# RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQs

In this chapter, we provide RD Sharma Solutions for Chapter 14 Quadrilaterals MCQS for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Chapter 14 Quadrilaterals MCQS Maths pdf, free RD Sharma Solutions for Chapter 14 Quadrilaterals MCQS Maths book pdf download. Now you will get step by step solution to each question.

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

∵ 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
Let area of ∆ADP = 2×2
Then area of ∆ABD = 3×2

But area of AABD = 12 area AABC
∴ Area ∆ABC = 2 x area of ∆ABD
= 2 x 3x² = 6x²
= 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 = AreaAltitude = 9015 = 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
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 = h2
Now area of trap. ABFE = 12 (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 = 12 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) = 12 ar(||gm AQED)
⇒ 12 cm2 = 12 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)

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