RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS

In this chapter, we provide RD Sharma Solutions for Chapter 14 Quadrilaterals VSAQS 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 VSAQS Maths pdf, free RD Sharma Solutions for Chapter 14 Quadrilaterals VSAQS Maths book pdf download. Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 9
SubjectMaths
ChapterChapter 14
Chapter NameQuadrilaterals
ExerciseVSAQS

RD Sharma Solutions for Class 9 Chapter 14 Quadrilaterals VSAQS Download PDF

Question 1.
If ABC and BDE are two equilateral triangles such that D is the mid-ponit of BC, then find ar(∆ABC) : ar(∆BDE).
Solution:
ABC and BDE are two equilateral triangles and D is the mid-point of BC
RD Sharma Mathematics Class 9 Solutions Chapter 14 Quadrilaterals
Let each side of AABC = a
Then BD = a2
∴ Each side of triangle BDE will be a2
Solution Of Rd Sharma Class 9 Chapter 14 Quadrilaterals

Question 2.
In the figure, ABCD is a rectangle in which CD = 6 cm, AD = 8 cm. Find the area of parallelogram CDEF.
RD Sharma Math Solution Class 9 Chapter 14 Quadrilaterals
Solution:
In rectangle ABCD,
CD = 6 cm, AD = 8 cm
∴ Area of rectangle ABCD = CD x AD
= 6 x 8 = 48 cm2
∵ DC || AB and AB is produced to F and DC is produced to G
∴ DG || AF
∵ Rectangle ABCD and ||gm CDEF are on the same base CD and between the same parallels
∴ ar(||gm CDEF) = ar(rect. ABCD)
= 48 cm2

Question 3.
In the figure of Q. No. 2, find the area of ∆GEF.
RD Sharma Class 9 Questions Chapter 14 Quadrilaterals
Solution:
RD Sharma Class 9 Chapter 14 Quadrilaterals

Question 4.
In the figure, ABCD is a rectangle with sides AB = 10 cm and AD = 5 cm. Find the area of ∆EFG.
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals
Solution:
ABCD is a rectangle in which
AB = 10 cm, AD = 5 cm
RD Sharma Solutions Class 9 Chapter 14 Quadrilaterals
∵ ABCD is a rectangle
∴DC || AB,
DC is produced to E and AB is produced to G
∴DE || AG
∵ Rectangle ABCD and ||gm ABEF are on the same base AB and between the same parallels
∴ ar(rect. ABCD) = ar(||gm ABEF)
= AB x AD = 10 x 5 = 50 cm2
Now ||gm ABEF and AEFG are on the same
base EF and between the same parallels
∴ area ∆EFG = 12 ar(||gm ABEF)
= 12 x 50 = 25 cm2

Question 5.
PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then find or(∆RAS).
Solution:
In quadrant PLRM, rectangle PQRS is in scribed
RD Sharma Class 9 PDF Chapter 14 Quadrilaterals
Radius of the circle = 13 cm
A is any point on PQ
AR and AS are joined, PS = 5 cm
In right ∆PRS,
PR2 = PS2 + SR2
⇒ (132 = (5)2+ SR2
⇒ 169 = 25 + SR2
⇒ SR2 = 169 – 25 = 144 = (12)2
∴ SR = 12 cm
Area of rect. PQRS = PS x SR = 5x 12 = 60 cm2
∵ Rectangle PQRS and ARAS are on the same
base SR and between the same parallels
∴ Area ARAS = 12 area rect. PQRS 1
= 12 x 60 = 30 cm2

Question 6.
In square ABCD, P and Q are mid-point of AB and CD respectively. If AB = 8 cm and PQ and BD intersect at O, then find area of ∆OPB.
Solution:
In sq. ABCD, P and Q are the mid points of sides AB and CD respectively PQ and BD are joined which intersect each other at O
Side of square AB = 8 cm
Quadrilaterals Class 9 RD Sharma Solutions
∴ Area of square ABCD = (side)2
∵ Diagonal BD bisects the square into two triangle equal in area
∴ Area ∆ABD = 12 x area of square ABCD
= 12 x 64 = 32 cm2
∵ P is mid point of AB of AABD, and PQ || AD
∴ O is the mid point of BD
∴ OP = 12AD = 12 x 8 = 4 cm
and PB = 12 AB = 12 x 8 = 4 cm
∴ Area ∆OPB = 12PB x OP
= 12 x4x4 = 8 cm2

