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Practice each question, then open the answer or worked solution.
Bajinder runs ten times around a square track and covers $4 km$. Find the length of the track.
Ans: Given,length of track covered in 10 rounds $=4 km =4000 m [\therefore 1 km =1000 m ]$
So,length of track covered in 1 round $=\dfrac{4000}{10}=400 m$
Perimeter of square = $ 4 \times$ side
So length of track = $\dfrac{400}{4}=100m$
The lawn in front of Molly's house is $12 m \times 8 m$,whereas the lawn in front of Dolly's house is $15 m \times 5 m$. Bamboo fencing is built around both the lawns. How much fencing is required for both?
Ans: Perimeter of the lawn in front of Molly's house
$=2 \times(\text { L }+\text {B })$
$2 \times(12+8) =2 \times 20 =40 m$
Perimeter of the lawn in front of Dolly's house
$=2 \times(\text { L }+\text { B })$
$2 \times(15+5) =2 \times 20 =40 m$
So,Fencing is required for both $=40 +40 =80 m$
The perimeter of a regular pentagon is $1540 cm$. How long is it's every side?
Ans: Given,the perimeter of a regular pentagon=$1540 cm$
We know Perimeter = sum of all sides
Perimeter of the regular pentagon=$5 \times$ length of sides $=1540 cm$
Sides $=\dfrac{1540}{5} cm =308 cm$
So,length of the side of the regular pentagon is $308 cm$.
The perimeter of a triangle is $28 cm$. One of its sides is $8 cm$. Write all the sides of the possible isosceles triangles with these measurements.
Ans: Given,the perimeter of a triangle =$28 cm$
Let sides of a triangle are x,y and z
than $x= y$,$z=8 cm$
We know Perimeter = sum of all sides =$28 cm$
$x+y+z=28$
$x+x+8=28$
$2x=28-8$
$x=\dfrac{20}{2}$
$x=10cm$
Sides are $10cm,10cm$ and $8cm$
Let sides of a triangle are x,y and z
And $x=y=8cm$
We know Perimeter = sum of all sides =$28 cm$
$x+y+z=28$
$8+8+z=28$
$16+z=28$
$z=28-16$
$z=12cm$
Sides are $8cm,8cm$ and $12cm$
The length of an Aluminium strip is $40 cm$. If the lengths in $cm$ are measured in natural numbers, write the measurement of all the possible rectangular frames which can be made out of it. (For example, a rectangular frame with $15 cm$ length and $5 cm$ breadth can be made from this strip.)
Ans: Perimeter of rectangular frame $=$ Length of the aluminum strip
$2 \times($ length $+$ breadth $)=40 cm$
Length + Breadth = 20 cm
∴ The possible measurement of rectangular frames are 1 cm $\times$ 19 cm, 2 cm $\times$ 18 cm, 3 cm $\times$ 17 cm, 4 cm $\times$ 16 cm, 5 cm $\times$ 15 cm, 6 cm $\times$ 14 cm, 7 cm $\times$ 13 cm, 8 cm $\times$ 12 cm, 9 cm $\times$ 11 cm, 10 cm $\times$ 10 cm
The base of a tent is a regular hexagon of perimeter $60 cm$. What is the length of each side of the base?
Ans: Perimeter of hexagon $=6 \times$ sides $=60 cm$
$\text { Side }=\dfrac{60}{6} =10 cm$
So, each side of the base is $10 cm$.
In an exhibition hall, there are 24 display boards each of length $1 m 50 cm$ and breadth $1 m$. There is a $100 m$ long aluminium strip, which is used to frame these boards.
How many boards will be framed using this strip? Find also the length of the aluminium strip required for the remaining boards.
Ans: Length $=1 m 50 cm =1.5 m$
Breadth $=1 m$
Than,perimeter of one display board= $2 \times($ length $+$ breadth $)$
$2 \times(1.5 m +1 m )$
$2 \times 2.5 m =5 m$
So,perimeter of the display board =$5 \times 24=120 m$
Number of boards framed $=\dfrac{100}{5}=20$
And length of the aluminum strip required for the remaining boards $=(120-100) m =20 m$
In the above question, how many square meters of cloth are required to cover all the display boards? What will be the length in $m$ of the cloth used, if its breadth is $120 cm$ ?
Ans: Length $=1 m 50 cm =1.5 m$
Breadth $=1 m$
Area of the display board
$24 \times($ length $\times$ breadth $)=24 \times 1.5 m \times 1 m =36 m ^{2}$
$36 m ^{2}$ cloths are needed to cover all the display boards.
Now, breadth of the cloth $=120 cm =\dfrac{120}{100} m =1.2 m$
$[\therefore 1 m =100 cm ]$
So,length of the cloth $=\dfrac{\text { area }}{\text { breadth }}=\dfrac{36 m ^{2}}{1.2 m }=30 m$
What is the length of the outer boundary of the park shown in Figure, what will be the total cost of fencing it at the rate of Rs 20 per meter? There is a rectangular flower bed in the centre of the park. Find the cost of manuring the flower bed at the rate of Rs $5 0$ per square meter.
Ans: We know Perimeter = sum of all sides
Perimeter $=(200+300+80+300+200+260) m=1340 m$
Than,rate of fencing it at the rate of Rs 20 per meter
$=$ Rs. $(1340 \times 20)=$ Rs. 26800
Area of the flower bed $=(100 \times 80) m ^{2}=8000 m ^{2}$
Also,rate of manuring the flower bed per square metre= Rs 50 per square meter
Than,rate of manuring the flower bed per 8000 square meter $=$ Rs. $(50 \times 8000)=$ Rs. 400000
Total cost of fencing the park shown in Figure is Rs 55000. Find the cost of fencing per meter.
Ans: We know Perimeter = sum of all sides
Perimeter $=150+100+120+180+270+280$=$1100 m$
Now cost of fencing the per meter$=\dfrac{\text { Total cost of fencing }}{\text { Perimeter of the park }}=\text { Rs. }\left(\dfrac{55000}{1100}\right)=\text { Rs.50 }$