ML Aggarwal Solution Class 9 Chapter 16 Mensuration EXERCISE 16.2
EXERCISE 16.2
Question 1
(i) Find the area of quadrilateral whose one diagonal is 20 cm long and the perpendiculars to this diagonal from other vertices are of length 9 cm and 15 cm.
Sol :
Let ABCD be quadrilateral in which AC=20 cm
⟂BY=9 cm
⟂DY=15 cm
(a) Area of quadrilateral ABCD
=Area of ΔABC+Area of ΔACD
$=\frac{1}{2} \times$ base $\times$ height $+\frac{1}{2} \times$ base $\times$
height
$=\frac{1}{2} \times \mathrm{AC} \times \mathrm{BX}+\frac{1}{2} \times \mathrm{AC} \times \mathrm{DY}$
$=\left(\frac{1}{2} \times 20 \times 9\right)+\left(\frac{1}{2} \times 20 \times 15\right) \mathrm{cm}^{2}$
=(10×9+10×15) cm2
=(90+150) cm2
=240 cm2
(ii) Find the area of a quadrilateral whose diagonals are of length 18 cm and 12 cm and they intersect each other at right angles.
Sol :
Let ABCD be a quadrilateral in which diagonals AC and BD intersect each other at M at right angles AC=18 cm and BD=12 cm
Area of quadrilateral ABCD
$=\frac{1}{2} \times \text { diagonal } \mathrm{AC} \times \text { diagonal } \mathrm{BD}$
$=\frac{1}{2} \times 18 \times 12 \mathrm{~cm}^{2}$
=9×12 cm2
=108 cm2
Question 2
Find the area of the quadrilateral field ABCD whose sides AB = 40 m, BC = 28 m, CD = 15 m, AD = 9 m and ∠A = 90°
Sol :
A quadrilateral ABCD in which AB=40 m
BC=28 m, CD=15 m, AD=9 m
In ∠BAD, ∠A=90°
By Pythagorass theorem
⇒BD2=BA2+AD2
⇒BD2=(40)2+(9)2
⇒BD2=1600+81=1681
⇒BD=41
Area of quadrilateral ABCD=Area of ΔBAD+Area of ΔBDC
$=\frac{1}{2} \times$ Base $\times$ height $+$ Area of $\triangle \mathrm{BDC}$
$=\frac{1}{2} \times 40 \mathrm{~m} \times 9 \mathrm{~m}+$ Area of $\triangle \mathrm{BDC}$
=20 m×9m+Area of ΔBDC
=180m2+Area of ΔBDC
Now to find area of ΔBDC
Let a=BD=41 m, b=CD=15 m, c=BC=28 m
$\mathrm{S}=\frac{a+b+c}{2}=\frac{41+15+28}{2}=42 \mathrm{~m}$
Area of ΔBDC$=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{42(42-41)(42-15)(42-28)}$
$=\sqrt{42 \times 1 \times 27 \times 14}$
$=\sqrt{2 \times 3 \times 7 \times 3 \times 3 \times 3 \times 2 \times 7}$
=2×7×3×3=126 m2
Area of quadrilateral ABCD
=180m2+Area of ΔBDC
=180m2+126m2=306 m2
Question 3
Find the area of the quadrilateral ABCD in which ∠BCA= 90°, AB = 13 cm and ACD is an equilateral triangle of side 12 cm.
Sol :
Quadrilateral ABCD in which ∠BCA=90°
AB=13 cm ,
ΔACD is equilateral in which AC=CD=AD=12 cm
In right angled ΔABC
By Pythagoras theorem,
⇒AB2=AC2+BC2
⇒(13)2=(12)2+BC2
⇒BC2=(13)2-(12)2
⇒BC2=169-144
⇒BC2=25
⇒BC=√25=5 cm
Area of quadrilateral ABCD=Area of ΔABC+Area of ΔACD
$=\frac{1}{2} \times$ base $\times$ height $+\frac{\sqrt{3}}{4} \times(\text { side })^{2}$
$=\frac{1}{2} \times \mathrm{AC} \times \mathrm{BC}+\frac{\sqrt{3}}{4} \times(12)^{2} \mathrm{~cm}^{2}$
$=\frac{1}{2} \times 12 \times 5+\frac{\sqrt{3}}{4} \times 12 \times 12 \mathrm{~cm}^{2}$
=6×5+√3×3×12 cm2
=30+36√3
=30+36×1.732=30+62.28
=92.28 cm2
Question 4
Find the area of quadrilateral ABCD in which ∠B = 90°, AB = 6 cm, BC = 8 cm 13 and CD = AD = 13 cm.
Sol :
A quadrilateral ABCD in which AB=6 cm,
BC=8 cm, CD=13 cm and AD=13 cm
In ΔABC ,∠B=90°
By Pythagoras theorem
⇒AC2=AB2+BC2
⇒AC2=(6)2+(8)2
⇒AC2=36+64=100
⇒AC2=102
⇒AC=10 cm
Area of quadrilateral ABCD
=Area of ΔABC+Area of ΔACD
$=\frac{1}{2} \times$ Base $\times$ height $\times$ Area of $\Delta \mathrm{ACD}$
$=\frac{1}{2} \times \mathrm{AB} \times \mathrm{BC}+$ Area of $\Delta \mathrm{ACD}$
$=\frac{1}{2} \times 6 \times 8 \mathrm{~cm}^{2}+$ Area of $\Delta \mathrm{ACD}$
$=3 \times 8 \mathrm{~cm}^{2}+$ Area of $\Delta \mathrm{ACD}$
$=24 \mathrm{~cm}^{2}+$ Area of $\Delta \mathrm{ACD}$...(1)
Now, to find area of ΔACD
∴let a=AC=10 cm, b=CD=13 cm, c=AD=13 cm
$s=\frac{a+b+c}{2}=\frac{10+13+13}{2} \mathrm{~cm}$
$=\frac{10+26}{2} \mathrm{~cm}=\frac{36}{2} \mathrm{~cm}=18 \mathrm{~cm}$
Area of $\Delta \mathrm{ACD}
$=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{18(18-10)(18-13)(18-13)}$
$=\sqrt{12 \times 8 \times 5 \times 5}$
$=\sqrt{6 \times 3 \times 8 \times 5 \times 5}$
$=\sqrt{3 \times 2 \times 3 \times 2 \times 2 \times 2 \times 5 \times 5}$
$=3 \times 2 \times 2 \times 5=60 \mathrm{~cm}^{2}$
∴From (1)
Area of quadrilateral ABCD
$=24 \mathrm{~cm}^{2}+$ Area of $\Delta \mathrm{ACD}$
$=24 \mathrm{~cm}^{2}+60 \mathrm{~cm}^{2}$
=84 cm2
Question 5
The perimeter of a rectangular cardboard is 96 cm ; If its breadth is 18 cm, find the length and the area of the cardboard.
Sol :
Le ABCD be a rectangle Let length be l cm and breadth =18 cm
Perimeter =96 cm
⇒2(l+b)=96 cm
⇒(l+18)=$\frac{96}{2}$ cm
⇒l=48-18=30 cm
Area of the rectangular cardboard
=l×b
=3×18 cm
=540 cm2
Question 6
The length of a rectangular hall is 5 m more than its breadth, If the area of the hall is 594 m2, find its perimeter.
Sol :
Let ABCD be rectangular field
Let Breadth= x m
then length=(x+5) m
Area of rectangular field=l×b
⇒594=x(x+5)
⇒594=x2+5x
⇒0=x2+5x-594
⇒x2+5x-594=0
⇒x2+27x-22x-594=0
⇒x(x+27)-22(x+27)=0
⇒(x-22)(x+27)=0
Either x-22=0 or x+27=0
x=22 m , x=-27 (not possible)
Breadth=22 m
Length=(x+5)=(22+5)=27 m
Perimeter=2(l+b)
=2(27+22) m
=2×49=98 m
Question 7
(a) The diagram (i) given below shows two paths drawn inside a rectangular field 50 m long and 35 m wide. The width of each path is 5 metres. Find the area of the shaded portion.
(b) In the diagram (ii) given below, calculate the area of the shaded portion. All measurements are in centimetres.
Sol :
(a) Area of shaded portion=Area of rectangle ABCD+Area of rectangle PQRS-Area of square LMNO
$=50 \times 5 \mathrm{~m}^{2}+5 \times 35 \mathrm{~m}^{2}-5 \times 5 \mathrm{~m}^{2}$
$=250 \mathrm{~m}^{2}+175 \mathrm{~m}^{2}-25 \mathrm{~m}^{2}=250 \mathrm{~m}^{2}+150 \mathrm{~m}^{2}=400 \mathrm{~m}^{2}$
(b) Area of shaded portion=Area of ABCD-5×Area of any small square
=l×b-5×side×side
=(8×6-5×2×2) cm2
=(48-20) cm2
=28 cm2
Question 8
A rectangular plot 20 m long and 14 m wide is to be covered with grass leaving 2 m all around. Find the area to be laid with grass.
Sol :
Let ABCD be the plot Length of P lot =20 m
Breadth of Plot =14 m
Let PQRS be the grassy plot
Length of grassy Lawn =20 m-2×2 m
=20 m-4m=16 m
Breadth of grassy Lawn =14 m-2×2 m=14 m-4 m=10 m
Area of grassy Lawn = Length × Breadth =16×10 m2
=160 m2
Question 9
The shaded region of the given diagram represents the lawn in front of a house. On three sides of the lawn there are flower beds of width 2 m.
(i) Find the length and the breadth of the lawn.
(ii) Hence, or otherwise, find the area of the flower – beds.
