## RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS

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

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

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

Let each side of AABC = a

Then BD = \(\frac { a }{ 2 }\)

∴ Each side of triangle BDE will be \(\frac { a }{ 2 }\)

Question 2.

In the figure, ABCD is a rectangle in which CD = 6 cm, AD = 8 cm. Find the area of parallelogram CDEF.

Solution:

In rectangle ABCD,

CD = 6 cm, AD = 8 cm

∴ Area of rectangle ABCD = CD x AD

= 6 x 8 = 48 cm^{2}

∵ 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 cm^{2}

Question 3.

In the figure of Q. No. 2, find the area of ∆GEF.

Solution:

Question 4.

In the figure, ABCD is a rectangle with sides AB = 10 cm and AD = 5 cm. Find the area of ∆EFG.

Solution:

ABCD is a rectangle in which

AB = 10 cm, AD = 5 cm

∵ 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 cm^{2}

Now ||gm ABEF and AEFG are on the same

base EF and between the same parallels

∴ area ∆EFG = \(\frac { 1 }{ 2 }\) ar(||gm ABEF)

= \(\frac { 1 }{ 2 }\) x 50 = 25 cm^{2}

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

Radius of the circle = 13 cm

A is any point on PQ

AR and AS are joined, PS = 5 cm

In right ∆PRS,

PR^{2} = PS^{2} + SR^{2}

⇒ (13^{2} = (5)^{2}+ SR^{2}

⇒ 169 = 25 + SR^{2}

⇒ SR^{2} = 169 – 25 = 144 = (12)^{2}

∴ SR = 12 cm

Area of rect. PQRS = PS x SR = 5x 12 = 60 cm^{2}

∵ Rectangle PQRS and ARAS are on the same

base SR and between the same parallels

∴ Area ARAS = \(\frac { 1 }{ 2 }\) area rect. PQRS 1

= \(\frac { 1 }{ 2 }\) x 60 = 30 cm^{2}

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

∴ Area of square ABCD = (side)^{2}

∵ Diagonal BD bisects the square into two triangle equal in area

∴ Area ∆ABD = \(\frac { 1 }{ 2 }\) x area of square ABCD

= \(\frac { 1 }{ 2 }\) x 64 = 32 cm^{2}

∵ P is mid point of AB of AABD, and PQ || AD

∴ O is the mid point of BD

∴ OP = \(\frac { 1 }{ 2 }\)AD = \(\frac { 1 }{ 2 }\) x 8 = 4 cm

and PB = \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) x 8 = 4 cm

∴ Area ∆OPB = \(\frac { 1 }{ 2 }\)PB x OP

= \(\frac { 1 }{ 2 }\) x4x4 = 8 cm^{2}

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 cm^{2}, 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 cm^{2}

FD is joined

∵ D is mid point of BC

∴ AD is the median and median divides the triangle into two triangles equal in area

area ∆ADC = \(\frac { 1 }{ 2 }\) ar(∆ABC)

= \(\frac { 1 }{ 2 }\) x 16 = 8 cm^{2}

Similarly, E is mid point of DC

∴ area (∆ADE) = \(\frac { 1 }{ 2 }\) ar(∆ADC)

= \(\frac { 1 }{ 2 }\) x 8 = 4 cm^{2}

∵ F is mid point of AE of ∆ADE

∴ ar(∆DEF) = \(\frac { 1 }{ 2 }\)area (∆ADE)

= \(\frac { 1 }{ 2 }\) x 4 = 2 cm^{2}

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 cm^{2}

∆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 cm^{2}

∴ ar(∆ASR) = 17 cm^{2}

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

CQ : QP = 3 : 1

ar(∆PBQ) = 10 cm^{2}

In ||gm ABCD,

BD is its diagonal

∴ ar(∆ABD) = ar(∆BCD) = \(\frac { 1 }{ 2 }\) 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 cm^{2}

∴ ar(∆PBC) = 30 + 10 = 40 cm^{2}

Now ∆ABD and ∆PBC are between the

same parallel but base PB = \(\frac { 1 }{ 2 }\) AB

∴ ar(∆ABD) = 2ar(∆PBC)

= 2 x 40 = 80 cm^{2}

But ar(||gm ABCD) = 2ar(∆ABD)

= 2 x 80 = 160 cm^{2}

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 cm^{2}, 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 cm^{2}

AD and PE are joined

∵ D is mid point of BC

∴ AD is median

∴ ar(∆ABD) = ar(∆ACD)

= \(\frac { 1 }{ 2 }\) (∆ABC) = \(\frac { 1 }{ 2 }\) x 12 = 6 cm^{2} …(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 cm^{2}

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

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.