<\/span><\/h2>\nQuestion 1.
If ABC and BDE are two equilateral triangles such that D is the mid-ponit of BC, then find ar(\u2206ABC) : ar(\u2206BDE).
Solution:<\/p>\n
ABC and BDE are two equilateral triangles and D is the mid-point of BC
![\"RD](\"https:\/\/farm2.staticflickr.com\/1919\/44919889184_b6c6171fc1_o.png\")
Let each side of AABC = a
Then BD =\u00a0a<\/span>2<\/span><\/span><\/span><\/span><\/span>
\u2234 Each side of triangle BDE will be\u00a0a<\/span>2<\/span><\/span><\/span><\/span><\/span>
![\"Solution](\"https:\/\/farm2.staticflickr.com\/1924\/44919889414_2832c912b8_o.png\")
<\/p>\nQuestion 2.
In the figure, ABCD is a rectangle in which CD = 6 cm, AD = 8 cm. Find the area of parallelogram CDEF.
![\"RD](\"https:\/\/farm2.staticflickr.com\/1925\/44919889044_887bfdd85c_o.png\")
Solution:<\/p>\n
In rectangle ABCD,
CD = 6 cm, AD = 8 cm
\u2234 Area of rectangle ABCD = CD x AD
= 6 x 8 = 48 cm2<\/sup>
\u2235 DC || AB and AB is produced to F and DC is produced to G
\u2234 DG || AF
\u2235 Rectangle ABCD and ||gm CDEF are on the same base CD and between the same parallels
\u2234 ar(||gm CDEF) = ar(rect. ABCD)
= 48 cm2<\/sup><\/p>\nQuestion 3.
In the figure of Q. No. 2, find the area of \u2206GEF.
![\"RD](\"https:\/\/farm2.staticflickr.com\/1906\/44919888944_df9a1b2b90_o.png\")
Solution:
![\"RD](\"https:\/\/farm2.staticflickr.com\/1937\/44919888814_0f0a07c987_o.png\")
<\/p>\n
Question 4.
In the figure, ABCD is a rectangle with sides AB = 10 cm and AD = 5 cm. Find the area of \u2206EFG.
![\"RD](\"https:\/\/farm2.staticflickr.com\/1923\/44919888654_88d82fa963_o.png\")
Solution:
ABCD is a rectangle in which
AB = 10 cm, AD = 5 cm
![\"RD](\"https:\/\/farm2.staticflickr.com\/1935\/44919888484_31c49844e7_o.png\")
\u2235 ABCD is a rectangle
\u2234DC || AB,
DC is produced to E and AB is produced to G
\u2234DE || AG
\u2235 Rectangle ABCD and ||gm ABEF are on the same base AB and between the same parallels
\u2234 ar(rect. ABCD) = ar(||gm ABEF)
= AB x AD = 10 x 5 = 50 cm2<\/sup>
Now ||gm ABEF and AEFG are on the same
base EF and between the same parallels
\u2234 area \u2206EFG =\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span>\u00a0ar(||gm ABEF)
=\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span>\u00a0x 50 = 25 cm2<\/sup><\/p>\nQuestion 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(\u2206RAS).
Solution:
In quadrant PLRM, rectangle PQRS is in scribed
![\"RD](\"https:\/\/farm2.staticflickr.com\/1955\/44919888354_0790763d45_o.png\")
Radius of the circle = 13 cm
A is any point on PQ<\/p>\n
AR and AS are joined, PS = 5 cm
In right \u2206PRS,
PR2<\/sup>\u00a0= PS2<\/sup>\u00a0+ SR2<\/sup>
\u21d2 (132<\/sup>\u00a0= (5)2<\/sup>+ SR2<\/sup>
\u21d2 169 = 25 + SR2<\/sup>
\u21d2 SR2<\/sup>\u00a0= 169 \u2013 25 = 144 = (12)2<\/sup>
\u2234 SR = 12 cm
Area of rect. PQRS = PS x SR = 5x 12 = 60 cm2<\/sup>
\u2235 Rectangle PQRS and ARAS are on the same
base SR and between the same parallels
\u2234 Area ARAS =\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span>\u00a0area rect. PQRS 1
=\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span>\u00a0x 60 = 30 cm2<\/sup><\/p>\nQuestion 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 \u2206OPB.
Solution:<\/p>\n
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](\"https:\/\/farm2.staticflickr.com\/1955\/44919888224_d157bdbd7c_o.png\")
\u2234 Area of square ABCD = (side)2<\/sup>
\u2235 Diagonal BD bisects the square into two triangle equal in area
\u2234 Area \u2206ABD =\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span>\u00a0x area of square ABCD
=\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span>\u00a0x 64 = 32 cm2<\/sup>
\u2235 P is mid point of AB of AABD, and PQ || AD
\u2234 O is the mid point of BD
\u2234 OP =\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span>AD =\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span>\u00a0x 8 = 4 cm
and PB =\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span>\u00a0AB =\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span>\u00a0x 8 = 4 cm
\u2234 Area \u2206OPB =\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span>PB x OP
=\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span>\u00a0x4x4 = 8 cm2<\/sup><\/p>\nQuestion 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 \u2206ABC is 16 cm2<\/sup>, find the area of \u2206DEF.