Sol :
(i) Length of Lawn BCDE=BC
=AD-AB-CD=30 m-2m-2m
=30m-4m=26 m
Breadth of Lawn BCDE
=BE=AG-GH=12 m-2m=10m
(ii) Area of flower beds=Area of rectangle ADFG-Area of Lawn BCDE
=AD AG-BC BE
=30×12 m2-26×10m2
=360 m2-260 m2=100 m2
Question 10
A foot path of uniform width runs all around the inside of a rectangular field 50 m long and 38m wide. If the area of the path is 492 m². Find its width.
Sol :
Here ABCD be a rectangular field having length=50 m
and Breadth=38 m
Area of rectangular field ABCD
=l×b=50×38 m2
=1900 m2
Let the width of foot path all around the inside of a rectangular field =x m
Then length of rectangular field PQRS
=(50-x-x) m=(50-2x) m
Breadth of rectangular field PQRS
=(38-x-x) m
=(38-2x) m
Area of rectangular field PQRS=l×b
=(50-2x)(38-2x)
Area of foot path=Area of rectangular field ABCD-Area of rectangular field PQRS
⇒492=1900-(50-2x)(38-2x)
⇒492=1900-[50(38-2x)-2x(38-2x)]
⇒492=1900-(1900-100x-76x+4x2)
⇒492=1900-1900+100x+76x-4x2
⇒492=176x-4x2
⇒492=4(44x-x2)
⇒$\frac{492}{4}=44 x-x^{2}$
⇒123=44x-x2
⇒x2-44x+123=0
⇒x2-41x-3x+123=0
⇒x(x-41)-3(x-41)
⇒(x-3)(x-41)=0
Either x-3=0 or x-41=0
x=3 m, x=41 m (not possible)
Hence , width=3 m
Question 11
The cost of enclosing a rectangular garden with a fence all around at the rate of Rs. 15 per metre is Rs. 5400. If the length of the garden is 100 m And the area of the garden.
Sol :
Here ABCD be a rectangular garden length of the garden=100 m
Let Breadth of the garden=x m
Perimeter of the garden ABCD=2(l+b)
=2(100+x) m
=(200+2x) m
Cost of 1 m to enclosing a rectangular garden
=Rs 15
Cost for (200+2x) to enclosing a rectangular garden
=15(200+2x)
=3000+30x
But given cost=Rs 5400
Then , 3000+30x=5400
⇒30x=5400-3000
⇒$x=\frac{2400}{30}=80$ m
i.e. breadth of garden=80 m
⇒Area of rectangular field =l×b
⇒100×80 m2
⇒8000 m2
Question 12
A rectangular floor which measures 15 m x 8 m is to be laid with tiles measuring 50 cm x 25 cm find the number of tiles required further, if a carpet is laid on the floor so that a space of 1 m exists between its edges and the edges of the floor, what fraction of the floor is uncovered?
Sol :
Hence ABCD be a rectangular field given measurement of this field is 15 m×8 m
i.e. length of rectangular field=15 m
and breadth of rectangular field=8 m
Area of rectangular field=l×b
=15×8=120 m2
measurement of tiles=50 cm×25 cm
i.e. length of tile=50 cm
$=\frac{50}{100} \mathrm{~m}=\frac{1}{2} \mathrm{~m}$
Breadth of tile=25 cm
$=\frac{25}{100}=\frac{1}{4} \mathrm{~m}$
Area of one tile$=\frac{1}{2} \times \frac{1}{4} \mathrm{~m}^{2}=\frac{1}{8} \mathrm{~m}^{2}$
Then number of required tiles
$=\frac{\text { Area of rectangular field }}{\text { Area of one tile }}$
$=\frac{(120)}{\left(\frac{1}{8}\right)}=\frac{120 \times 8}{1}$
=960 tiles
Now, length of carpet=15m-1m-1m
=15m-2m=13m
Breadth of carpet=8m-1m-1m=8m-2m=6 m
Area of carpet=l×b
=13×6 m2
=78 m2
Area of floor which is uncovered by carpet=Area of floor-Area of carpet
=120-78=42 m2
Fraction $=\frac{\text { Area of floor which is uncovered by carpet }}{\text { Area of floor }}$
$=\frac{42}{120}=\frac{7}{20}$
Question 13
The width of a rectangular room is $\frac{3}{5}$ of its length x metres. If its perimeter isy metres, write an equation connecting.vandy. Find the floor area of the room if its perimeter is 32 m.
Sol :
Given that the length of rectangular room =x m and width of rectangular room $=\frac{3}{5}$ of its length
$=\frac{3}{5} \times x \mathrm{~m}=\frac{3 x}{5} \mathrm{~m}$
Perimeter=2(l+b)
⇒$y=2\left(x+\frac{3}{5} x\right)$
⇒$y=2\left(\frac{5 x+3 x}{5}\right)$
⇒$y=2 \times \frac{8 x}{5}$
⇒$y=\frac{16 x}{5}$
⇒5y=16x
⇒16x=5y....(1)
Which is required relation between x and y
Now, given perimeter=32 m
i.e. the value of y=32 m
Substituting the value of y in equation (1), we get
⇒16x=5×32
⇒$x=\frac{5 \times 32}{16}$
⇒$x=\frac{5 \times 2}{1}$
⇒x=10 m
Breadth(width)
$=\frac{3}{5} \times x \mathrm{~m}=\frac{3}{5} \times 10 \mathrm{~m}$
=3×2 m=6 m
Floor area of the room= l×b=10×6 m2
=60 m2
Question 14
A rectangular garden 10 m by 16 m is to be surrounded by a concrete walk of uniform width. Given that the area of the walk is 120 square metres, assuming the width of the walk to be x, form an equation in x and solve it to find the value of x.
Sol :
Here ABCD be a rectangular garden having length=10 m and breadth=16 m
Then Area of ABCD=l×b=10 m×16 m=160 m2
Width of the walk to be= x m
Then, length of rectangular garden PQRS
=(10-x-x) m
=(10-2x) m
Breadth of rectangular garden PQRS=(16-x-x) m
=(16-2x) m
Question 15
A rectangular room is 6 m long, 4.8 m wide and 3.5 m high. Find the inner surface area of the four walls.
Sol :
Here length of rectangular room=6 m
Breadth of rectangular room $=4.8 \mathrm{~m}$ and height of rectangular room $=3.5 \mathrm{~m}$ Then, inner surface area of four wall:
$=2(\ell+b) \times h=2(6+4.8) \times 3.5 \mathrm{~m}^{2}$
$=2 \times 10.8 \times 3.5 \mathrm{~m}^{2}=21.6 \times 3.5 \mathrm{~m}^{2}=75.6 \mathrm{~m}^{2}$
Question 16
A rectangular plot of land measures 41 metres in length and 22.5 metres in width. A boundary wall 2 metres high is built all around the plot at a distance of 1.5 m from the plot. Find the inner surface area of the boundary wall.
Sol :
Length of rectangular plot=41 metre
Breadth of rectangular plot =22.5 metre
and height of boundary wall =2 metre
But boundary wall is built at a distance of 1.5 m
Then new length =(41+1.5+1.5) m
=(41+3.0) m
=44 m
New Breadth =22.5 m+1.5 m+1.5 m
=22.5 m+3.0 m
=25.5 m
Now, The inner surface area of the boundary wall
$=2(l+b) \times h=2(44+25.5) \times 2 \mathrm{~m}^{2}$
$=2 \times 69.5 \times 2 \mathrm{~m}^{2}=2 \times 139 \mathrm{~m}^{2}=278 \mathrm{~m}^{2}$
Question 17
(a) Find the perimeter and area of the figure
(i) given below in which all corners are right angled.
(b) Find the perimeter and area of the figure
(ii) given below in which all corners are right angles.
(c) Find the area and perimeter of the figure
(iii) given below in which all corners are right angles and all measurement in centimetres.
Sol :
(a) Given that
AB=2 m, BE=4 m, FE=4 m and FG=1.5 m
Now, BD=4 m+1.5 m=5.5 m
Also AC=BD=5.5 m
CG=(4+2) m=6 m
Perimeter of figure (i)
=AC+CG+GF+FE+EB+BA
=5.5 m+6 m+1.5 m+4 m+4 m+2 m=23 m
Area of given figure=Area of ABEDC+Area of FEDG
$=$ Length $\times$ Breadth $+$ Length $\times$ Breadth
$=2 \mathrm{~m} \times 5.5 \mathrm{~m}+4 \mathrm{~m} \times 1.5 \mathrm{~m}$
$=11 \mathrm{~m}^{2}+6.0 \mathrm{~m}^{2}=17 \mathrm{~m}^{2}$
(b) In the figure (ii) $\mathrm{AB}=\mathrm{CD}=3 \mathrm{~m}$
$\mathrm{HI}=\mathrm{AC}=7 \mathrm{~m}, \mathrm{JF}=\mathrm{BE}=5 \mathrm{~m}, \mathrm{GF}=\mathrm{DE}=2 \mathrm{~m}$
$\mathrm{DG}=\mathrm{EF}=8 \mathrm{~m}, \mathrm{GH}=\mathrm{J}=2 \mathrm{~m}$
Also, $\mathrm{CH}=\mathrm{CD}+\mathrm{DG}+\mathrm{GH}=3 \mathrm{~m}+8 \mathrm{~m}+2 \mathrm{~m}=13 \mathrm{~m}$
Perimeter of given figure $=\mathrm{AB}+\mathrm{AC}+\mathrm{CH}+\mathrm{HI}+\mathrm{IJ}+\mathrm{JF}+\mathrm{FE}+\mathrm{BE}$
=3 m+7 m+13 m+7 m+2 m+5 m+8 m+5 m=50 m
Area of given figure=Area of 1st figure+Area of 2nd figure+Area of 3 rd figure
$=7 \mathrm{~m} \times 3 \mathrm{~m}+8 \mathrm{~m} \times 2 \mathrm{~m}+7 \mathrm{~m} \times 2 \mathrm{~m}$
$=21 \mathrm{~m}^{2}+16 \mathrm{~m}^{2}+14 \mathrm{~m}^{2}=51 \mathrm{~m}^{2}$
(c)
Here given and from , it is clear that
$\mathrm{AB}=12 \mathrm{~cm}, \mathrm{AL}=\mathrm{BC}=7 \mathrm{~cm}, \mathrm{JK}=\mathrm{DE}=5 \mathrm{~cm}$
$\mathrm{HI}=\mathrm{GF}=3 \mathrm{~cm}, \mathrm{LK}=\mathrm{HG}=\mathrm{CD}=2 \mathrm{~cm}$
Now, Perimeter of given figure
=AB+BC+CD+DE+EF+FG+GH+HI+IJ+JK+KL+LA
$=12 \mathrm{~cm}+7 \mathrm{~cm}+2 \mathrm{~cm}+5 \mathrm{~cm}+3 \mathrm{~cm}+3 \mathrm{~cm}$
$+2 \mathrm{~cm}+3 \mathrm{~cm}+3 \mathrm{~cm}+5 \mathrm{~cm}+2 \mathrm{~cm}+7 \mathrm{~cm}$
=54 cm
Area of given figure =Area of 1st part+Area of 2nd part+Area of 3rd part+Area of 4th part+Area of 5th part $=7 \mathrm{~cm} \times 2 \mathrm{~cm}+2 \mathrm{~cm} \times 3 \mathrm{~cm}+(2 \mathrm{~cm}+3 \mathrm{~cm}) \times 2 \mathrm{~cm}$
$+2 \mathrm{~cm} \times 3 \mathrm{~cm}+7 \mathrm{~cm} \times 2 \mathrm{~cm}$
$=14 \mathrm{~cm}^{2}+6 \mathrm{~cm}^{2}+10 \mathrm{~cm}^{2}+6 \mathrm{~cm}^{2}+14 \mathrm{~cm}^{2}$
$=50 \mathrm{~cm}^{2}$
Question 18
The length and the breadth of a rectangle are 12 cm and 9 cm respectively. Find the height of a triangle whose base is 9 cm and whose area is one third that of rectangle.
