Test: Heron's Formula- 1 - Class 9 MCQ

are of triangle = 1/2 x base x height
                       = 1/2 x 8 x 10
                       =  40

It is given that the sides of a triangle are in the ratio 3 : 4 : 5.

Let the length of sides are 3x, 4x and 5x.

It's perimeter is 36 cm.

3x+4x+5x = 36

12x = 36

x = 3

The value of x is 3. The length of sides are 9, 12, 15.

It is an right angled triangle because the sum of squares of two smaller sides is equal to the square of larger sides.

9² + 12² = 15²

81 + 144 = 225

Here, the length of hypotenuse is 15 cm.

The area of triangle is

1/2 x 9 x12

54 cm²

area of a triangle = base x height/2

for a right angled isosceles triangle, base = height

so area = base²/2

8 x2 = base²

base = height = 4 cm

hypotnuese = √base² + height² = √2base²

= √2 x 8 x 2

= √32 cm

Let the height of triangular field be h metres.

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


 

Solution:

• Given that the base of the right triangle is 8 cm and the hypotenuse is 10 cm.


• Using the Pythagorean theorem, we can find the height of the triangle:


• Height = sqrt(hypotenuse² - base²) = sqrt(10² - 8²) = sqrt(100 - 64) = sqrt(36) = 6 cm


• Now, we can calculate the area of the triangle using the formula: Area = 0.5 * base * height


• Area = 0.5 * 8 * 6 = 24 cm²


 

Therefore, the area of the right triangle is 24 cm², which corresponds to option A.

Solution :- Area of a triangle of sides a , b & c is : √( s(s-a)(s-b)(s-c) )

where s=(a+b+c)/2

ln given triangle , s=(20+12+16)/2=48/2=24

Thus area of given triangle = √(24(24-20)(24-12)(24-16))

=√(24×4×12×8)

=√9216

=96

Thus area of triangle of sides 20cm ,12cm and 16cm is 96cm²

Area of  triangle:

Perimeter of Triangle:

Perimeter of a closed figure is the sum of lengths of all of its sides.
Given, sides of a triangle are 3 cm, 4 cm and 5 cm.
Then, its perimeter = (3 + 4 + 5) cm
= 12 cm

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

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

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.

and, equating this with the basic area formula,

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