Test: Heron's Formula- 1


25 Questions MCQ Test Mathematics (Maths) Class 9 | Test: Heron's Formula- 1


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This mock test of Test: Heron's Formula- 1 for Class 9 helps you for every Class 9 entrance exam. This contains 25 Multiple Choice Questions for Class 9 Test: Heron's Formula- 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Heron's Formula- 1 quiz give you a good mix of easy questions and tough questions. Class 9 students definitely take this Test: Heron's Formula- 1 exercise for a better result in the exam. You can find other Test: Heron's Formula- 1 extra questions, long questions & short questions for Class 9 on EduRev as well by searching above.
QUESTION: 1

The area of a triangle with base 8 cm and height 10 cm is

Solution:
QUESTION: 2

The sides of a triangle are in the ratio of 3 : 4 : 5. If its perimeter is 36 cm, then what is its area?

Solution: We know that area of parallelogram = base × height = 60 × 24 cm =1440 cm^2
QUESTION: 3

An isosceles right triangle has area 8 cm2. The length of its hypotenuse is

Solution:
QUESTION: 4

The cost of turfing a triangular field at the rate of Rs. 45 per 100 m2 is Rs. 900. If the double the base of the triangle is 5 times its height, then its height is

Solution:

Let the height of triangular field be h metres.

It is given that 2 x (base) = 5 × (Height)


 

QUESTION: 5

The area of the the triangle having sides 1 m, 2 m and 2 m is :

Solution:
Through Heron’s formula =
S=Perimeter/2
S=5/2
Let a=1m, b=2m ,c=2m

Formulae= Root(s(s-a)(s-b)(s-c))
=root(5/2(5/2 - 1)(5/2 - 2)(5/2 - 2))
=root(5/2*3/2*1/2*1/2)
=root(15/16)
=root(15)/4
QUESTION: 6

The base of a right triangle is 8 cm and hypotenuse is 10 cm. Its area will be

Solution:
QUESTION: 7

The area of a triangle whose sides are 12 cm, 16 cm and 20 cm is

Solution: S = 12+16+20/2
= 24
Area of a triangle = √s(s-a)(s-b)(s-c)
= √24(24-12)(24-16)(24-20)
= √24 x 12 x 8 x 4
=6 x 4 x 4
= 96 cm^2
QUESTION: 8

The difference of semi-perimeter and the sides of △ABC are 8, 7 and 5 cm respectively. Its semi-perimeter ‘s’ is

Solution:

QUESTION: 9

The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then the area of the triangle is

Solution:
QUESTION: 10

The sides of a triangular flower bed are 5 m, 8 m and 11 m. the area of the flower bed is

Solution:
QUESTION: 11

A triangle ABC in which AB = AC = 4 cm and ∠A = 90o, has an area of

Solution:
QUESTION: 12

The perimeter and area of a triangle whose sides are of lengths 3 cm, 4 cm and 5 cm respectively are

Solution:
QUESTION: 13

The perimeter of a rhombus is 20 cm. One of its diagonals is 8 cm. Then area of the rhombus is

Solution:

Given, Perimeter of a rhombus = 20 cm
Perimeter of a rhombus = 4*side
Hence, side = 20/4 = 5 cm.
Now, we know that the diagonals of a rhombus bisect each other at right angles (90 degree).
Hence 'a right angled triangle can be visualised with 'side' as the hypotenuse'.

diagonal length = 8 cm
Half the length (since diagonal bisects each other) = 8/2 = 4 cm
(d/2)2 + (d1/2)^2 = 52
4^2 + (d1/2)2 = 52
(d1/2)2 = 9
d1/2 = 3
d1 = 3*2 = 6 cm
Hence other diagonal = 6 cm.

Area = 1/2 * d1*d = 1/2 * 8 * 6 = 24 cm2

QUESTION: 14

The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is

Solution:

QUESTION: 15

The length of the sides of a triangle are 5 cm, 7 cm and 8 cm. Area of the triangle is :

Solution:
QUESTION: 16

If the side of an equilateral triangle is 4 cm, then its area is

Solution:
QUESTION: 17

The area of a triangle whose sides are 15 cm, 8 cm and 19 cm is

Solution:
QUESTION: 18

If one side and one diagonal of a rhombus and 20 m and 24 m, then its area =

Solution:

The rhombus has a perimeter of 80m.
A rhombus has 4 equal sides, each would be 20m.
One of the diagonals is 24m...this is bisected by the other diagonal into 12m segs.
The diagonals of a rhombus intersect at right angles. Use the Pythagorean Theorem to find the length of the other diagonal...
12² + b² = 20²
144 + b² = 400
b² = 256
b = 16
This is of course the length of just one segment, the diagonal is 32m.
Now we can use our area formula with the diagonals.
A = ½d¹d²
A = ½(24)(32)
A = 384 m²

QUESTION: 19

The edges of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting it at the rate of 9 paise per cmis

Solution:
QUESTION: 20

Area of an equilateral triangle of side 2 cm is :

Solution:
QUESTION: 21

If the perimeter of an equilateral triangle is 24 m, then its area is

Solution:

Given:
Area of equilateral ∆ = 16√3 m²
Let the side of equilateral ∆ = a m
Area of equilateral ∆ = (√3/4)a²
16√3 = (√3/4)a²
16√3 × (4/√3) = a²
64 = a²
a = √64
a =√ 8 × 8 = 8 m
Side of equilateral ∆ = 8 m
Perimeter of equilateral ∆ = 3 × side
Perimeter of equilateral ∆ = 3 × 8 = 24 m.
Hence, the perimeter of equilateral ∆ is 24 m.

QUESTION: 22

Each of the equal sides of an isosceles triangle is 2 cm greater than its height. If the base of the triangle is 12 cm, then its area is

Solution:

and, equating this with the basic area formula,

i.e. A = 1/2.h.b

QUESTION: 23

The product of difference of semi-perimeter and respective sides of △ABC are given as 13200 m3. The area of △ABC, if its semi-perimeter is 132m, is given by

Solution:
QUESTION: 24

The area of equilateral triangle of side ‘a’ is 4√3 cm2. Its height is given by

Solution:
QUESTION: 25

The base of a right triangle is 8 cm and hypotenuse is 10 cm. Its area will be :

Solution:
Given, 

base= 8cm

hypotenuse= 10cm  

First, let us find the height of the triangle:  

Let height = h  

By applying pythagorean theorem  

h 2 + 8 2 = 10 2

h 2 + 64 = 100

h 2 = 100 - 64

h 2 = 36

h = 6cm  

Now let us find the area,  

Area = 1/2 x b x h    

= 1/2 x 8 x 6  

= 1/2 x 48  

= 24cm 2

Therefore, area of the right triangle = 24cm 2

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