Sol :
The given length of a rectangle $=12 \mathrm{~cm}$ and Breadth of a rectangle $=9 \mathrm{~cm}$ Area of rectangle $=\ell \times b=12 \mathrm{~cm} \times 9 \mathrm{~cm}=108 \mathrm{~cm}^{2}$
By given condition,
Area of $\Delta \mathrm{ABC}=\frac{1}{3} \times$ Area of rectangle
$=\frac{1}{3} \times 108 \mathrm{~cm}^{2}=36 \mathrm{~cm}^{2}$
Let height of $\Delta \mathrm{ABC}=h \mathrm{~cm}$
Now, we know that Area of $\Delta \mathrm{ABC}=\frac{1}{2} \times$ base $\times$ height
$36 \mathrm{~cm}^{2}=\frac{1}{2} \times 9 \mathrm{~cm} \times h \mathrm{~cm}$
$\Rightarrow \quad 36=\frac{1}{2} \times 9 \times h \Rightarrow 36 \times 2=9 \times h$
$\Rightarrow \quad h=\frac{36 \times 2}{0} \Rightarrow h=4 \times 2 \Rightarrow h=8 \mathrm{~cm}$
Hence , height of ΔABC=8 cm
Question 19
The area of a square plot is 484 mV Find the length of its one side and the length of its one diagonal.
Sol :
Let $\mathrm{ABCD}$ be any square plot whose area is $484 \mathrm{~m}^{2}$, then sides of square is $\mathrm{AB}, \mathrm{BC}, \mathrm{CD}$ and $\mathrm{AD}$
Now, Area of square $=$ side $\times$ side
$\Rightarrow \quad 484=(\text { side })^{2} \Rightarrow(\text { side })^{2}=484$
$\Rightarrow$ side $=\sqrt{484} \Rightarrow$ side $=22 \mathrm{~m}$
i.e. $\mathrm{AB}=\mathrm{BC}=22 \mathrm{~m}$
In $\Delta \mathrm{ABC}$ (By Pythagoras theorm)
$\Rightarrow A C^{2}=(22)^{2}+(22)^{2} \Rightarrow A C^{2}=484+484$
$\Rightarrow A C^{2}=968 \Rightarrow \quad A C=\sqrt{968}$
$\Rightarrow A C=\sqrt{484 \times 2} \Rightarrow \quad A C=22 \times \sqrt{2}$
$\Rightarrow A C=22 \times 1.414 \quad(\because \sqrt{2}=1.414)$
$\Rightarrow \quad A C=31 \cdot 11 m$
Hence, length of side $=22 \mathrm{~m}$ and length of diagonal $=31.11 \mathrm{~m}$
Question 20
A square has the perimeter 56 m. Find its area and the length of one diagonal correct upto two decimal places.
Sol :
Here $\mathrm{ABCD}$ is a square let its side $=x \mathrm{~m}$
Then, perimeter of square =4×side
56=4x
In $\triangle \mathrm{ABC}$, By Pythagoras theorem,
$\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}$
$\Rightarrow \quad A C^{2}=(14)^{2}+(14)^{2} \Rightarrow A C^{2}=196+196$
$\Rightarrow A C^{2}=392 \Rightarrow A C=\sqrt{392}$
$\Rightarrow \quad A C=\sqrt{196 \times 2} \Rightarrow \quad A C=14 \sqrt{2}$
$\Rightarrow \quad A C=14 \times 1.41 .4 \quad(\because \sqrt{2}=1.414)$
$\Rightarrow \quad A C=19.80 \mathrm{~m}$
Hence, side of square $=14 \mathrm{~m}$ and diagonal $=19.80 \mathrm{~m}$
Question 21
A wire when bent in the form of an equilateral triangle encloses an area of 36√3 cm2. Find the area enclosed by the same wire when bent to form:
(i) a square, and
(ii) a rectangle whose length is 2 cm more than its width.
Sol :
Area of equilateral triangle=36√3 cm2
Let side of equilateral triangle = x cm
Then , Area $=\frac{\sqrt{3}}{4}(\text { side })^{2}$
⇒$36 \sqrt{3}=\frac{\sqrt{3}}{4} \times(x)^{2}$
⇒$36 \sqrt{3}=\frac{\sqrt{3}}{4} \times x^{2}$
⇒$\frac{\sqrt{3}}{4} x^{2}=36 \sqrt{3}$
⇒$x^{2}=\frac{36 \sqrt{3} \times 4}{\sqrt{3}}$
⇒$x^{2}=36 \times 4$
⇒$x=\sqrt{36 \times 4}$
⇒x=6×2
⇒x=12 cm
Perimeter of equilateral triangle =3×side
=3×12 cm=36 cm
Perimeter of equilateral triangle=Perimeter of square
⇒36=4×side
⇒4×side=36
⇒side $=\frac{36}{4} \mathrm{~cm}$
⇒side=9 cm
i.e. side of square=9 cm
Area of square=side×side=9×9
=81 cm2
(ii) Perimeter of triangle=Perimeter of rectangle...(1)
But, given condition in rectangle, The length is $2 \mathrm{~cm}$ more than its width Let width of rectangle $=x \mathrm{~cm}$ Length of rectangle $=(x+2) \mathrm{cm}$
Then, perimeter of rectangle $=2(\ell+b)$ $=2[(x+2)+x]=2(2 x+2)=4 x+4$
But, from equation (1),
$4 x+4=$ Perimeter of triangle
$\Rightarrow \quad 4 x+4=36 \Rightarrow 4 x=36-4 \Rightarrow 4 x=32$
$\Rightarrow \quad x=\frac{32}{4} \Rightarrow x=8 \mathrm{~cm}$
i.e. Length of rectangle $=8 \mathrm{~cm}+2 \mathrm{~cm}=10 \mathrm{~cm}$ Breadth of rectangle $=8 \mathrm{~cm}$ Area of rectangle = length $\times$ Breadth $=10 \mathrm{~cm} \times 8 \mathrm{~cm}=80 \mathrm{~cm}^{2}$
Question 22
Two adjacent sides of a parallelogram are: 15 cm and 10 cm. If the distance between the longer sides is 8 cm, find the area of the parallelogram. Also find the distance between shorter sides.
Sol :
Here $\mathrm{ABCD}$ is a parallelogram in which longer side $\mathrm{AB}=15 \mathrm{~cm}$ and shorter side $=10 \mathrm{~cm}$
Distance between longer side $=\mathrm{DM}=8 \mathrm{~cm}$ (given)
Let $\mathrm{DN}$ is the distance between the shorter side
Now, Area of parallelogram $\mathrm{ABCD}=$ Base $\times$ height $=\mathrm{AB} \times \mathrm{DM}=15 \mathrm{~cm} \times 8 \mathrm{~cm}=120 \mathrm{~cm}^{2}$
Now, when base is AD
Then, Area of parallelogram $=\mathrm{AD} \times \mathrm{DN}$
$\Rightarrow \quad 120=10 \times \mathrm{DN} \Rightarrow 10 \times \mathrm{DN}=120$
$\Rightarrow \quad \mathrm{DN}=\frac{120}{10} \Rightarrow \mathrm{DN}=12 \mathrm{~cm}$
Question 23
ABCD is a parallelogram with sides AB = 12 cm, BC = 10 cm and diagonal AC = 16 cm. Find the area of the parallelogram. Also find the distance between its shorter sides.
