆ABC) = or(∆ABD)

Subtracting ar(∆AOB)

ar(∆ABC) – ar(∆AOB) = ar(∆ADB) – ar(∆AOB)

⇒ ar(∆BOC) = ar(∆AOD)

ar(∆AOD) = ar(∆BOC) (c)

Question 16.

ABCD is a trapezium in which AB || DC. If ar(∆ABD) = 24 cm² and AB = 8 cm, then height of ∆ABC is

(a) 3 cm

(b) 4 cm

(c) 6 cm

(d) 8 cm

Solution:

In trapezium ABCD, AB || DC

AC and BD are joined

ar(∆ABD) = 24 cm2

AB = 8 cm,

Question 17.

ABCD is a trapezium with parallel sides AB = a and DC = b. If E and F are mid-points of non-parallel sides AD and BC respectively, then the ratio of areas of quadrilaterals ABFE and EFCD is

(a) a : b

(b) (a + 3b) : (3a + b)

(c) (3a + b) : (a + 3b)

(d) (2a + b) : (3a + b)

Solution:

In quadrilateral ABCD, E and F are the mid points of AD and BC

AB = a, CD = b

Let h be the height of trapezium ABCD then height of each quadrilateral

ABFE = altitude of quadrilateral EFCD = \(\frac { h }{ 2 }\)

Now area of trap. ABFE = \(\frac { 1 }{ 2 }\) (sum of parallel sides) x altitude

Question 18.

ABCD is a rectangle with O as any point in its interior. If or(∆AOD) = 3 cm2 and ar(∆BOC) = 6 cm2, then area of rectangle ABCD is

(a) 9 cm2

(b) 12 cm2

(c) 15 cm2

(d) 18 cm2

Solution:

In rectangle ABCD, O is any point

ar(∆AOD) = 3 cm2

and ar(∆BOC) = 6 cm2

Join OA, OB, OC and OD

We know that if O is any point in ABCD Then ar(AOB) + ar(COD) = ar(AOB) + ar(BOC)

= 3 + 6 = 9 cm

∴ ar(rect. ABCD) = 2 x 9 = 18 cm (d)

Question 19.

The mid-points of the sides of a triangle ABC along with any of the vertices as the fourth point make a parallelogram of area equal to

Solution:

P,Q and R the mid points of the sides of a ∆ABC then area of any parallelogram formed by the mid points and one vertex of the given triangle has area = \(\frac { 1 }{ 2 }\) area ∆ABC (b)

Question 20.

In the figure, ABCD and FECG are parallelograms equal in area. If ar(∆AQE) = 12 cm2, then ar(||gm FGBQ) =

(a) 12 cm2

(b) 20 cm2

(c) 24 cm2

(d) 36 cm2

Solution:

In the figure, ABCD and EFCG are parallelograms equal in area and ar(∆AQE) = 12 cm2

In ||gm AQED, AE is the diagonal

∴ ar(∆AQE) = \(\frac { 1 }{ 2 }\) ar(||gm AQED)

⇒ 12 cm2 = \(\frac { 1 }{ 2 }\) ar(||gm AQED)

∴ ar(||gm AQED) = 24 cm2

∵ ar ||gm ABCD = ar ||gm FECG

⇒ ar(||gm ∆QED) + ar(|| gm QBCE)

= ar(||gm QBCE) + ar(||gm FGBQ)

⇒ ar(||gm ∆QED) = ar(||gm FGBQ)

= 24 cm2 (c)

Hope given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS are helpful to complete your math homework.

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