Question 7.
ABC is a triangle in which D is the mid-point of BC. E and F are mid-points of DC and AE respectively. If area of ∆ABC is 16 cm2, find the area of ∆DEF.
Solution:
In ∆ABC, D is mid point of BC. E and F are the mid points of DC and AE respectively area of ∆ABC = 16 cm2
FD is joined
RD Sharma Class 9 Solution Chapter 14 Quadrilaterals
∵ D is mid point of BC
∴ AD is the median and median divides the triangle into two triangles equal in area
area ∆ADC = 12 ar(∆ABC)
= 12 x 16 = 8 cm2
Similarly, E is mid point of DC
∴ area (∆ADE) = 12 ar(∆ADC)
= 12 x 8 = 4 cm2
∵ F is mid point of AE of ∆ADE
∴ ar(∆DEF) = 12area (∆ADE)
= 12 x 4 = 2 cm2

Question 8.
PQRS is a trapezium having PS and QR as parallel sides. A is any point on PQ and B is a point on SR such that AB || QR. If area of ∆PBQ is 17 cm2, find the area of ∆ASR.
Solution:
In trapezium PQRS,
PS || QR
A and B are points on sides PQ and SR
Such that AB || QR
area of ∆PBQ = 17 cm2
Class 9 RD Sharma Solutions Chapter 14 Quadrilaterals
∆ABQ and ∆ABR are on the same base AB and between the same parallels
∴ ar(∆ABQ) = ar(∆ABR) …(i)
Similarly, ∆ABP and ∆ABS are on the same base and between the same parallels
∴ ar(ABP) = ar(∆ABS) …(ii)
Adding (i) and (ii)
ar( ∆ABQ) + ar( ∆ABP)
= ar(∆ABR) + ar(∆ABS)
⇒ ar(∆PBQ) = ar(∆ASR)
Put ar(PBQ) = 17 cm2
∴ ar(∆ASR) = 17 cm2

Question 9.
ABCD is a parallelogram. P is the mid-point of AB. BD and CP intersect at Q such that CQ : QP = 3 : 1. If ar(∆PBQ) = 10 cm2, find the area of parallelogram ABCD.
Solution:
In ||gm ABCD, P is mid point on AB,
PC and BD intersect each other at Q
Class 9 Maths Chapter 14 Quadrilaterals RD Sharma Solutions
CQ : QP = 3 : 1
ar(∆PBQ) = 10 cm2
In ||gm ABCD,
BD is its diagonal
∴ ar(∆ABD) = ar(∆BCD) = 12 ar ||gm ABCD
∴ ar(||gm ABCD) = 2ar(∆ABD) …(i)
In ∆PBC CQ : QP = 3 : 1
∵ ∆PBQ and ∆CQB have same vertice B
∴ 3 x area ∆PBQ = ar(∆CBQ)
⇒ area(∆CBQ) = 3 x 10 = 30 cm2
∴ ar(∆PBC) = 30 + 10 = 40 cm2
Now ∆ABD and ∆PBC are between the
same parallel but base PB = 12 AB
∴ ar(∆ABD) = 2ar(∆PBC)
= 2 x 40 = 80 cm2
But ar(||gm ABCD) = 2ar(∆ABD)
= 2 x 80 = 160 cm2

Question 10.
P is any point on base BC of ∆ABC and D is the mid-point of BC. DE is drawn parallel to PA to meet AC at E. If ar(∆ABC) = 12 cm2, then find area of ∆EPC.
Solution:
P is any point on base of ∆ABC
D is mid point of BC
DE || PA drawn which meet AC at E
ar(∆ABC) = 12 cm2
AD and PE are joined
RD Sharma Book Class 9 PDF Free Download Chapter 14 Quadrilaterals
∵ D is mid point of BC
∴ AD is median
∴ ar(∆ABD) = ar(∆ACD)
= 12 (∆ABC) = 12 x 12 = 6 cm2 …(i)
∵ ∆PED and ∆ADE are on the same base DE and between the same parallels
∴ ar(∆PED) = ar(∆ADE)
Adding ar(∆DCE) to both sides,
ar(∆PED) + ar(∆DCE) = ar(∆ADE) + ar(∆DCE)
ar(∆EPC) = ar(∆ACD)
⇒ ar(∆EPC) = ar(∆ABD) = 6 cm2 [From (i)]
∴ ar(∆EPC) = 6 cm2

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