Sol :
Here $A B C D$ be a parallelogram with sides $\mathrm{AB}=12 \mathrm{~cm}, \mathrm{BC}=10 \mathrm{~cm}$ and $\mathrm{AC}=16 \mathrm{~cm}$
Now, for Area of $\Delta \mathrm{ABC}$
$\mathrm{BC}=a=10 \mathrm{~cm}, \mathrm{AC}=b=16 \mathrm{~cm}, \mathrm{AB}=c=12 \mathrm{~cm}$
$s=\frac{a+b+c}{2}=\frac{10+16+12}{2}=\frac{38}{2} \mathrm{~cm}=19 \mathrm{~cm}$
Area of $\Delta \mathrm{ABC}=\sqrt{s(\mathrm{~s}-a)(s-b)(s-c)}$
$=\sqrt{19 \times(19-10)(19-16)(19-12)}$
$=\sqrt{19 \times 9 \times 3 \times 7}=\sqrt{19 \times 3 \times 3 \times 3 \times 7}$
$=3 \sqrt{19 \times 3 \times 7}=3 \sqrt{19 \times 21}=3 \sqrt{399} \mathrm{~cm}^{2}$
Area of parallelogram $=2 \times$ Area of $\triangle \mathrm{ABC}$
$=2 \times 3 \sqrt{399} \mathrm{~cm}^{2}=6 \times \sqrt{399} \mathrm{~cm}^{2}=6 \times 19.96 \mathrm{~cm}^{2}$
$=119.8 \mathrm{~cm}^{2}$
Let DM be the distance between the shorter lines. we take base $=\mathrm{AD}=\mathrm{BC}=10 \mathrm{~cm}$
Area of parallelogram $=\mathrm{AD} \times \mathrm{DM}$
$(\because$ Area $=$ base $\times$ height $)$
$\Rightarrow 119.8=10 \times \mathrm{DM} \Rightarrow 10 \times \mathrm{DM}=119.8$
$\Rightarrow \mathrm{DM}=\frac{119,8}{10} \Rightarrow \mathrm{DM}=11.98 \mathrm{~cm}$
Hence, distance between shorter lines
=11.98 cm
Question 24
Diagonals AC and BD of a parallelogram ABCD intersect at O. Given that AB = 12 cm and perpendicular distance between AB and DC is 6 cm. Calculate the area of the triangle AOD.
Sol :
Here $\mathrm{ABCD}$ be a parallelogram $\mathrm{AC}$ and $\mathrm{BD}$ be its diagonals which is intersect at O
$\mathrm{AB}=12 \mathrm{~cm}$ and $\mathrm{DM}=6 \mathrm{~cm}$
Area of parallelogram ABCD=AB×DM
(∵Base×Height)
=12 cm×6 cm=72 cm2
Area of $\Delta \mathrm{AOD}=\frac{1}{4} \times$ Area of parallelogram
$\mathrm{ABCD}=\frac{1}{4} \times 72 \mathrm{~cm}^{2}=18 \mathrm{~cm}^{2}$
Question 25
ABCD is a parallelogram with side AB = 10 cm. Its diagonals AC and BD are of length 12 cm and 16 cm respectively. Find the area of the parallelogram ABCD.
Sol :
Here $\mathrm{ABCD}$ be a parallelogram,
$A B=10 \mathrm{~cm}, A C=12 \mathrm{~cm}$
$A O=C O=\frac{12 \mathrm{~cm}}{2}=6 \mathrm{~cm}$
BD=16 cm
$\mathrm{BO}=\mathrm{OD}=\frac{16 \mathrm{~cm}}{2}=8 \mathrm{~cm}$
In $\Delta \mathrm{AOB}$
$a=10 \mathrm{~cm}, b=\mathrm{AO}=6 \mathrm{~cm}, c=\mathrm{BO}=8 \mathrm{~cm}$
$s=\frac{a+b+c}{2}=\frac{10 \mathrm{~cm}+6 \mathrm{~cm}+8 \mathrm{~cm}}{2}=\frac{24 \mathrm{~cm}}{2}=12 \mathrm{~cm}$
Area of $\triangle \mathrm{AOB}=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{12 \times(12-10)(12-6)(12-8)} \mathrm{cm}^{2}$
$=\sqrt{12 \times 2 \times 6 \times 4}$
$=\sqrt{12 \times 12 \times 4} \mathrm{~cm}^{2}=12 \times 2 \mathrm{~cm}^{2}=24 \mathrm{~cm}^{2}$
Area of parallelogram $\mathrm{ABCD}=4 \times$
Area of $\Delta \mathrm{AOB}$. $=4 \times 24 \mathrm{~cm}^{2}=96 \mathrm{~cm}^{2}$
Question 26
The area of a parallelogram is p cm2 and its height is q cm. A second parallelogram has equal area but its base is ‘r’ cm more than that of the first. Obtain an expression in terms of p, q and r for the height h of the second parallelogram.
Sol :
Given area of first parallelogram $=p \mathrm{~cm}^{2}$
height of first parallelogram $=q \mathrm{~cm}$
Then, Area of parallelogram $=$ Base $\times$ height
$\Rightarrow p=$ Base $\times q \Rightarrow$ Base $=\frac{p}{q}$
Now, Base of second parallelogram
$=\left(\frac{p}{q}+r\right)$ or $=\frac{p+q r}{q} \mathrm{~cm}$
Also, Area of second parallelogram = Area of first parallelogram.
Area of the second parallelogram $=p \mathrm{~cm}^{2}$
$\Rightarrow \quad$ Base $\times$ height $=p \mathrm{~cm}^{2} $
$\Rightarrow\left(\frac{\mathrm{P}+q r}{q}\right) \times h=\mathrm{P}$
$\Rightarrow h=\frac{\mathrm{P} \times q}{(\mathrm{P}+q r)} \mathrm{cm}$
$ \Rightarrow h=\frac{\mathrm{P} q}{\mathrm{P}+q r} \mathrm{~cm}$
Hence, Height of second parallelogram
$=\frac{p q}{P+q r} \mathrm{~cm}$
Question 27
What is the area of a rhombus whose diagonals are 12 cm and 16 cm ?
Sol :
Here $\mathrm{ABCD}$ be a rhombus
$\mathrm{BD}=12 \mathrm{~cm}$ and $\mathrm{AC}=16 \mathrm{~cm}$ are diagonals
Then Area of rhombus $\mathrm{ABCD}$
$=\frac{1}{2} \times \mathrm{AC} \times \mathrm{BD}$
$=\frac{1}{2} \times 16 \mathrm{~cm} \times 12 \mathrm{~cm}$
$=8 \mathrm{~cm} \times 12 \mathrm{~cm}=96 \mathrm{~cm}^{2}$
Question 28
The area of a rhombus is 98 cm². If one of its diagonal is 14 cm, what is the length of the other diagonal?
Sol :
Area of rhombus $=98 \mathrm{~cm}^{2}$
one diagonal $=14 \mathrm{~cm}$
We know that,
Area of rhombus $=\frac{1}{2} \times$ product of diagonals
$\Rightarrow \quad 98=\frac{1}{2} \times$ one diagonal $\times$ other diagonal.
$\Rightarrow \quad 98=\frac{1}{2} \times 14 \times$ diagonal
$\Rightarrow \quad$ other diagonal $=\frac{98 \times 2}{14} \mathrm{~cm}$
$\Rightarrow$ other diagonal $=7 \times 2 \mathrm{~cm}=14 \mathrm{~cm}$
Hence, other diagonal $=14 \mathrm{~cm}$
Question 29
The perimeter of a rhombus is 45 cm. If its height is 8 cm, calculate its area.
Sol :
Here $\mathrm{ABCD}$ be a rhombus
Let each side $=x \mathrm{~cm}$
Given perimeter $=45 \mathrm{~cm}$ i.e. $\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{AD}=45 \mathrm{~cm}$
$\Rightarrow \quad x+x+x+x=45 \Rightarrow 4 x=45$
$\Rightarrow \quad x=\frac{45}{4} \mathrm{~cm}$
Given height $=8 \mathrm{~cm}$
Then area of rhombus $=$ base $\times$ height
$=\frac{45}{4} \times 8 \mathrm{~cm}^{2}=45 \times 2 \mathrm{~cm}^{2}=90 \mathrm{~cm}^{2}$
Note : For base take here any side of rhombus.
Question 30
PQRS is a rhombus. If it is given that PQ = 3 cm and the height of the rhombus is 2.5 cm, calculate its area.
Sol :
Here given that PQRS is a rhombus
$\mathrm{PQ}=3 \mathrm{~cm}$ (given) and height $=2.5 \mathrm{~cm}$
Here $\mathrm{PQ}$ is base of rhombus $\mathrm{PQRS}$.
Also $\mathrm{SM}=2.5 \mathrm{~cm}$
i.e. height of rhombus
Area of rhombus $\mathrm{PQRS}=$ base $\times$ height
$=3 \mathrm{~cm} \times 2.5 \mathrm{~cm}=7.5 \mathrm{~cm}^{2}$
Question 31
If the diagonals of a rhombus are 8 cm and 6 cm, find its perimeter.
Sol :
Let $\mathrm{ABCD}$ be any rhombus
$\mathrm{AC}$ and $\mathrm{BD}$ are two diagonals.
Then, $A C=8 \mathrm{~cm}$ and $B D=6 \mathrm{~cm}$
Here, $A O=4 \mathrm{~cm} \& B O=3 \mathrm{~cm}$
In $\Delta \mathrm{ABC}$
By Pythagoras theorem,
$\Rightarrow \quad A B^{2}=A O^{2}+B O^{2}$
$\Rightarrow A B^{2}=(4)^{2}+(3)^{2}$
$\Rightarrow A B^{2}=16+9$
$\Rightarrow A B^{2}=25$
$\Rightarrow A B=\sqrt{25} \Rightarrow \quad A B=5 \mathrm{~cm}$
i.e. side of rhombus $\mathrm{ABCD}=5 \mathrm{~cm}$
Perimeter of rhombus $=4 \times$ side $=4 \times 5 \mathrm{~cm}=20 \mathrm{~cm}$
Question 32
If the sides of a rhombus are 5 cm each and one diagonal is 8 cm, calculate
(i) the length of the other diagonal, and
(ii) the area of the rhombus.