Solution:
In \u2206ABC, D is mid point of BC. E and F are the mid points of DC and AE respectively area of \u2206ABC = 16 cm2<\/sup>
FD is joined
![\"RD](\"https:\/\/farm2.staticflickr.com\/1933\/44919888134_23db3d28c6_o.png\")
\u2235 D is mid point of BC
\u2234 AD is the median and median divides the triangle into two triangles equal in area
area \u2206ADC =\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span>\u00a0ar(\u2206ABC)
=\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span>\u00a0x 16 = 8 cm2<\/sup>
Similarly, E is mid point of DC
\u2234 area (\u2206ADE) =\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span>\u00a0ar(\u2206ADC)
=\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span>\u00a0x 8 = 4 cm2<\/sup>
\u2235 F is mid point of AE of \u2206ADE
\u2234 ar(\u2206DEF) =\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span>area (\u2206ADE)
=\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span>\u00a0x 4 = 2 cm2<\/sup><\/p>\nQuestion 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 \u2206PBQ is 17 cm2, find the area of \u2206ASR.
Solution:
In trapezium PQRS,
PS || QR
A and B are points on sides PQ and SR
Such that AB || QR
area of \u2206PBQ = 17 cm2<\/sup>
![\"Class](\"https:\/\/farm2.staticflickr.com\/1980\/44919887944_94fd19affd_o.png\")
\u2206ABQ and \u2206ABR are on the same base AB and between the same parallels
\u2234 ar(\u2206ABQ) = ar(\u2206ABR) \u2026(i)
Similarly, \u2206ABP and \u2206ABS are on the same base and between the same parallels
\u2234 ar(ABP) = ar(\u2206ABS) \u2026(ii)
Adding (i) and (ii)
ar( \u2206ABQ) + ar( \u2206ABP)
= ar(\u2206ABR) + ar(\u2206ABS)
\u21d2 ar(\u2206PBQ) = ar(\u2206ASR)
Put ar(PBQ) = 17 cm2<\/sup>
\u2234 ar(\u2206ASR) = 17 cm2<\/sup><\/p>\nQuestion 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(\u2206PBQ) = 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](\"https:\/\/farm2.staticflickr.com\/1955\/44919887804_41e1703c9f_o.png\")
CQ : QP = 3 : 1
ar(\u2206PBQ) = 10 cm2<\/sup>
In ||gm ABCD,
BD is its diagonal
\u2234 ar(\u2206ABD) = ar(\u2206BCD) =\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span>\u00a0ar ||gm ABCD
\u2234 ar(||gm ABCD) = 2ar(\u2206ABD) \u2026(i)
In \u2206PBC CQ : QP = 3 : 1
\u2235 \u2206PBQ and \u2206CQB have same vertice B
\u2234 3 x area \u2206PBQ = ar(\u2206CBQ)
\u21d2 area(\u2206CBQ) = 3 x 10 = 30 cm2<\/sup>
\u2234 ar(\u2206PBC) = 30 + 10 = 40 cm2<\/sup>
Now \u2206ABD and \u2206PBC are between the
same parallel but base PB =\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span>\u00a0AB
\u2234 ar(\u2206ABD) = 2ar(\u2206PBC)
= 2 x 40 = 80 cm2<\/sup>
But ar(||gm ABCD) = 2ar(\u2206ABD)
= 2 x 80 = 160 cm2<\/sup><\/p>\nQuestion 10.
P is any point on base BC of \u2206ABC and D is the mid-point of BC. DE is drawn parallel to PA to meet AC at E. If ar(\u2206ABC) = 12 cm2<\/sup>, then find area of \u2206EPC.
Solution:
P is any point on base of \u2206ABC
D is mid point of BC
DE || PA drawn which meet AC at E
ar(\u2206ABC) = 12 cm2<\/sup>
AD and PE are joined
![\"RD](\"https:\/\/farm2.staticflickr.com\/1969\/44919887664_9cc9678034_o.png\")
\u2235 D is mid point of BC
\u2234 AD is median
\u2234 ar(\u2206ABD) = ar(\u2206ACD)
=\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span>\u00a0(\u2206ABC) =\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span>\u00a0x 12 = 6 cm2<\/sup>\u00a0\u2026(i)
\u2235 \u2206PED and \u2206ADE are on the same base DE and between the same parallels
\u2234 ar(\u2206PED) = ar(\u2206ADE)
Adding ar(\u2206DCE) to both sides,
ar(\u2206PED) + ar(\u2206DCE) = ar(\u2206ADE) + ar(\u2206DCE)
ar(\u2206EPC) = ar(\u2206ACD)
\u21d2 ar(\u2206EPC) = ar(\u2206ABD) = 6 cm2 [From (i)]
\u2234 ar(\u2206EPC) = 6 cm2<\/sup><\/p>\nWe have included all the information regarding\u00a0CBSE<\/a> RD Sharma Class 9 Solutions Chapter 14 VSAQS<\/span>. If you have any query feel free to ask in the comment section.\u00a0<\/span><\/p>\n
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-14-VSAQS.jpg\")
<\/p>\n