RD Sharma Class 9 Solutions Chapter 14 MCQS (Updated for 2024)

RD Sharma Class 9 Solutions Chapter 14 MCQs are available here. RD Sharma Class 9 Solutions Chapter 14 MCQs are created by our team of experts in a detailed manner. 

• RD Sharma Class 9 Maths Solutions

Access RD Sharma Class 9 Solutions Chapter 14 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)

A triangle and a parallelogram are on the same base and between the same parallels. The ratio of the areas of the triangle and parallelogram is
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 3
Solution:
A triangle and a parallelogram are on the same base and between the same parallels, then area of a 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  ∆ABC, D, E, and 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 midpoints 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 = 12 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)

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