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Practice each question, then open the answer or worked solution.
Four regular hexagons are drawn so as to form the design as shown in Fig. 6.11 If the perimeter of the design is $28 cm$, find the length of each side of the hexagon.
Ans: Given, the perimeter of figure $=28cm$
No. of Sides=14
$14 \times \text { side }=28 cm$
$\text { side }=\dfrac{28}{14} cm =2 cm$
So, the length of each side of the hexagon is 2cm.
The perimeter of an isosceles triangle is $50 cm$. If one of the two equal sides is $18 cm$, find the third side.
Ans:Given,perimeter=50cm
We know Perimeter = sum of all sides
$50=a+b+c$
$50=18+18+c$
$50=36+c$
$50-36=c$
$14cm=c$
Third side = 14cm.
The length of a rectangle is three times its breadth. The perimeter of the rectangle is $40 cm$. Find its length and breadth.
Ans: Given,perimeter of the rectangle = $40 cm$
Let $L$ and $B$ are the length and breadth of rectangle
According to the question $L =3 B$
Perimeter of the rectangle $=2 \times( L + B )$
$40 =2 \times(3 B + B )$
$40=2 \times 4 B =8 B$
$\dfrac{40}{8}$=B
$5 cm=B$
And length=3$\times 5=15$
There is a rectangular lawn $10 m$ long and $4 m$ wide in front of Meena's house (Fig. 6.12). It is fenced along the two smaller sides and one longer side leaving a gap of $1 m$ for the entrance. Find the length of fencing.
Ans:In rectangular lawn length=10m
And breath = $4 m$
Than the length of fenced along the two smaller sides of lawn
$4 m +4 m =8 m$
Total fencing = $8m+10m-1m=17m$
The region given in Fig. 6.13 is measured by taking a rectangle ($\square $) as a unit. What is the area of the region?
Ans: There are 13 rectangular shapes in the figure
So, area of the region $=13$ sq units
Tahir measured the distance around a square field as 200 rods (lathi). Later he found that the length of this rod was $140 cm$. Find the side of this field in meters.
Ans: Given,the length of the road $=140 cm$
Perimeter = the distance around a square field
$4 \times$ side $=$ length of 200 rods
$4 \times \text { side }=200 \times 140 cm$
$\text { Side }=\dfrac{200 \times 140}{4}$
$=\dfrac{28000}{4} cm =7000 cm$
So,length of the side $=70 m$
The length of a rectangular field is twice its breadth. Jamal jogged around it four times and covered a distance of $6 km$. What is the length of the field?
Ans: Let. $L$ and $B$ are the length and breadth of the rectangle.
Given, $L =2 B$
Perimeters $=2 \times( L + B )$
According to the question
$4 \times[2 \times( L + B )]=6 km$
$4 \times[2( L + B )]=6 km$
$8 \times(2 B + B )=6 km$
$8 \times 3 B =6 km$
$24 B =6 \times 1000 m =6000 m$
$B =\dfrac{6000}{24} m =250 m$
So,the length of the field $=2 \times 250 m =500 m$
Three squares are joined together as shown in Fig. 6.14. Their sides are $4 cm ; 10 cm$ and $3 cm $. Find the perimeter of the figure.
Ans:
We know Perimeter = sum of all sides
Perimeter $=10+6+4+4+4+10+3+3+3+7=54cm$
In Fig. 6.15 all triangles are equilateral and $A B=8$ units. Other triangles have been formed by taking the midpoints of the sides. What is the perimeter of the figure.
Ans: Given, all the triangle are equilateral and $AB =8$ unit
We know Perimeter = sum of all sides
Perimeter $=4+2+1+1+1+2+4+4+2+1+1+1+2+4+4+2+1+1+1+2+4$=$45$ units
The length of a rectangular field is $250 m$ and the width is $150 m$. Anuradha runs around this field 3 times. How far did she run? How many times should she run around the field to cover a distance of $4 km$ ?
Ans: Given, length $(l)=250 m$
and width $(b)=150 m$
Than,perimeter of the field $=2(l+b)=2(250+150) m =2 \times 400 m =800 m$
Because distance covered in one round $=$ Perimeter $=800 m$
So,distance covered in three rounds $=3 \times 800=2400 m$
Now, number of rounds to cover $4 km$ =$\dfrac{4000}{800}=5$
$[\because 1 km =1000 m ]$
So, she should run 5 times around the field to cover the distance of $4 km$.