Sol :
Here ABCD be a rhombus
$\mathrm{AB}, \mathrm{BC}, \mathrm{CD}$ and $\mathrm{AD}$ are the sides of rhombus Then, $A B=B C=C D=A D=5 \mathrm{~cm}$
Also, $A C=8 \mathrm{~cm}$
Then, $\mathrm{AO}=4 \mathrm{~cm}$
In $\Delta \mathrm{AOB}$,
By Pythagoras theorem, $\mathrm{AB}^{2}=\mathrm{AO}^{2}+\mathrm{BO}^{2}$
$\Rightarrow(5)^{2}=(4)^{2}+\mathrm{BO}^{2} \Rightarrow 25=16+\mathrm{BO}^{2}$
$\Rightarrow \mathrm{BO}^{2}=25-16 \Rightarrow \mathrm{BO}^{2}=9$
$\Rightarrow \mathrm{BO}=\sqrt{9}=3 \mathrm{~cm}$
$\because \quad \mathrm{BD}=2 \times \mathrm{BO}=2 \times 3 \mathrm{~cm}=6 \mathrm{~cm}$
Hence, length of other diagonal $=6 \mathrm{~cm}$
Area of rhombus $=\frac{1}{2} \times$ Product of diagonals
$=\frac{1}{2} \times 8 \mathrm{~cm} \times 6 \mathrm{~cm}$
$=4 \mathrm{~cm} \times 6 \mathrm{~cm}=24 \mathrm{~cm}^{2}$
Question 33
(a) The diagram (t) given below is a trapezium. Find the length of BC and the area of the trapezium Assume AB = 5 cm, AD = 4 cm, CD = 8 cm
(b) The diagram (ii) given below is a trapezium Find (i) AB (ii) area of trapezium ABCD.
(c) The cross-section of a canal is shown in figure (iii) given below. If the canal is 8 m wide at the top and 6 m wide at the bottom and the area of the cross-section is 16.8 m², calculate its depth
Sol :
(a) Here ABCD is a trapezium
$\mathrm{AB}=5 \mathrm{~cm}, \mathrm{AD}=4 \mathrm{~cm}$ and $\mathrm{CD}=8 \mathrm{~cm}$
Also Draw BN $\perp$ CD
Then, $\mathrm{BN}=4 \mathrm{~cm}$
$\mathrm{CN}=\mathrm{CD}-\mathrm{ND}$
$\mathrm{CN}=\mathrm{CD}-\mathrm{AO}$
$\mathrm{CN}=8 \mathrm{~cm}-5 \mathrm{~cm}=3 \mathrm{~cm}$
Now, in $\Delta \mathrm{BCN}$
By Pythagoras theorem,
$\mathrm{BC}^{2}=\mathrm{BN}^{2}+\mathrm{CN}^{2}$
$\Rightarrow \quad \mathrm{BC}^{2}=(4)^{2}+(3)^{2}$
$\Rightarrow \mathrm{BC}^{2}=16+9$
$\Rightarrow \quad B C^{2}=25 \Rightarrow B C=\sqrt{25} \Rightarrow B C=5 \mathrm{~cm}$
Hence , Length of BC=5 cm
Area of trapezium $=\frac{1}{2}$ (sum of parallel sides) $\times$ height
$=\frac{1}{2}(\mathrm{AB}+C D) \cdot A \mathrm{D}=\frac{1}{2}(5+8) \times 4 \mathrm{~cm}^{2}$
$=\frac{1}{2} \times 13 \times 4 \mathrm{~cm}^{2}=13 \times 2 \mathrm{~cm}^{2}=26 \mathrm{~cm}^{2}$
Hence, Area of trapezium=26
(b) In diagram (ii) Given that
$A D=8$ unit, $B C=2$ unit, $C D=10$ unit
Draw CN $\perp \mathrm{AD}$
Then, $\mathrm{AN}=2 \mathrm{units}$
$\mathrm{DN}=\mathrm{AD}-\mathrm{DN}$
$=8$ units $-6$ units $=2$ units
In $\Delta \mathrm{CDN}$
By Pythagoras theorem
$\mathrm{CD}^{2}=\mathrm{DN}^{2}+\mathrm{NC}^{2}$
$\Rightarrow \quad(10)^{2}=(6)^{2}+\mathrm{NC}^{2}$
$\Rightarrow \quad \mathrm{NC}^{2}=(10)^{2}-(6)^{2} \Rightarrow \mathrm{NC}^{2}=100-36$
$\Rightarrow \mathrm{NC}^{2}=64 \Rightarrow \mathrm{NC}=\sqrt{64}$
$\Rightarrow \quad \mathrm{NC}=8$ units
Also from figure $\mathrm{NC}=\mathrm{AB}=8$ units
Area of trapezium $=\frac{1}{2}$ (sum of parallel sides) $\times$ height
$=\frac{1}{2}(\mathrm{BC}+\mathrm{AD}) \times \mathrm{AB}=\frac{1}{2}(2+8) \times 8$ sq. units
$=\frac{1}{2} \times 10 \times 8 \mathrm{sq} . \mathrm{units}=5 \times 8 \mathrm{sq} .$ units
$=40 \mathrm{sq} . \mathrm{units}$
(c) Let $\mathrm{ABCD}$ be the cross section of canal in the shape of trapezium. $\mathrm{AB}=6 \mathrm{~m}, \mathrm{DC}=8 \mathrm{~m}$
Let AL be the depth of canal
Area of cross-section
$=16.8 \mathrm{~m}^{2}$
$\frac{1}{2} \times($ sum of parallel sides $) \times$ depth $=72$
$\Rightarrow \quad \frac{1}{2} \times(\mathrm{AB}+\mathrm{DC}) \times \mathrm{AL}=72$
$\Rightarrow \frac{1}{2} \times(6+8) \times \mathrm{AL}=16.8$
$\Rightarrow \frac{1}{2} \times 14 \times \mathrm{AL}=16.8 \Rightarrow \mathrm{AL}=\frac{16.8 \times 2}{14} \mathrm{~m}$
$\Rightarrow \quad \mathrm{AL}=\frac{16.8 \times 1}{7} \mathrm{~m} \Rightarrow \mathrm{AL}=2.4 \mathrm{~m}$
Question 34
The distance between parallel sides of a trapezium is 12 cm and the distance between mid-points of other sides is 18 cm. Find the area of the trapezium.
Sol :
Let $\mathrm{ABCD}$ be the triapezium in which $\mathrm{AB} \| \mathrm{DC}$.
Height $C L=12 \mathrm{~cm}$
Let $E$ and $F$ be the mid-points of sides AD and BC respectivity, then $\mathrm{EF}=18 \mathrm{~cm}$
$\because \mathrm{E}$ and $\mathrm{F}$ are the mid-points of sides $\mathrm{AD}$ and $\mathrm{BC}$
$\therefore E F=\frac{1}{2}(A B+D C)=18 \mathrm{~cm}$
Now area of trap. $\mathrm{ABCD}$
$=\frac{1}{2}(\mathrm{AB}+\mathrm{DC}) \times$ height $=18 \mathrm{~cm} \times 12 \mathrm{~cm}$
$=216 \mathrm{~cm}^{2}$
Question 35
The area of a trapezium is 540 cm². If the ratio of parallel sides is 7 : 5 and the distance between them is 18 cm, find the length of parallel sides.
Sol :
Area of trapezium $=540 \mathrm{~cm}^{2}$
ratio of parallel sides $=7: 5$
Let one parallel side $=7 x \mathrm{~cm}$
Then other parallel side $=5 x \mathrm{~cm}$ distance between parallel sides $=18 \mathrm{~cm}$
i.e. height $=18 \mathrm{~cm}$
Area of trapezium $=\frac{1}{2} \times($ sum of parallel sides $) \times$ height
$\Rightarrow \quad 540=\frac{1}{2} \times(7 x+5 x) \times 18$
$\Rightarrow \quad 540=\frac{1}{2} \times 12 x \times 18 $
$\Rightarrow 540=6 x \times 18$
$\Rightarrow \quad 540=108 x \Rightarrow 108 x=540 \Rightarrow x=\frac{540}{108}=5$
First parallel side $=7 x=7 \times 5=35 \mathrm{~cm}$
and second parallel side $=5 x=5 \times 5=25 \mathrm{~cm}$
Question 36
The parallel sides of an isosceles trapezium are in the ratio 2 : 3. If its height is 4 cm and area is 60 cm2, find the perimeter.
Sol :
Here ABCD is an isosceles trapezium
Where BC=AD
Height=4 cm
Let CD=2x
Then , AB=3x
Area of trapezium $=\frac{1}{2}$ (sum of parallel sides) $\times$ height
$\Rightarrow \quad 60=\frac{1}{2} \times(2 x+3 x) \times 4 \Rightarrow 60=\frac{1}{2} \times 5 x \times 4$
$\Rightarrow \quad 60=5 x \times 2 \Rightarrow 60=10 x \Rightarrow 10 x=60$
$\Rightarrow \quad x=\frac{60}{10} \Rightarrow x=6$
i.e. $\mathrm{CD}=2 x=2 \times 6 \mathrm{~cm}=12 \mathrm{~cm}$
$\mathrm{AB}=3 x=3 \times 6 \mathrm{~cm}=18 \mathrm{~cm}$
Now, $\mathrm{AN}=\mathrm{BM}$
Also $A N=A B-B N$
$\Rightarrow \quad A N=A B-(M N+B M)$
$\Rightarrow \quad A N=A B-(C D+B M) \quad \quad(\because M N=C D)$
$\Rightarrow \quad A N=A B-(C D+A N) \quad(\because B M=A N)$
$\Rightarrow \quad A N=18-(12+A N)$
$\Rightarrow \mathrm{AN}=18-12-\mathrm{AN} \Rightarrow \mathrm{AN}+\mathrm{AN}=6$
$\Rightarrow \quad 2 \mathrm{AN}=6 \Rightarrow \mathrm{AN}=\frac{6}{2} \Rightarrow \mathrm{AN}=3$
In $\Delta \mathrm{AND}$
By Pythagoras theorem,
$\mathrm{AD}^{2}=\mathrm{DN}^{2}+\mathrm{AN}^{2}$
$\Rightarrow \mathrm{AD}^{2}=(4)^{2}+(3)^{2}$
$\Rightarrow \mathrm{AD}^{2}=16+9 \Rightarrow \mathrm{AD}^{2}=25$
$\Rightarrow \mathrm{AD}=\sqrt{25} \Rightarrow \mathrm{AD}=5 \mathrm{~cm}$
then $\mathrm{AD}=\mathrm{BC}=5 \mathrm{~cm}$
Perimeter of trapezium $=\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{AD}$
$=18 \mathrm{~cm}+5 \mathrm{~cm}+12 \mathrm{~cm}+5 \mathrm{~cm}=40 \mathrm{~cm}$
Question 37
The area of a parallelogram is 98 cm². If one altitude is half the corresponding base, determine the base and the altitude of the parallelogram.
