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Tamilnadu 6th Standard Social Science T3 - குடிமையியல் - உள்ளாட்சி அமைப்பு - ஊரகமும் நகர்ப்புறமும் Important Questions And Answers Study Material - QB365 Set B
Published on: 07/09/2019
Term 3 Perimeter and Area
Download Tamil Nadu 6th Standard Maths question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
Questions + Answers key
Take MCQ Maths Test1.
Fill in the blanks.
i) 5 cm2 = _____ mm2
ii) 26 m2 = _____ cm2
iii) 8 km2 = _____ m2
2.
Find the perimeter of the given figure.
3.
Find the area of a square of side 15 cm
4.
Find the number of tiles required to fill the are a of following ligures
(i)
(ii)
(iii)
(iv)
5.
Find the perimeter and area of the following shapes
6.
A square park has 40 m as its perimeter. What is the length of its side? Also find its area.
7.
The area of a rectangular shaped photo is 820 sq. cm. and its width is 20 cm. What is its length? Also find its perimeter
8.
Find the perimeter and the area of a right angled triangle whose sides are 6 feet, 8 feet and 10 feet.
9.
The length and breadth of a rectangular sheet of a paper are 15 cm and 12 cm respectively. A rectangular piece is cut from one of its corners. Which of the following statement is correct for the remaining sheet?
Perimeter remains the same but the area changes
Area remains the same but the perimeter changes
There will be a change in both area and perimeter
Both the area and perimeter remains the same
10.
The side of a square is 10 cm. If its side is tripled, then by how many times will its perimeter increase?
2 times
4 times
6 times
3 times
11.
If every side of a rectangle is doubled, then its area becomes _____ times
2
3
4
6
12.
If two identical rectangles of perimeter 30 cm are joined together, then the perimeter of the new shape will be
equal to 60 cm
less than 60 cm
greater than 60 cm
equal to 45 cm
13.
The following figures are of equal area. Which figure has the least perimeter?
1.
i) 5 cm2 = 500 mm2
ii) 26 m2 = 2,60,000 cm2
iii) 8 km2 = 80,00,000 m2
2.
Perimeter = Total length of the boundary
= (6 + 2 + 10 + 3 + 2 + 1 + 3 + 4 + 2 + 6 + 9) cm
= 48 cm
3.
Side of the square, s = 15 cm
Area of the square, A = (s \(\times\) s) sq. units.
= 15 \(\times\) 15
= 225 sq. cm. (or) 225 cm2
4.
(i) 4 square tiles horizontally and 4 square tiles vertically.
\(\Rightarrow\) 4 \(\times\) 4 = 16 sq. units.
No. of tiles tilled = 7
No. of tiles required = 16 -7 = 9
(ii) 4 square tiles vertically and 3 square tiles horizontally .
Area \(\Rightarrow\) 4 \(\times\) 3 = 12 sq. units.
No. of tiles filled = 6
(iii) 4 square tiles vertically and 3 square files horizontally.
Area = 4 \(\times\) 3 = 12 sq. units.
No. of tiles filled = 6
Required tiles = 12 - 6 = 6
(iv) 4 square tiles horizontally and 4 square tiles vertically.
Area = 4 \(\times\) 4 = 16 sq. units.
No. of tiles filled = 8
Required tiles = 16 - 8 = 8
5.
(i) Given :
Perimeter = 12 times 4 cm
= 12 \(\times\) 4 = 48 cm
Area \(\Rightarrow\) Given = 5 squares
= 5 \(\times\) area of square
= 5 \(\times\) (sider)2
= 5 \(\times\) 42
\(\therefore\) A = 5 \(\times\) 16 = 80 Cm2
\(\therefore\) P = 48 cm, A = 80 cm2
(ii) Given :
Perimeter = 4 times 4 cm + 4 times 5cm
= 4 (4) + 4 (5)
= 16 + 20 = 36 cm
Area = 4 triangles + 1 square
= \(4\left( \frac { 1 }{ 2 } bh \right) +1\left( s \right) ^{ 2 }\)
= \(4\left( \frac { 1 }{ 2 } \left( 5 \right) \left( 4 \right) \right) +1\left( { 3 }^{ 2 } \right) \)
= 4 \(\times\) 5 \(\times\) 2 + 9
= 40 + 9 = 49 cm2
(iii) Given :
Perimeter = (50 + 12 + l3 + 40 + 10 +10 + 10 + 5) cm
= 150 cm
= \(\left( lb+\cfrac { 1 }{ 2 } bh+{ s }^{ 2 } \right) { cm }^{ 2 }\)
= \(\left( 50\times 5 \right) +\cfrac { 1 }{ 2 } \times 12\times 5+{ 10 }^{ 2 }\)
= 250 + 30 + 100
= 380 cm2
\(\therefore\) P = 150 cm, A = 380 cm2
6.
Perimeter (P) = 40 m
Side (8) =?
Area (A) =?
4S = P
\(4S=40\Rightarrow S=\cfrac { 40 }{ 4 } =10m\)
A = S2 = 102 = 100m2
\(\therefore\) Side (S) = 10 m, Area (A) = 100 m2
7.
Area of a rectangle = 820 sq. cm
Width (b) = 20 cm
Length (l) =?
Perimeter =?
\(I=\cfrac { A }{ b } \)
= \(\cfrac { 820 }{ 20 } =41\\ \)
= =2(41+20)
= 2 (61) = 122 cm
\(\therefore\) I = 41 cm , P = 122 cm
8.
Sides are = 6 feet, 8 feet and 10 feet
Perimeter = ( a + b + c) feet
= (6+8+10) = 24 feet.
Area = \(\cfrac { 1 }{ 2 } \)bh.sq.feet
= \(\cfrac { 1 }{ 2 } \times 6\times 8\)
= 3 \(\times\) 8 = 24 feet
\(\therefore\) P = 24 feet, A = 24 sq. feet
9.
(c)
There will be a change in both area and perimeter
10.
(d)
3 times
11.
(c)
4
12.
(b)
less than 60 cm
13.
(b)
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