mensuration imp questions

88. Three circles of radius 3.5 cm each
are placed in such a way that each
touches the other two. The area
of the portion enclosed by the
circles is
Radius of each circle = 3.5 cm
From the figure.
DABC will be an equilateral triangle
of side 7 cm each.
Now, the required area
= Area of DABC – 3× (Area of a
sector of angle 60° in a circle of
radius 3.5 cm)
=√3/4(7)2-3[(60/360) x 22/7 x (3.5)2 (cm)2

92 The area of the largest triangle,
that can be inscribed in a semicircle of radius r cm, is
The largest triangle inscribed in
a semi-circle will have base equal
to 2r cm and height equal to r cm
as shown in figure
Area =1/2 × base × height
= 1/2 × 2r × r = r2 cm2 
trisection=1/3


94a tringle is formed on the  half OF CIRCLE sUCH THAT ITS BASE IS DIAMETER OF THE CIRCLE .what will be the area of rest part. (1/2circle- traingle)
Let Radius of circle = a units
 Area of semi circle
=⊼a2/ 2sq.units
Both triangles D ABC and D BCD
are isosceles and equal.
Area of each triangle = 1/2 (a)2
Area of both triangles
= 2 (1/2) (a)2  =(a)2 sq.units
 Area of shaded region
=⊼(a)2/ 2-(a)2 sq.units


98. A 7 m wide road runs outside
around a circular park, whose
circumference is 176 m. The area
of the road is :
If the radius of the circular park
be r metre, then 2 ⊼r = 176
 2 × (22/7) x r = 176
 r =176 x 7/22 =28 metre
Radius of the park with road
= 28 + 7 = 35 metre
Area of the road
=22/7[(35)2 – (28)2]
=22/7× 63 ×7 = 1386 m2 

101 The area of a circle is increased
by 22 cm2 when its radius is increased by 1 cm. The original
radius of the circle is 
⊼(r + 1)2 – ⊼r2 = 22




Three circles of radius a, b, ctouch each other externally. The
area of the triangle formed by
joining their centre is
x= AB = a + b
y = BC = b + c
z = CA = a + c
 s =AB+BC+CA / 2
= a + b + c
Area of D ABC
= √s(s - x)(s - y)(s - z)
= √(a + b + c)abc




112
The area of the incircle of an
equilateral triangle of side 42 cm
is

Radius of in circle=a/2√3

=> 42/2√3

=>7√3Area of in circle = 22/7∗49∗3=462cm2

113. The ratio of the area of the in-circle and the circum-circle of a
square is Let the side of the square be 2x. Then radius of incircle = a Radius of circum-circle √[(a)2 + (a)2] = √2a

Ratio of area = ⊼(a)2:⊼ (√2a)2 = 1 : 2

114.The area of an equilateral triangle
inscribed in a circle is 4 3 cm2.
The area of the circle is
Area of tABC =√3/4× (side)2= 43
=>side = √16 = 4 cm
<BOD = 60°
< sin 60° =BD/OB
=>√3/2=2 /OB
OB=4/√3
 Area of circle = ⊼r2
=(16/3)⊼ cm2 

119 The sides of a triangle are 6 cm,
8 cm and 10 cm. The area of the
greatest square that can be inscribed in it, is
Here, 62+82 = 102
Hence, DABC is right angled
BD is perpendicular to AC
1/ 2 × AB × BC
 =1 / 2 × AC × BD
=>1/2 × 6 × 8 = 1/2× 10 × BD
=>BD =48/10=24/5
BD = diagonal of square
=>Area of square =24 x 24/ 2x5x5
´=576/50
cm2  

The area of a circle inscribed in
a square of area 2 m2 is
1st side nikalo incircle ka phir area
Side of square = √2 metre
Radius of in-circle
=  =√2/2metre
Area of the circle = ⊼r2
= ⊼/2sq. metre

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For the equilateral triangle of
side a,
In radius =
a/ 2√3
Circum-radius =a/√3

127. A circle is inscribed in an equilateral triangle and a square is
inscribed in that circle. The ratio of the areas of the triangle
and the square is
Let AB = BC = CA = x units, then
AD =√ [(x)2 -(x)2/4]=(√3x/2)
=>OD =(1/3) AD =x/ 2√3 = radius ofcircle
=> Diagonal of square=  (2x/2√3) = x/√3
 Triangle : Square
=(3/√4) (x)2 : (x)2/ (2x3)

133The perimeter of a rectangle is
160 metre and the difference of
two sides is 48 metre. Find the
side of a square whose area is
equal to the area of this rectangle

Let the length and breadth of
rectangle are a and b respectively.
According to the question,
2(a + b) = 160
Þ a + b = 80 ...(i)
a – b = 48 ...(ii)
2a = 128 (On adding)
 a =128/2 = 64m
From equation (i),
b = 80 – 64 = 16 m
 Area of rectangle
= 64 × 16 m2
Area of square
= 64 × 16 m2
=>(side)2 = 64 × 16
 side = 8 × 4 = 32 m