வெல்லுங்கள் CSAT 2025 - 16: Q and A on Perimeter & Area

Questions & Answers - (Multiple Choice) - Perimeter & Area
கொள்குறி வினா- விடைகள் - சுற்றளவு & பரப்பு

1. Two sides of the triangle are 12cm and 16cm. If the height of the triangle corresponding to the base 16cm. is 6cm. Find the height of the triangle corresponding to 12cm.as base
side.
a. 6cm. b. 8cm.
c. 9cm. d. 10cm.

2. The sides of the triangle are 16cm, 12cm and 20cm. Find the area of the triangle.
a. 48 sqcm. b. 64 sqcm.
c. 96 sqcm. d. 128 sqcm.

3. The base and height of the triangle are in the ratio 5 : 6. If the area of the triangle is 135 sq.cm, then find the sum of the base and height of the triangle.
a. 27cm. b. 22cm.
c. 35cm. d. 33cm.

4. Find the area of the right angled triangle whose hypotenuse is 20cm and one of the sides is 12cm.
a. 96 sqcm. b. 84 sqcm.
c. 48 sqcm. d. 108 sqcm.

5. The length of a rectangle is 28cm and its perimeter is equal to the perimeter of a square with side 22cm. Find the ratio of the area and perimeter of the rectangle.
a. 36 : 11 b. 14 : 11
c. 56 : 11 d. 56 : 33

6. If the area of a square is numerically equal to its perimeter, find the area of the square whose side is double the given square.
a. 16 sq.units b. 32 sq.units
c. 36 sq.units d. 64 sq.units

7. A path of uniform width, 1.5m, runs around the inside of a rectangular field 32m by 25m. Find the area of the path.
a. 216 sq.units b. 162 sq.units
c. 324 sq.units d. 264 sq.units

8.A rectangular field is by 30m in length and 22m in breadth. Two mutually perpendicular roads each 2.5m wide, are drawn inside field so that one road is parallel to the length of the field and the other is parallel to its breadth. Find the area of the crossroads.
a. 123.75 sq.m. b. 130 sq.m.
c. 136.25 sq.m. d. 242.1 sq.m

9. A horse is tied to a peg at one corner of a square shaped grass field of side 20 m by means of a 14 m long rope. The area of the inside part of the field in which the horse can graze is
a. 308 sq.m. b. 130 sq.m.
c. 231 sq.m. d. 154 sq.m.

10. The radii of two circles are 19 cm and 9 cm respectively. The diameter of the circle which has circumference equal to the sum of the circumference of these two circles is
a. 28 cm b. 56 cm
c. 42 cm d. 63 cm

11. The area of the circle that can be inscribed in a square of side 28 cm, is
a. 308 sqcm. b. 1232 sqcm.
c. 616 sqcm. d. 2464 sqcm.

12. The area of the sector of a circle with radius 12 cm and of angle 120°at centre is
a. 48π sqcm. b. 24π sqcm.
c. 36π sqcm. d. 72π sqcm

13. A man walked along the diagonal path across a square field instead of not along the edges of the field. Approximately how much percentage of distance did he save in this way?
a. 20% b. 22%
c. 26.2% d. 29.3%

14. The length of a rectangle is trebled and the breadth is halved. What is the change in area in percentage?
a. 150% increase b. 50% increase
c. 250% increase d. 200% increase

15. What is the least number of square tiles needed to pave a floor of 9m 2cm long and 7m 38cm broad?
a. 41 b. 82
c. 99 d. 140

Answers

1. b 2. c 3. d 4. a 5. c

6. c 7. b 8. a 9. d 10. b

11. c 12. a 13. d 14. b 15. c

Explanation to answers

1.The area of a triangle is half of the product of base and height.
Here area = (1/2)×16×6
Let the required height be 'h'
So (1/2)×12×h = (1/2)×16×6
=> h = 8 cm.

2.The area of triangle
= √[s(s - a)(s - b)(s - c)]
Where s is the semi circumference of the triangle and a, b and c are sides of the given triangle.
Here s = (1/2)(a+b+c)
= (1/2)(16+12+20) = 24cm.
So [s(s - a)(s - b)(s - c)]
= 24(24 - 16)(24 - 12)(24 - 20)
= 24×8×12×4 = 24 ×24 ×4×4 = 96×96
Hence the area is 96 sq.cm.

3. Let the base and height of the triangle be 5l and 6l respectively.
Here (1/2)(5l)(6l) = 135
=> l^2 = 9 => l = 3
Sum of the base and height
= 5l + 6l = 11l = 11×3 = 33 cm.

4. Other side =
√[(hypotenuse)^2 - (given side)^2]
= √[(20^2) - (12^2)] = √256 = 16cm.
Area = (1/2)×(16)×(12) = 96sqcm.

5. Let b be the berth of the rectangle.
Here 2(28 + b) = 4×22
=> b = 16
Perimeter of the rectangle
= 2(28 + 16) = 88
Area of the rectangle = 28×16 = 448
Required ratio 448 : 88 = 56 : 11

6. If 'a' is the side of square, then
Perimeter = 4a and Area = a^2
Given a^2 = 4a => a = 4 units
Side of the required square = 8 units
Required area = 8×8 = 64 sq.units.

7. Outer dimesion of the field 32×25
Inner dimesion of the field
(32 - 2×1.5) × (25 - 2×1.5)
The area of the path
= Area of big rectangle -
Area of small rectangle
= 32×25 - 29×22
= 800 - 638 = 162 sq.m.

8. Area of cross roads =
Area of rectangle with length and wide
+Area of rectangle with breadth and wide
- Area of square formed by wide
= 30×2.5 + 22×2.5 - 2.5×2.5
= 75 + 55 - 6.25 = 123.75 sq.m.

9.Area of a circle is π(r^2)
Here r = 14
The grazing area must be only
one-fourth area of the circle with
radius 14cm.
So required area = (1/4)(22/7)(14×14)
= 154 sq.m

10. The circumference of a circle of
radius 'r' is = 2πr
Here 2πR = 2π(19+9) where R is the
radius of the new circle.
=> R = 28
But diameter = 2R = 2×28 = 56 cm.

11. Here the diameter of the circle is the side of the square.
So the radius of the circle
= 28/2 = 14 cm.
Area of the circle is
= (22/7)×14×14 = 616sq.cm.

12. The area of sector of a circle
= (D/360)×(π)(r^2)
Here D = 120 ; r = 12
So required area
= (120/360)×12×12×π
= 48π sq.cm.

13. Let the edge of the square field be 'a' units.
The distance along the edges = 2a units.
But the distance walked along the diagonal is =√2 a = 1.414 a
Distance saved = 2a - 1.414 a = 0.586 a
Required percent
= (0.586a)/(2a) ×100 = 29.3 %

14. Length increased by 200%
Breadth decreased by 50%
Net area change in %
= +200 - 50 + (+200)(-50)/(100)
= 200 - 50 - 100 = +50% increase

15.Length = 902 cm
Breadth = 738 cm.
To find HCF of 902, 738
902 =2 × 451 = 2 × 11 × 41
738 = 2× 369 = 2 × 9 × 41
So HCF = 2×41 = 82
The least number of square tiles needed to pave the floor
= (902×738)/(82×82) = 11×9 = 99.

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