Sol :
The given area of a parallelogram $=98 \mathrm{~cm}^{2}$
given condition that one altitude is half the corresponding base
Let base $=x \mathrm{~cm}$
Then corresponding altitude $=\frac{x}{2} \mathrm{~cm}$
Area of parallelogram $=$ Base $\times$ Altitude
(where Base is correspoonding base)
$\Rightarrow \quad 98=x \mathrm{~cm} \times \frac{x}{2} \mathrm{~cm} \Rightarrow 98=\frac{x^{2}}{2}$
$\Rightarrow \quad 98 \times 2=x^{2} \Rightarrow x^{2}=196 \Rightarrow x=\sqrt{196}$
$\Rightarrow \quad x=14 \mathrm{~cm}$
i.e. Base $=14 \mathrm{~cm}$
and Altitude $=\frac{14 \mathrm{~cm}}{2}=7 \mathrm{~cm}$
Question 38
The length of a rectangular garden is 12m more than its breadth. The numerical value of its area is equal to 4 times the numerical value of its perimeter. Find the dimensions of the garden
Sol :
Let the breadth of rectangular garden $=x \mathrm{~m}$
Then length of rectangular garden $=(x+12) \mathrm{m}$ Area $=\ell \times b=(x+12) \times x \mathrm{~m}^{2}=\left(x^{2}+12 x\right) \mathrm{m}^{2}$
Perimeter $=2(\ell+b)=2[(x+12)+x] \mathrm{m}$
$=2[x+12+x] \mathrm{m}=2(2 x+12) \mathrm{m}=4 x+24 \mathrm{~m}$
According to question,
Numerical value of Area $=4 \times$ numerical value of perimeter
$\Rightarrow \quad x^{2}+12 x=4 \times(4 x+24)$
$\Rightarrow \quad x^{2}+12 x=16 x+96$
$\Rightarrow \quad x^{2}+12 x-16 x-96=0 \Rightarrow x^{2}-4 x-96=0$
$\Rightarrow \quad x^{2}-12 x+8 x-96=0$
$\Rightarrow \quad x(x-12)+8(x-12)=0$
$\Rightarrow \quad(x+8)(x-12)=0$
Either $x+8=0$ or $x-12=0$
$x=-8$ (not possible) or $x=12$ Hence, Breadth of rectangular garden $=12 \mathrm{~m}$ Length of rectangular garden $=12 \mathrm{~m}+12 \mathrm{~m}=24 \mathrm{~m}$
Question 39
If the perimeter of a rectangular plot is 68 m and length of its diagonal is 26 m, find its area.
Sol :
Given that perimeter of a rectangular plot $=68 \mathrm{~m}$
and length of its diagonal $=26 \mathrm{~m}$ Here, $A B C D$ be the rectangular plot let length of rectangular plot $=x \mathrm{~m}$
and Breadth of rectangular plot $=y \mathrm{~m}$
Then, Perimeter $=2$ (Length + Breadth)
$\begin{array}{ll}\Rightarrow & 68=2(x+y) \Rightarrow \frac{68}{2}=x+y \\ \Rightarrow & 34=x+y \Rightarrow x+y=34 \\ \Rightarrow & x=(34-y)\end{array}$...(1)
Also, in $\Delta \mathrm{ABC}$
By Pythagoras theorem, $A C^{2}=A B^{2}+B C^{2} \Rightarrow(26)^{2}=x^{2}+y^{2}$
$(\because \mathrm{AC}=$ diagonal of rectangular plot)
$\Rightarrow \quad x^{2}+y^{2}=676$
Substituting the value of $x$ from (1), we get
$\Rightarrow(34-y)^{2}+y^{2}=676 \Rightarrow 1156+y^{2}-68 y+y^{2}=676$
$\Rightarrow \quad 2 y^{2}-68 y+1156-676=0$
$\Rightarrow \quad 2 y^{2}-68 y-480=0 \Rightarrow 2\left(y^{2}-34 y-240\right)=0$
$\Rightarrow y^{2}-34 y-240=0 \Rightarrow y^{2}-24 y-10 y-240=0$
$\Rightarrow \quad y(y-24)-10(y-24)=0$
$\Rightarrow \quad(y-10)(y-24)=0$
Either, $\quad y-10=0 \quad$ or $\quad y-24=0$
$y=10 \mathrm{~m} \quad$ or $y=24 \mathrm{~m}$
Substituting the value of $y$ in equation (1), we get when $y=10 \mathrm{~m}, x=(34-10) \mathrm{m}=24 \mathrm{~m}$
Either $y=24 \mathrm{~m}$
$x=(34-24) \mathrm{m}=10 \mathrm{~m}$
the required Area in both the cases $=x y$ $=24 \mathrm{~m} \times 10 \mathrm{~m}$ or $10 \mathrm{~m} \times 24 \mathrm{~m}=240 \mathrm{~m}$ or $240 \mathrm{~m}$
Hence, area of the rectangular block =240 m
Question 40
A rectangle has twice the area of a square. The length of the rectangle is 12 cm greater and the width is 8 cm greater than 2 side of a square. Find the perimeter of the square.
Sol :
Let the side of a square $=x \mathrm{~cm}$
Then length of rectangle $=(x+12) \mathrm{cm}$ Breadth of rectangle $=(x+8) \mathrm{cm}$
Area of square $=$ side $\times$ side $=x \mathrm{~cm} \times x \mathrm{~cm}=x^{2} \mathrm{~cm}^{2}$
Area of rectangle $=$ length $\times$ Breadth $=(x+12) \mathrm{cm} \times(x+8) \mathrm{cm}=(x+12)(x+8) \mathrm{cm}^{2}$
According to question Area of rectangle $=2 \times$ Area of square
$\Rightarrow \quad(x+12)(x+8)=2 \times x^{2}$
$\Rightarrow \quad x(x+8)+12(x+8)=2 x^{2}$
$\Rightarrow \quad x^{2}+8 x+12 x+96=2 x^{2}$
$\Rightarrow \quad x^{2}-2 x^{2}+8 x+12 x+96=0$
$\Rightarrow \quad-x^{2}+20 x+96=0 \Rightarrow-\left(x^{2}-20 x-96\right)=0$
$\Rightarrow \quad x^{2}-20 x-96=0 \Rightarrow x^{2}-24 x+4 x-96=0$
$\Rightarrow \quad x(x-24)+4(x-24)=0$
$\Rightarrow \quad(x+4)(x-24)=0$
Either $(x+4)=0 \quad$ or $\quad x-24=0$
$\therefore$ side of square $=24 \mathrm{~cm}$
Perimeter of square $=4 \times$ side
$=4 \times 24 \mathrm{~cm}=96 \mathrm{~cm}$
Question 41
The perimeter of a square is 48 cm. The area of a rectangle is 4 cm2 less than the area of the square. If the length of the rectangle is 4 cm greater than its breadth, find the perimeter of the rectangle.
Sol :
Perimeter of square $=48 \mathrm{~cm}$
$\therefore$ Side $=\frac{\text { Perimeter }}{4}=\frac{48}{4}=12 \mathrm{~cm}$
Area $=(\text { side })^{2}=(12)^{2}=144 \mathrm{~cm}^{2}$
$\therefore$ Area of rectangle $=144-4=140 \mathrm{~cm}^{2}$
Let breadth of rectangle $=x \mathrm{~cm}$
Then length $=x+4 \mathrm{~cm}$
$\therefore$ Area $=(x+4) \times x \mathrm{~cm}^{2}$
$\therefore(x+4) x=140 \Rightarrow x^{2}+4 x-140=0$
$\Rightarrow x^{2}+14 x-10 x-140=0$
$\Rightarrow x(x+14)-10(x+14)=0 \Rightarrow(x+14)(x-10)=0$
Either $x+14=0$, then $x=-14$
or $x-10=0$, then $x=10$
$\therefore$ Breadth $=10 \mathrm{~cm}$
Then length $=10+4=14 \mathrm{~cm}$
$\therefore$ Perimeter $=2(l+b)=2(14+10)=2 \times 24 \mathrm{~cm}=48 \mathrm{~cm}$
Question 42
In the adjoining figure, ABCD is a rectangle with sides AB = 10 cm and BC = 8 cm. HAD and BFC are equilateral triangles; AEB and DCG are right angled isosceles triangles. Find the area of the shaded region and the perimeter of the figure.
Sol :
$\mathrm{ABCD}$ is a rectangle and $\mathrm{AB}=10 \mathrm{~cm}$
$\mathrm{BC}=8 \mathrm{~cm}$
$\Delta \mathrm{HAD}$ and $\Delta \mathrm{BFC}$ and equilateral triangles whose each side is $8 \mathrm{~cm}$.
$\Delta \mathrm{AEB}$ and $\Delta \mathrm{DCG}$ are right angled isosceles triangles whose each hypotenues $=10 \mathrm{~cm}$.
