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Maths 5 marks

T3 CH3 Perimeter and Area · Tamil Nadu Board STD 6 (Tamil - English Medium). 21 questions.

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21 questions · timed · auto-graded

Question 15 Marks
The length of a rectangle is three times its breadth. If its perimeter is 64 cm, find the sides of the rectangle
Answer
Given perimeter of a rectangle $=64 cm$ Also given length is three times its breadthLet the breadth of
the rectangle $=b c m$
$\therefore$ Length $=3 \times b c m$ Perimeter $=64 m$
i.e. $2 \times(I+b)=64 m$
$2 \times(3 b+b)=64 m$
$2 \times 4 b=64 m$
$4 b=\frac{64}{2}=32 m$
$b=\frac{32}{4}=8 m$
$1=3 \times b$
$=3 \times 8$
$=24 m$
$\therefore$ Breadth of the rectangle $=8 m$ Length of the rectangle $=24 m$
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Question 25 Marks
A rectangle has length 40 cm and breadth 20 cm. How many squares with side 10 cm can be formed from it
Answer
Area of rectangle $=($ length $\times$ breadth $)$ units ${ }^2$ Length $=40 cm$ Breadth $=20 cm$
$\therefore$ Area $=(40 \times 20) cm ^2$
$=800 cm ^2$ Area of rectangle $=800 cm ^2$ Area of square $=($ side $\times$ side $)$ units ${ }^2$
side $=10 cm$ Area of square $=(10 \times 10) cm ^2$
$=100 cm ^2$ Required number of squares $=\frac{\text { Area of Rectangle }}{\text { Area of } 1 \text { square }}$
$=\frac{800 cm ^2}{100 cm ^2}$
$=8$
8 squares can be formed.
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Question 35 Marks
The length of a rectangular park is 14 m more than its breadth. If the perimeter of the park is 200 m, what is its length? Find the area of the park
Answer
Given length of rectangular park is $14 m$ more than its breadth. Let the breadth be ' $b^{\prime}{ }^{\prime} m$
$\therefore$ Length of the park will be $l=b+14 m$ Given perimeter $=200 m$
$2 \times(1+b)=200 m$
$2 \times(b+14+b)=200 m \ldots[\because 1=b+14]$
$2 \times(2 b+14)=200 m$
$2 b+14=\frac{200}{2} m$
$2 b+14=100 m$
$2 b=100-14 m$
$2 b=86 m$
$b=\frac{86}{2} m$
$b=43 m$
$=(57 \times 43) m^2$
$=2.451 m ^2 \quad$ Area of the park $=2,451 m ^2$
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Question 45 Marks
Rahim and Peter go for a morning walk, Rahim walks around a square path of side 50 m and Peter walks around a rectangular path with length 40 m and breadth 30 m. If both of them walk 2 rounds each, who covers more distance and by how much?
Answer
Perimeter of the square = (4 × side) units
Side of the square = 50 m
∴ Perimeter = 4 × 50 m = 200 m
Rahim walked twice around the park
∴ Total distance walked by Rahim = 2 × 200 m
Total distance covered by Rahim = 400 m
Perimeter of a rectangle = 2(length + breadth) unit
Length = 40 m
Breadth = 30 m
∴ Perimeter = 2 × (40 + 30) m
= (2 × 70) m
= 140 m
Peter walked around twice
∴ Distance covered by Peter = 2 × 140 m
= 280 m
∴ Distance covered by Peter = 280 m
400 m > 280 m
Difference = 480 – 280
= 120 m
∴ Rahim covers 120 m more than Peter
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Question 55 Marks
Find the approximate area of the flower in the given square grid.
Image
Answer
No. of full squares $=11 No$. of half squares $=9$ Area of $11$ full squares
$=11 \times 1 cm ^2$
$=11 cm ^2$
$=9 \times \frac{1}{2} cm ^2$
$=4.5 cm ^2 $
$=15.5 cm ^2$
Area of $9$ half squares
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Question 65 Marks
Look at the picture of the house given and find the total area of the shaded portion.
Image
Answer
$\text { Area of a square }=a \times \text { a sq.units }$
$=6 \times 6 cm ^2$
$=36 cm ^2 \quad \text { Area of a rectangle }=1 \times b \text { sq-units }$
$=9 \times 6 cm ^2$
$=54 cm ^2 \quad \text { Area of a triangle }=\frac{1}{2} \times b \times h \text { sq.units }$
$=\frac{1}{2} \times 4^2 \times 6 cm ^2$
$=12 cm ^2$
$=(36+54+12) cm ^2$
$=102 cm ^2$
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Question 75 Marks
Two plots have the same perimeter. One is a square of side 10 m and another is a rectangle of breadth 8 m. Which plot has the greater area and by how much?
