QNO9.The area of a ∆ whose sides are 8cm , 15cm and 17 cm
120 cm2
30 cm2
60 cm2
None of these
QNO10.The area of ∆ whose sides are 3 cm , 4cm and 5cm is
12
13
42
None of these
QNO11.The area of an equilateral ∆ whose perimeter is 30 cm is
3
5
25
None of these
QNO12.The perimeter of an isosceles right angled having area 100 cm2 is
10 (
100(
10 (
None of these
QNO13.If length, breadth and height of a cuboid is 5 cm, 3 cm and 4cm .Then its total surface area will be
94 cm2
49 cm2
98 cm 2
None of these
QNO14 If length of one edge of a cube is 10 cm .Total surface area will be
6 cm2
60 cm 2
600 cm2
None of these
QNO15.If the curved surface area of a right circular cylinder of height 14 cm is 88 cm2, and then diameter of base of cylinder is
2 cm
3 cm
4cm
None of these
QNO16.Formula to calculate T.S.A of a cone is
None of these
QNO17.If the height of a cone is 16 cm and radius of base is 12 cm, then slant height is
20 cm
30 cm
40 cm
None of these
QNO18.If h = 10 cm and radius of base is 24 m, then height will be
62 cm
26 cm
65 cm
None of these
QNO19.If radius of a cone is 7cm and height is 24cm, then slant height will be
25cm
52cm
205cm
None of these
QNO 20.The curved surface area of a cone of slant height 10cm and radius of base 7 cm will be
220 cm2
202 cm2
340 cm2
None of these
Answers
Ques: The area of a ∆ whose sides are 8cm , 15cm and 17 cm.
Solution:-
s = 8+15+17/2
s = 40/2
s = 20
Semi-perimeter is 20 cm.
Area of triangle :
➝ √20(20 - 8)(20 - 15)(20 - 17)
➝ √20 × 12 × 5 × 3
➝ √2×2×5×2×2×3×5×3
➝ 2 × 2 × 5 × 3
➝ 60
Therefore,
Area of triangle is 60 cm².
________________________________
Ques: The area of ∆ whose sides are 3 cm , 4cm and 5cm is
Solution:-
s = 3+4+5/2
s = 12/2
s = 6
Semi-perimeter is 6 cm.
Area of triangle :
➝ √6(6 - 3)(6 - 4)(6 - 5)
➝ √6 × 3 × 2 × 1
➝ √3 × 2 × 3 × 2 × 1
➝ 2 × 3
➝ 6
Therefore,
Area of triangle is 6 cm².
_________________________________
Ques: The area of an equilateral ∆ whose perimeter is 30 cm is
Solution:-
If Perimeter is 30 cm.
Let, all sides of triangle be x. [All sides measure will be x because triangle is an equilateral triangle.]
Perimeter of triangle = Sum of all sides
➝ 30 = x + x + x
➝ 30 = 3x
➝ 30/3 = x
➝ x = 10
All sides measure will be 10 cm.
So,
s = 30/2
s = 15
Semi-perimeter is 15 cm.
Area of triangle :
➝ √30 (30 - 10)(30 - 10)(30 - 10)
➝ √30 × 20 × 20 × 20
➝ √3 × 2 × 5 × 2 × 2 × 5 × 2 × 2 × 5 × 2 × 2 × 5
➝ 2 × 2 × 2 × 5 × 5 × √3 × 2
➝ 200 × √6
➝ 200 × 2.4494
➝ 489.89
Therefore,
Area of triangle is 489.89 cm² .
________________________________
Ques : The perimeter of an isosceles right angled having area 100 cm2 is.
Solution:-
If it is an isosceles right angle triangle then it's height and base will be of same measurements.
Let, Base and height of triangle be x and x.
Area of triangle = 1/2 × base × height
➝ 100 = 1/2 × x × x
➝ 100 = x²/2
➝ 100 × 2 = x²
➝ 200 = x²
➝ x = √200
➝ x = 14.14
Two sides base and height are of 14.14 cm .
By Pythagoras theorem,
Hypotenuse² = Base² + Perpendicular²
➝ Hypotenuse² = (14.14)² + (14.14)²
➝ Hypotenuse² = 199.94 + 199.94
➝ Hypotenuse² = 399.88
➝ Hypotenuse = √399.88
➝ Hypotenuse = 19.99
Perimeter = 19.99 + 14.14 + 14.14
= 48.27
Therefore,
Perimeter of triangle is 48.27 cm.
_______________________________
Ques : .If length, breadth and height of a cuboid is 5 cm, 3 cm and 4cm .Then its total surface area
Solution:-
Total surface area:
➝ 2×[(5 × 3) + (3 × 4)+(4 × 5)]
➝ 2 × [15 + 12 + 20]
➝ 30 + 24 + 40
➝ 94
Therefore,
Total surface area of cuboid is 94 cm².