Let $A E=E B=x c m$
Now in $\Delta \mathrm{ABE}, \mathrm{AE}^{2}+\mathrm{EB}^{2}=\mathrm{AB}^{2}$
$\Rightarrow x^{2}+x^{2}(10)^{2} \quad \Rightarrow 2 x^{2}=100$
$\Rightarrow x^{2}=\frac{100}{2}=50$
$\therefore \quad x=\sqrt{50}=\sqrt{25 \times 2}=5 \sqrt{2} \mathrm{~cm}$
Now area of $\Delta \mathrm{AEB}=\Delta \mathrm{GCD}$
$=\frac{1}{2} x \times x=\frac{1}{2} x^{2} \mathrm{~cm}^{2}=\frac{1}{2} \times 50=25 \mathrm{~cm}^{2}$
and area of $\Delta \mathrm{HAD}=$ area of $\mathrm{BFC}$
$=\frac{\sqrt{3}}{4} \times(8)^{2} \mathrm{~cm}^{2}=\frac{\sqrt{3}}{4} \times 64=16 \sqrt{3} \mathrm{~cm}^{2}$
Area of shaded portion $=$ Area of rect. $\mathrm{ABCD}$ $+2$ area of $\Delta \mathrm{AEB}+2$ area of $\Delta \mathrm{BFC}$
$=(10 \times 8+2 \times 25+2 \times 16 \sqrt{3}) \mathrm{cm}^{2}$
$=(80+50+32 \sqrt{3}) \mathrm{cm}^{2}=(130+32 \sqrt{3}) \mathrm{cm}^{2}$
Perimeter of the figure $=\mathrm{AE}+\mathrm{EB}+\mathrm{BF}$
$+\mathrm{FC}+\mathrm{CD}+\mathrm{GD}+\mathrm{DH}+\mathrm{HA}$
$=4 \mathrm{AE}+4 \mathrm{BF}=(4 \times 5 \sqrt{2}+4 \times 8) \mathrm{cm}$
$=(20 \sqrt{2}+32) \mathrm{cm} .=(32+20 \sqrt{2}) \mathrm{cm}$
Question 43
(a) Find the area enclosed by the figure (i) given below, where ABC is an equilateral triangle and DGFG is an isosceles trapezium.
All measurements are in centmetces.
(b) Find the area enclosed by the figure (ii) given below. AH measurements are in centimetres.
(c) In the figure (iii) given below, from a 24. cm x 24 cm piece of cardboard, a block in the shape of letter M is cut off. Find the area of the cardboard left over, all measurements are in centimetres.
Sol :
(a) $\triangle \mathrm{ABC}$ is an equilateral triangle and $\mathrm{DEFG}$ is an isosceles trapezium in which
$\mathrm{EF}=\mathrm{GD}=5 \mathrm{~cm}$
$\mathrm{DE}=6 \mathrm{~cm}$
and $\mathrm{GF}=\mathrm{GB}+\mathrm{BC}+\mathrm{CF}=3+6+3=12 \mathrm{~cm}$
AB=AC=BC=6 cm
Join $\mathrm{BD}$ and $\mathrm{CE}$
In right $\Delta \mathrm{CEF}, \mathrm{CE}^{2}=\mathrm{EF}^{2}-\mathrm{CF}^{2}$
$=5^{2}-3^{2}=25-9=16$
$\therefore \mathrm{CE}=\sqrt{16}=4 \mathrm{~cm}$
Now area of $\triangle \mathrm{ABC}$
$=\frac{\sqrt{3}}{4} \times(6)^{2}=36 \times \frac{\sqrt{3}}{4} \mathrm{~cm}^{2}=9 \sqrt{3} \mathrm{~cm}^{2}$
and area of trap. DEFG
$=\frac{1}{2}(D E+G F) \times C E=\frac{1}{2}(6+12) \times 4 \mathrm{~cm}^{2}$
$=\frac{1}{2} \times 18 \times 4=36 \mathrm{~cm}^{2}$
$\therefore$ Area of the figure $=(9 \sqrt{3}+36) \mathrm{cm}^{2}$
$=9 \times 1 \cdot 732+36=15 \cdot 59+36 \mathrm{~cm}^{2}$
$=51 \cdot 59 \mathrm{~cm}^{2}$
(b) Length of rectangle $=2+2+2+2=8 \mathrm{~cm}$
and $\quad$ width $=2 \mathrm{~cm}$.
$\therefore$ Area $=l+b=8 \times 2=16 \mathrm{~cm}^{2}$
Area of each trap $=\frac{1}{2}(2+2) \times(6-2)$
$=\frac{1}{2} \times 4 \times 4=8 \mathrm{~cm}^{2}$
$\therefore$ Total area $=$ area of react.+ area of 2 tapezium.
$=16+8+8=32 \mathrm{~cm}^{2}$
(c) Length of each rectangle $=24 \mathrm{~cm}$
width=6 cm
$\therefore$ Area of each rectangle $=l \times b=24 \times 6$
$=144 \mathrm{~cm}^{2}$
Base of each parallelogram $=8 \mathrm{~cm}$.
and $\quad$ height $=6 \mathrm{~cm}$.
$\therefore$ Area of each parallelogram $=8 \times 6$
$=48 \mathrm{~cm}^{2}$
Now area of the $M$ -shaped figure $=2 \times 144$$+2 \times 48=\mathrm{cm}^{2}$
$=288+96=384 \mathrm{~cm}^{2}$
and area of the square cardboard $=24 \times 24$
$=576 \mathrm{~cm}^{2}$
$\therefore$ Area of the removing cardboard
$=576-384=192 \mathrm{~cm}^{2}$
Question 44
(a) The figure (i) given below shows the cross-section of the concrete structure with the measurements as given. Calculate the area of cross-section.
(b) The figure (ii) given below shows a field with the measurements given in metres. Find the area of the field.
(c) Calculate the area of the pentagon ABCDE shown in fig. (iii) below, given that AX = BX = 6 cm, EY = CY = 4 cm, DE = DC = 5cm,DX = 9cmand DX is perpendicular to EC and AB.
Sol :
(a) In figure (i)
AB=1.8 m, CD=0.6 m , DE=1.2 m
EF=0.3 m , AF=2.4 m
Produce $D E$ to meet $A B$ in $G$ then
$\angle \mathrm{FEG}=\angle \mathrm{GAF} 90^{\circ}$
$\therefore$ AGEF is a rectangle Area of given figure
$=$ Area of rectangle $\mathrm{AGEF}$
+ Area of trapezium GBCD
$=\ell \times b+\frac{1}{2}($ sum of parallel sides $\times$ height $)$
$=\mathrm{AF} \times \mathrm{AG}+\frac{1}{2}(\mathrm{~GB}+\mathrm{CD}) \times \mathrm{DG}$
$=2.4 \mathrm{~m} \times 0.3 \mathrm{~m}+\frac{1}{2}$
$[(\mathrm{AB}-\mathrm{AG})+\mathrm{CD}] \times(\mathrm{DE}+\mathrm{EG})$
$(\because A G=F E$ and using $E G=A F)$
$=0.72 \mathrm{~m}^{2}+\frac{1}{2}[(1.8 \mathrm{~m}-0.3 \mathrm{~m})+0.6 \mathrm{~m}] \times(1.2 \mathrm{~m}+2.4 \mathrm{~m})$
$=0.72 \mathrm{~m}^{2}+\frac{1}{2}[1.5 \mathrm{~m}+0.6 \mathrm{~m}] \times 3.6 \mathrm{~m}$
$=0.72 \mathrm{~m}^{2}+\frac{1}{2} \times 2.1 \mathrm{~m} \times 3.6 \mathrm{~m}$
$=0.72 \mathrm{~m}^{2}+2.1 \mathrm{~m} \times 1.8 \mathrm{~m}$
$=0.72 \mathrm{~m}^{2}+3.78 \mathrm{~m}^{2}=4.50 \mathrm{~m}^{2}=4.5 \mathrm{~m}^{2}$
(b) ABCD is a pentagonal field in which
$A X=12 m, B X=30 m, X Z=15 m, C Z=25 m$
$D Z=10 m, A D=12 m+15 m+10 m=37 m$
$E Y=20 m$
Area of pentagonal field $\mathrm{ABCDE}=$ Area of $\Delta \mathrm{ABX}$$+$ Area of trapezium $\mathrm{BCZX}+$ Area of $\Delta \mathrm{CDZ}+$ Area of $\Delta \mathrm{AED}$
$=\frac{1}{2} \times$ base $\times$ height $+\frac{1}{2}$ (sum of parallel sides)
$\times$ height $+\frac{1}{2} \times$ base $\times$ height $+\frac{1}{2} \times$ base $\times$ height
$=\frac{1}{2} \times \mathrm{B} \mathrm{X} \times \mathrm{AX}+\frac{1}{2}(\mathrm{BX}+\mathrm{CZ}) \times \mathrm{XZ}+\frac{1}{2} \times \mathrm{CZ} \times$
$\mathrm{DZ}+\frac{1}{2} \times \mathrm{AD} \times \mathrm{EY}$
$=\frac{1}{2} \times 30 \mathrm{~m} \times 12 \mathrm{~m}+\frac{1}{2}(30 \mathrm{~m}+25 \mathrm{~m}) \times 15 \mathrm{~m}+\frac{1}{2}$
$\times 25 \mathrm{~m} \times 10 \mathrm{~m}+\frac{1}{2} \times 37 \mathrm{~m} \times 20 \mathrm{~m}$
$=15 \mathrm{~m} \times 12 \mathrm{~m}+7.5 \mathrm{~m} \times 55 \mathrm{~m}+25 \mathrm{~m} \times 5 \mathrm{~m}+37 \mathrm{~m} \times 10 \mathrm{~m}$
$=180 \mathrm{~m}^{2}+412.5 \mathrm{~m}^{2}+125 \mathrm{~m}^{2}+370 \mathrm{~m}^{2}$
$=1087.5 \mathrm{~m}^{2}$
(c) Here ABCDE is the pentagon
Given that
AX=BX=6 cm, EY=CY=4 cm
$\mathrm{DE}=\mathrm{DC}=5 \mathrm{~cm}, \mathrm{DX}=9 \mathrm{~cm}$
And $\mathrm{DX} \perp$ to $\mathrm{EC}$ and $\mathrm{AB}$
In $\Delta \mathrm{DEY}$
By Pythagoras theorem,
$\mathrm{DE}^{2}=\mathrm{DY}^{2}+\mathrm{EY}^{2} \Rightarrow(5)^{2}=\mathrm{DY}^{2}+(4)^{2}$
$\Rightarrow \quad 25=\mathrm{DY}^{2}+16 \Rightarrow \mathrm{DY}^{2}=25-16=9$
$\Rightarrow \quad \mathrm{DY}=\sqrt{9}=3 \mathrm{~cm}$
Area of pentagonal field $\mathrm{ABCDE}=$ Area of $\Delta \mathrm{DEY}$ + Area of $\Delta \mathrm{DCY}+$ Area of trapezium EYXA + Area of trapezium CYXB.