Answer

Given perimeter of square $=$ perimeter of rectangle
$4 \times$ side $=2($ length + breadth $)$
$(4 \times 10) m=2(1+8) m$
$\frac{4 \times 10}{2}=1+8$
$20=1+8$
$1=20-8$
$1=12 m$
$\therefore$ length of the rectangle $=12 mArea$ of the square plot
$=$ side $\times$ side
$=10 \times 10 m ^2$ Area of the rectangular plot
$=100 m ^2$
$=$ length $\times$ breadth
$=(12 \times 8) m^2$
$=96 m^2$
$100 m^2>96 m^2$
$\therefore$ Square plot has greater area by $4 m^2$
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Question 85 Marks
The table given below contains some measures of the triangle. Find the unknown values.
Side 1 Side 2 Side 3 Perimeter
6 cm 5 cm 2 cm ?
Answer
Perimeter of a triangle = sum of three sides
Perimeter = 6 + 5 + 2 cm
= 13 cm
p = 13 cm
Tabulating the unknowns.
Side 1Side 2Side 3Perimeter
6 cm5 cm2 cm13 cm
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Question 95 Marks
The table given below contains some measures of the triangle. Find the unknown values.
Side 1 Side 2 Side 3 Perimeter
? 8 m 3 m 17 m
Answer
Perimeter of a triangle = sum of three sides
Perimeter = (side 1 + side 2 + side 3) m
17 = (side 1 + 8 + 3) m
17 m = (side 1 + 11) m
side 1 = 17 – 11
= 6 m
Tabulating the unknowns.
Side 1Side 2Side 3Perimeter
6 m8 m3 m17 m
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Question 105 Marks
The table given below contains some measures of the triangle. Find the unknown values.
Side 1 Side 2 Side 3 Perimeter
11 feet ? 9 feet 28 feet
Answer
Perimeter of a triangle = sum of three sides
Perimeter = side 1 + side 2 + side 3
28 feet = 11 feet + side 2 + 9 feet
28 feet = 20 feet + side 2
28 – 20 = side 2
side = 8 feet
Tabulating the unknowns.
Side 1Side 2Side 3Perimeter
11 feet8 feet9 feet28 feet
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Question 115 Marks
The table given below contains some measures of the right angled triangle. Find the unknown values.
Base Height Area
20 cm 40 cm ?
Answer
Area of the right triangle $=\frac{1}{2} \times($ base $\times$ height $)$ unit $^2$
$b=20 cm$
$h=40 cm$
$\text { Area }=\frac{1}{2}(b \times h) cm ^2$
$=\frac{1}{2} \times 20 \times 40$
$=400 cm ^2$
$A=400 cm ^2$
Tabulating the unknown values
Base Height Area
20 cm 40 cm 400 cm^(2)
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Question 125 Marks
The table given below contains some measures of the right angled triangle. Find the unknown values.
Base Height Area
5 feet ? 20 sq.feet
Answer
Area of the right triangle $=\frac{1}{2} \times($ base $\times$ height $)$ unit ${ }^2$ $b=20 cm$
$h=40 cm$
$\text { Area }=\frac{1}{2}(b \times h) cm ^2$
$=\frac{1}{2} \times 20 \times 40$
$=400 cm ^2$
$A=400 cm ^2$
Tabulating the unknown values
Base Height Area
20 cm 40 cm 400 cm^(2)
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Question 135 Marks
The table given below contains some measures of the right angled triangle. Find the unknown values.
Base Height Area
? 12 m 24 sq.m
Answer
Area of the right triangle $=\frac{1}{2} \times($ base $\times$ height $)$ unit 2
$24=\frac{1}{2} \times b \times 12 m^2 \text { Base }=\frac{24 \times 2}{12} m$
$=4 m \text { Base }=4 m$
Tabulating the unknown values
Base Height Area
4m 12 m 24 sq*m
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Question 145 Marks
The table given below contains some measures of the square. Find the unknown values.
Side Perimeter Area
6 cm ? ?
Answer
Perimeter of a square $=(4 \times$ side $)$ units Area of a square $=($ side $\times$ side $)$ unit $^2$ $s=6 cm$
Perimeter $=4 \times 5$ unit
$=4 \times 6 cm$
$=24 cm$
$P=24 cm$
$\text { Area }=s \times s \text { unit }^2$
$=6 \times 6 cm^2$
$=36 cm^2$
$A=36 cm^2$
Side Perimeter Area
6 cm 24 cm 36 cm$^2$
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Question 155 Marks
The table given below contains some measures of the square. Find the unknown values.