_______________________________
Ques : If length of one edge of a cube is 10 cm .Total surface area will be :
Solution:-
Total surface area of cube:
➝ 6 × (10)²
➝ 6 × 100
➝ 600
Therefore,
Total surface area of cube is 600 cm².
________________________________
Ques: .If the curved surface area of a right circular cylinder of height 14 cm is 88 cm2, and then diameter of base of cylinder is
Solution:-
Curved surface area = 2 × 22/7 × r × 14
➝ 88 = 44 × 2 × r
➝ 88 = 88 × r
➝ r = 88/88
➝ r = 1
Radius = 1 cm.
Diameter = 1 × 2 = 2
Therefore,
Diameter of base of cylinder is 2 cm.
________________________________
Ques : Formula to calculate T.S.A of a cone is
Answer :-
Total surface area of cone = πr² + πrl
Where,
- r is radius.
- l is slant height.
_______________________________
Ques : If the height of a cone is 16 cm and radius of base is 12 cm, then slant height is
Solution:-
Slant height of cone :
➝ √(16)² + (12)²
➝ √256 + 144
➝ √400
➝ 20
Therefore,
Slant height of cone is 20 cm
________________________________
Ques : If h = 10 cm and radius of base is 24 m, then height will be
Solution:-
➝ √(24)² + (10)²
➝ √576 + 100
➝ 26
Therefore,
Slant height of cone is 26 cm.
________________________________
Ques : If radius of a cone is 7cm and height is 24cm, then slant height will be
Solution:-
Slant height of cone :
➝ √(7)² + (24)²
➝ √49 + 576
➝ √625
➝ 25
Therefore,
Slant height of cone is 25 cm.
________________________________
Ques : The curved surface area of a cone of slant height 10cm and radius of base 7 cm will be
Solution:-
Curved surface area of cone :
➝ 22/7 × 7 × 10
➝ 22 × 10
➝ 220
Therefore,
Curved surface area of cone is 220 cm².
________________________________
Formulas used :
• Heron's formula that is
Area of triangle = √s(s -a)(s - b)(s - c)
Where,
- s is semi-perimeter.
- a, b, and c are sides of triangle
• Semi-perimeter = Perimeter of triangle/2
• Perimeter of triangle = Sum of all sides of triangle.
• Area of triangle = 1/2 × base × height.
• Surface area of cube = 6(edge)²
• Curved surface area of cylinder = 2 πrh
Where, r is radius and h is height of cylinder.
• Diameter = 2 × radius
• Slant height of cone = √ r² + h²
• Curved surface area of cone = πrl
Where, r is radius and l is slant height of cone.
Answer:
Solution :-
❶
Firstly let's find its semiperimeter by herons formula
Area = √s(s-a)(s-b)(s-c)
Area = √20(20-8)(20-15)(20-17)
Area = √20 × 12×5×3
Area = 2×2×5×3
Area = 60 cm²
❷
Semiperimeter = 3+4+5/2
Semiperimeter = 12/2
Semiperimeter = 6
Area = √6(6-3)(6-4)(6-5)
Area = √6 ×3×2×1
Area = 2 × 3
Area = 6 cm²
❸
Let x be all sides
x + x + x = 30
3x = 30
x = 30/3
x = 10
Now,
Let's find semiperimeter
semiperimeter = 30/2
Semiperimeter = 15
Area = √15(15-10)(15-10)(15-10)
Area = √3.87 × 5 × 5 × 5
Area = 245 (approx)
❹
Let the height and base be x
100 = ½ × x × x
100 = x²/2
100 × 2 = x²
200 = x²
√200 = x
x = 14.14 cm
Now,
By using Pythagoras theorem
H² = (14.14)²+ (14.14)²
H² = 399.88
H = √399.88
H = 19.99
Perimeter = 14.14 + 14.14 + 19.99 = 48.27
❺
TSA = 2(lb+bh+lh)
TSA = 2(5×3+3×4+5×4)
TSA = 2(15+12+20)
TSA = 2 × 47
TSA = 94
❻
TSA = 6s²
TSA = 6 × 10²
TSA = 6 × 100
TSA = 600
❼
CSA = πrl
88 = 22/7 × r × 14
88 = 22 × 2r
88 = 44r
r = 2
Diameter = 2r = 2× 2 = 4cm
❽
TSA (cone) = πr² + πrl
❾
Slanght height = √16² + 12²
Slanght height = √256 + 144
Slanght height = √400 = 20
❶⓿
Slanght height = √24² + 10²
Slanght height = √576 + 100
Slanght height = 26
❶❶
Slanght height = √7²+24²
Slanght height = √49 + 576
Slanght height = √625 = 25
❶❷
CSA = πrl
CSA = 22/7 × 7 × 10
CSA = 22 × 10
CSA = 220 cm