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Test: Heron's Formula- 2 - Class 9 MCQ


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25 Questions MCQ Test Mathematics (Maths) Class 9 - Test: Heron's Formula- 2

Test: Heron's Formula- 2 for Class 9 2024 is part of Mathematics (Maths) Class 9 preparation. The Test: Heron's Formula- 2 questions and answers have been prepared according to the Class 9 exam syllabus.The Test: Heron's Formula- 2 MCQs are made for Class 9 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Heron's Formula- 2 below.
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Test: Heron's Formula- 2 - Question 1

Find the length of each side of an equilateral triangle having area of 9 root 3 cm square

Detailed Solution for Test: Heron's Formula- 2 - Question 1

Given area of equilateral triangle = 9√3 cm²

Also,

Area of the equilateral triangle = √3/4 × (side)²

So,

√3/4 × (side)² = 9√3

side² = (9√3 × 4)/√3

side² = 36

side = √36

side = 6cm

Test: Heron's Formula- 2 - Question 2

If the height of a parallelogram having 500 cm2 as the area is 20 cm, then its base is of length

Detailed Solution for Test: Heron's Formula- 2 - Question 2
  • The area (A) of a parallelogram is calculated using the formula: A = base (b) x height (h).
  • Here, the area is 500 cm² and the height is 20 cm.
  • To find the base, rearrange the formula: b = A / h.
  • Substituting the values: b = 500 cm² / 20 cm.
  • This simplifies to b = 25 cm.
  • Thus, the length of the base is 25 cm.
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Test: Heron's Formula- 2 - Question 3

The area of an isosceles right angled triangle of equal side 30 cm, is given as

Detailed Solution for Test: Heron's Formula- 2 - Question 3

as it is right angle triangle so
area =1/2 x base x height
        =1/2 x 30 x 30
        = 450 cm2

Test: Heron's Formula- 2 - Question 4

If one side of a scalene △ is doubled then area would be increased by

Detailed Solution for Test: Heron's Formula- 2 - Question 4

Test: Heron's Formula- 2 - Question 5

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

Detailed Solution for Test: Heron's Formula- 2 - Question 5

Let's denote the legs of the isosceles right triangle as aaa (both legs are equal) and the hypotenuse as c.

Test: Heron's Formula- 2 - Question 6

If the area of an equilateral triangle is 36√3 cm2, then the perimeter of the triangle is

Detailed Solution for Test: Heron's Formula- 2 - Question 6

√3/4 a2 = 36√3
a2 = 36 x4
a = 6 x 2
a = 12
so P = 3a = 3 x 12 = 36

Test: Heron's Formula- 2 - Question 7

Given the product of diagonals of a rhombus ABCD is 2500 cm2, its area is

Detailed Solution for Test: Heron's Formula- 2 - Question 7

The area (A) of a rhombus can be calculated as:
A = (1/2) * d1 * d2
where d1 and d2 are the lengths of the diagonals of the rhombus.
Given that the product of the diagonals (d1 * d2) is 2500 cm², we can use the formula directly:
A = (1/2) * d1 * d2
Substitute the given value into the formula:
A = (1/2) * 2500
A = 2500 / 2
A = 1250 cm²
Therefore, the area of the rhombus is 1250 cm².

Test: Heron's Formula- 2 - Question 8

The area and length of one diagonal of a rhombus are given as 200 cm2 and 10 cm respectively. The length of other diagonal is

Test: Heron's Formula- 2 - Question 9

In each side of a △ is halved then its perimeter will be decreased by

Detailed Solution for Test: Heron's Formula- 2 - Question 9

Test: Heron's Formula- 2 - Question 10

In ∆ABC, angle A = 30°, side 'b' = 4 units, side 'c' = 6 units. Find the area of ∆ABC,

Detailed Solution for Test: Heron's Formula- 2 - Question 10

Area (∆ABC) = 1/2 × bc × sin A

= 1/2 × 4 × 6 × sin 30º

= 12 × 1/2 (since sin 30º = 1/2)

Area = 6 square units.

Test: Heron's Formula- 2 - Question 11

The edges of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting it at the rate of 70 paise per cm2 is

Test: Heron's Formula- 2 - Question 12

The area of a right angled triangle if the radius of its circumcircle is 3 cm and altitude drawn to the hypotenuse is 2 cm.

Detailed Solution for Test: Heron's Formula- 2 - Question 12

Test: Heron's Formula- 2 - Question 13

The area of one triangular part of a rhombus ABCD is given as 125 cm2. The area of rhombus ABCD is

Test: Heron's Formula- 2 - Question 14

Length of perpendicular drawn on smallest side of scalene triangle is

Detailed Solution for Test: Heron's Formula- 2 - Question 14

Length of perpendicular drawn on smallest side of scalene triangle is always largest.

Test: Heron's Formula- 2 - Question 15

The sides of a triangle are in ratio 3 : 4 : 5. If the perimeter of the triangle is 84 cm, then area of the triangle is :

Test: Heron's Formula- 2 - Question 16

The perimeter of a rhombus is 20 cm. If one of its diagonals is 6 cm, then its area is

Test: Heron's Formula- 2 - Question 17

Area of an isosceles triangle ABC with AB = a = AC and BC = b is

Test: Heron's Formula- 2 - Question 18

Each side of an equilateral triangle measures 10 cm. Then the area of the triangle is

Test: Heron's Formula- 2 - Question 19

Length of perpendicular drawn on longest side of a scale △ is

Test: Heron's Formula- 2 - Question 20

Each side of an equilateral triangle is 2x cm. If x√3 = √48, then area of the triangle is :

Test: Heron's Formula- 2 - Question 21

The area of a rhombus of 96 cm2. If one of its diagonals is 16 cm, then the length of its side is

Detailed Solution for Test: Heron's Formula- 2 - Question 21

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Test: Heron's Formula- 2 - Question 22

The area of quadrilateral ABCD whose diagonals are perpendicular and of lengths 12 cm, 8 cm is

Test: Heron's Formula- 2 - Question 23

The area of a parallelogram whose base is 32 m and the corresponding altitude is 6 m is

Detailed Solution for Test: Heron's Formula- 2 - Question 23

area of parallelogram = base x height

=32 x 6 = 192 m2

Test: Heron's Formula- 2 - Question 24

Semiperimeter of scalene triangle of side k, 2k and 3k is

Test: Heron's Formula- 2 - Question 25

The sides of a triangle are x, y and z. If x + y = 7 m, y + z = 9 m, and z + x = 8 m, then area of the triangle is :

Detailed Solution for Test: Heron's Formula- 2 - Question 25

x + y = 7....(1)
y + z = 9....(2)
z + x = 8....(3)
So, 2(x+y+z) = 24
x+y+z = 12
Now, y + z = 9  then x = 3
z + x = 8 then y = 4
x + y = 7 then z = 5
We can see that, this becomes a right angled triangle.
Hence, Area of triangle = 1/2 * base * perpendicular
= 1/2 * 4 * 3
= 6 sq unit.

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