Side Perimeter Area
? 100 m ?
Answer
$\text { Perimeter of a square }=(4 \times \text { side }) \text { unitsArea of a square }=\left(\text { side } \times \text { side unit }^2 \text { Perimeter }=4 \times s ^{\text {unit }}\right.$
$100=(4 \times s) m$
$\frac{100}{4}=s$
$s=25 mArea =s \times s \text { unit }^2$
$=25 \times 25 m ^2$
$=625 m ^2$
$A=625 m ^2$
Completing the unknown values in the table
Side Perimeter Area
25 m 100 m 625m^(2)
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Question 165 Marks
The table given below contains some measures of the square. Find the unknown values.
Side Perimeter Area
? ? 49 sq.feet
Answer
Perimeter of a square $=(4 \times$ side $)$ units
Area of a square $=($ side $\times$ side $)$ unit $^2$
Area $=s \times s$ unit $^2$
$49=s \times s$ square feet
$s^2=7^2$
$s=7$ feet
Perimeter $=4 \times 5$ unit
$=4 \times 7$ feet
$=28$ feet
Perimeter $=28$ feet
Completing the unknown values in the table
Side Perimeter Area
7 feet 28 feet 49 sq.feet
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Question 175 Marks
The table given below contains some measures of the rectangle. Find the unknown values.
Length Breadth Perimeter Area
? 4 feet ? 20 sq.feet
Answer
$b=4 \text { feet }$
$\text { Area }=20 \text { sq.feet }$
$\text { Area }=1 \times b \text { sq.feet }$
$20=1 \times 4$
$I=\frac{20}{4} \text { feet }$
$I=5 \text { feet }$
$\text { Perimeter }=2(l+b) \text { units }$
$p=2(5+4) \text { feet }$
$=2 \times 9$
$p=18 \text { feet }$
Completing the unknown values in the table.
Length Breadth Perimeter Area
5 feet 4 feet 18 feet 20 sq.feet
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Question 185 Marks
The table given below contains some measures of the rectangle. Find the unknown values.
Length Breadth Perimeter Area
10 m ? ? 120 sq.m
Answer
$I=10 m$
$\text { Area }=120 sq \text { metre }$
$\text { Area }=I \times b \text { sq.m }$
$120=10 \times 6$
$b=\frac{120}{10}$
$b=12 m$
Perimeter $=2(l+b)$ units
$=2(10+12) \text { units }$
$=2 \times 22 m$
$p=44 m$
Completing the unknown values in the table.
Length Breadth Perimeter Area
10 m 12 m 44 m 120 sq. m
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Question 195 Marks
The table given below contains some measures of the rectangle. Find the unknown values.
Length Breadth Perimeter Area
5 cm 8 cm ? ?
Answer
Area of the rectangle $=$ (length $\times$ breadth) sq.unit
Perimeter of a rectangle $=2(1+b)$ units
$I=5 cm$
$b=8 cm$
$\therefore P=2(I+b) cm$
$=2(5+8) cm$
$=2 \times 13 cm$
$P=26 cm$
$\text { Area }=(I \times b) cm^2$
$=(5 \times 8) cm^2$
$A=40 cm^2$
Completing the unknown values in the table.
Length Breadth Perimeter Area
5 cm 8 cm 26 cm 40 cm$^2$
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Question 205 Marks
The table given below contains some measures of the rectangle. Find the unknown values.
Length Breadth Perimeter Area
13 cm ? 54 cm ?
Answer
$l=13 cm$
$p=54 cm$
Perimeter $=2(l+b)$ units
$54=2(13+b) cm$
$\frac{54}{2}=13+b$
$27=13+b$
$b=27-13$
$b=14 cm ^{-1}$
Area $=I \times b$ sq.unit
$=13 \times 14 cm ^2$
$A =182 cm ^2$
Completing the unknown values in the table.
Length Breadth Perimeter Area
13 cm 14 cm 54 cm 182 cm^(2)
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Question 215 Marks
The table given below contains some measures of the rectangle. Find the unknown values.
Length Breadth Perimeter Area
? 15 cm 60 cm ?
Answer
$b=15 cm$
$p=60 cm$
$p=2(l+b) \text { units }$
$60=2(l+15) cm$
$\frac{60}{2}=1+15$
$30=1+15$
$I=30-15$
$I=15 cm$
$\text { Area }=1 \times \text { bunit }^2$
$=15 \times 15 cm ^2$
$=225 cm ^2$
$A=225 cm ^2$
Completing the unknown values in the table.
Length Breadth Perimeter Area
15 cm 15 cm 60 cm 225 cm^(2)
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