$=\frac{1}{2} \times$ base $\times$ height $+\frac{1}{2} \times$ base $\times$ height $+\frac{1}{2}$ $\times($ sum of parallel sides $) \times$ height $+\frac{1}{2}$ $\times($ sum of parallel sides $) \times$ height
$=\frac{1}{2} \times E Y \times D Y+\frac{1}{2} \times C Y \times D Y+\frac{1}{2} x$ $(E Y+A X) \times(X Y)+\frac{1}{2} \times(C Y+B X) \times(X Y)$
$=\frac{1}{2} \times \dot{4} \mathrm{~cm} \times 3 \mathrm{~cm}+\frac{1}{2} \times 4 \mathrm{~cm} \times 3 \mathrm{~cm}+\frac{1}{2}$ $(4 \mathrm{~cm}+6 \mathrm{~cm}) \times(\mathrm{DX}-\mathrm{DY})+\frac{1}{2}(4 \mathrm{~cm}+6 \mathrm{~cm})$ $\times(\mathrm{DX}-\mathrm{DY})$
$=2 \mathrm{~cm} \times 3 \mathrm{~cm}+2 \mathrm{~cm} \times 3 \mathrm{~cm}+\frac{1}{2}(10 \mathrm{~cm}) \times$ $(9 \mathrm{~cm}-3 \mathrm{~cm})+\frac{1}{2} \times 10 \mathrm{~cm} \times(9 \mathrm{~cm}-3 \mathrm{~cm})$
$=6 \mathrm{~cm}^{2}+6 \mathrm{~cm}^{2}+5 \mathrm{~cm} \times 6 \mathrm{~cm}+5 \mathrm{~cm} \times 6 \mathrm{~cm}$
$=6 \mathrm{~cm}^{2}+6 \mathrm{~cm}^{2}+30 \mathrm{~cm}^{2}+30 \mathrm{~cm}^{2}$
$=72 \mathrm{~cm}^{2}$
Question 45
If the length and the breadth of a room are increased by 1 metre the area is increased by 21 square metres. If the length is increased by 1 metre and breadth is decreased by 1 metre the, area is decreased by 5 square metres. Find the perimeter of the room.
Sol :
Let the length of room= x m
and breadth of room= y m
Area of room $=\ell \times b$
$=x \mathrm{~m} \times y \mathrm{~m}=x y \mathrm{~m}^{2}$
Length is increased by $1 \mathrm{~m}$ then new length becomes $=(x+1) \mathrm{m}$
Breadth is increased by $1 \mathrm{~m}$ then new Breadth $=(y+1) \cdot \mathrm{m}$
Then new Area becomes $=$ new length $\times$ new breadth $=(x+1) \mathrm{m}(y+1) \mathrm{m}$
$(x+1)(y+1) \mathrm{m}^{2}$
According to question,
$x y=(x+1)(y+1)-21$
$\Rightarrow \quad x y=x(y+1)+1(y+1)-21$
$\Rightarrow \quad x y=x y+x+y+1-21$
$\Rightarrow \quad 0=x+y+1-21 \Rightarrow 0=x+y-20$
$\Rightarrow \quad x+y-20=0 \Rightarrow x+y=20$....(1)
Again, length is increased by 1 metre then new length becomes $=(x+1)$ metre Breadth is decreased by 1 metre new Breadth
becomes $=(y-1)$ metre
New Area $=$ new length $\times$ new breadth $=(x+1)$
$(y-1) \mathrm{m}^{2}$
Again, According to question
$x y=(x+1)(y-1)+5$
$\Rightarrow x y=x(y-1)+1(y-1)+5$
$\Rightarrow x y=x y-x+y-1+5$
$\Rightarrow 0=-x+y+4$
$\Rightarrow x-y=4$...(2)
From (1) and (2)
adding
$\begin{array}{l}x+y=20...(1) \\x-y=4...(2) \\\hline 2 x=24\end{array}$
$\Rightarrow \quad x=\frac{24}{2}=12 \mathrm{~m}$
Substituting the value of $x$ in equation (1), we get $12+y=20 \Rightarrow y=20-12 \Rightarrow y=8 \mathrm{~m}$
length of room $=12 \mathrm{~m}$,
Breadth of room $=8 \mathrm{~m}$
Perimeter $=2\{\ell+b)$
$=2(12 \mathrm{~m}+8 \mathrm{~m})=2 \times 20 \mathrm{~m}=40 \mathrm{~m}$
Question 46
A triangle and a parallelogram have the same base and same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Sol :
Sides of the triangle=26 cm, 28 cm and 30 cm
$\therefore \mathrm{s}=\frac{26+28+30}{2}=\frac{84}{2}=42$
Area of $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{42(42-26)(42-28)(42-30)}$
$=\sqrt{42 \times 16 \times 14 \times 12}$
$=\sqrt{7 \times 6 \times 4 \times 4 \times 7 \times 2 \times 6 \times 2}$
$=2 \times 4 \times 6 \times 7=336 \mathrm{~cm}$
$\therefore$ Area of parallelogram $=336 \mathrm{~cm}^{2}$
Base=28 cm
$\therefore$ Height $=\frac{\text { Area }}{\text { Base }}=\frac{336}{28}$
=12 cm
Question 47
A rectangle of area 105 cm² has its length equal to x cm. Write down its breadth in terms of x. Given that its perimeter is 44 cm, write down an equation in x and solve it to determine the dimensions of the rectangle.
Sol :
Area of rectangle $=105 \mathrm{~cm}^{2}$
given length of rectangle $=x \mathrm{~cm}$
Then, Area = length $\times$ Breadth
$\Rightarrow \quad 105=x \times$ Breadth
$\Rightarrow \quad$ Breadth $=\frac{105}{x} \mathrm{~cm}$
Given perimeter of rectangle $=44 \mathrm{~cm}$
$2(\ell+b)=44 \Rightarrow 2\left(x+\frac{105}{x}\right)=44$
$\Rightarrow \quad \frac{x^{2}+105}{x}=22 \Rightarrow x^{2}+105=22 x$
$\Rightarrow \quad x^{2}-22 x+105=0 \Rightarrow x^{2}-15 x-7 x+105=0$
$\Rightarrow \quad x(x-15)-7(x-15)=0$
$\Rightarrow \quad(x-7)(x-15)=0$
Either x-7=0 or x-15=0
x=7 or x=15 cm
When $x=7$, Breadth $=\frac{105}{7}=15 \mathrm{~cm}$
When $x=15$, Breadth $=\frac{105}{15}=7 \mathrm{~cm}$
Hence, required dimensions of rectangle $=15 \mathrm{~cm}$ $7 \mathrm{~cm}$
Question 48
The perimeter of a rectangular plot is 180 m and its area is 1800 m². Take the length of plot as x m. Use the perimeter 180 m to write the value of the breadth in terms of x. Use the value of the length, breadth and the area to,write an equation in x. Solve the equation to calculate the length and breadth of the plot.
Sol :
Given perimeter of a rectangle plot $=180 \mathrm{~m}$
and Area of a rectangle plot $=1800 \mathrm{~m}^{2}$
Taking length of rectangle $=x$ m
Perimeter $=2$ (length + breadth)
$\Rightarrow 180=2(x+$ Breadth $)$
$\Rightarrow \frac{180}{2}=x+$ Breadth $\Rightarrow 90=x+$ Breadth
$\Rightarrow x+$ Breadth $=90$
$\Rightarrow$ Breadth =(990-x) m
Area of rectangle $=$ Length $\times$ Breadth
$1800=x \mathrm{~m} \times(90-x) \mathrm{m}$
$\Rightarrow x(90-x)=1800$
$\Rightarrow 90 x-x^{2}=1800 \Rightarrow-\left(x^{2}-90 x\right)=1800 .$
$\Rightarrow x^{2}-90 x=-1800 \Rightarrow x^{2}-90 x+1800=0$
$\Rightarrow x^{2}-60 x-30 x+1800=0$
$\Rightarrow x(x-60)-30(x-60)=0$
$\Rightarrow(x-30)(x-60)=0$
$\Rightarrow$ Either $x-30=0$ or $x-60=0$
$x=30 \mathrm{~m}$ or $x=60 \mathrm{~m}$
When $x=30 \mathrm{~m}$ then Breadth $=(90-30) \mathrm{m}=60 \mathrm{~m}$
When $x=60 \mathrm{~m}$ then Breadth $=(90-60) \mathrm{m}=30 \mathrm{~m}$
Hence, required length of rectangle $=60 \mathrm{~m}$ and breadth of rectangle $=30 \mathrm{~m}$
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