Test: Introduction To Heron's Formula - Class 9 MCQ

Let coefficient of ratios be X
a = 3x
b = 4x
c = 5x

Using Heron's Formula,
s = (3+4+5)x/2 = 12x/2 = 6x



Now,
a = 5*3 = 15 cm
b = 5*4 = 20 cm
c = 5*5 = 25 cm

Perimeter = 15+20+25 = 60 cm

Let
a = 15cm
b = 25cm
c = 14cm
Semi-Perimeter = a + b + c / 2
= 15 + 25 + 14 / 2
= 54 / 2
= 27 cm.
Area Through Heron's Formula
= √s(s-a)(s-b)(s-c)
= √27(27-15)(27-25)(27-14)
= √27 × 12 × 2 × 13
= √3 × 3 × 3 × 2 × 2 × 3 × 2 × 13
= 3 × 3 × 2 × √2 × √13
= 18 √26

⇒ AB + BC + CA = 36 cm

⇒ 3x + 4x + 5x = 36 cm

⇒ 12x = 36

⇒ x = 3

∴ AB = 3x = 9

BC = 4x = 12

CA = 5x = 15

Now, (AB)² + (BC)²

=  (9)² + (12)²

=  81 + 144

=  255

= (AC)²

Δ ABC is a right angle triangle and ∠B is right angle.

Area of Δ ABC = 1/2 (AB x BC)

= 1/2 (9 x12)

= 54 cm²

s = a + b + c /2 = 12 + 9 + 15 /2 = 18cm

Area of PRS = √s(s - a)(s - b)(s - c) = 54 cm^2

s = a + b + c /2 = 7 + 6 + 9 /2 = 11 cm

Area of RPQ = √s(s - a)(s - b)(s - c) = 20.98 cm^2

Area of Quadrilateral = Area of RPQ + Area of PRS

= 54 + 20.98

= 74.98 cm^2

 Area of isosceles right angled triangle =  1/2 *b*h

Here if it is isosceles right angled triangle ,

Then perpendicular = base = x

So, area =1/2 *x*x

  400 = x²

  x² = 400

  x =  = 20 cm

  so   P = B = 20

  By Pythagoras theorem ,

  H² = P² +B²

  = (20)²+(20)²

  = 400 + 400

  = 800

H = 20 root 2 cm

Perimeter = 20 + 20 + 20 root 2

  = 40+20 root 2

  = 68.28 cm

area of actual triangle that he had to make

s = (5+5+6)/2

s= 8

a = √(8 x 3 x 3 x 2)

a₁= 12 sq units

area of triangle that the carpenter made

s = (5+5+8)/2

s = 9

a = √(9 x 4 x 4 x 1)

a₂ = 12 sq units

so difference a₁ - a₂ = 12 - 12 =0 sq units

Given that the sides of the triangle are in the ratio 5:12:13 and the perimeter is 60 cm, we can follow these steps:

1. Find the actual side lengths: Let the common ratio factor be xxx.
   So, the sides of the triangle are 5x, 12x, and 13x.

   The perimeter is the sum of the sides:

   5x+12x+13x=60   

           30x=60

           x=2

1. Therefore, the sides of the triangle are:

   5x=10 cm,12x=24 cm,13x=26 cm.

2. Check if it's a right-angled triangle: The sides are in the ratio 5:12:13, which is a well-known Pythagorean triplet, meaning it is a right-angled triangle.

3. Find the area: For a right-angled triangle, the area is:

   Area=12×Base×Height

   Here, the base is 10 cm, and the height is 24 cm:

   Area=12×10×24=120 cm^2

Thus, the correct answer is:

1. 120 cm².

The perimeter of a triangular field = 540m

Let the sides are 25x , 17x , 12 x

Perimeter of a ∆ = sum of three sides

25x + 17x + 12x = 540

 54x = 540

 x = 10 

1st side (a) - 25x = 25×10= 250m 

2nd side(b)= 17x = 17×10= 170m

3rd side (c)= 12x = 12 × 10 =120m

Semi - perimeter ( S) = a+b+c/2 

= (250 + 170+120)/2 = 540/2 = 270 m

 Area of the ∆= √ S(S - a)(S - b)(S - c)

[By Heron’s Formula]

= √ S(S - 250)(S - 170)(S - 120) 

= √ 270(270 - 250)(270 - 170)(270 - 120) 

= √ 270× 20×100×150 

= √ 81000000 

Area of the ∆= 9000 m²

Hence, the Area of the ∆= 9000 m²

let each side be a

so perimeter is 3a = 18x

a=6x

so area = √3a²/4

=√3*36x²/4

=9x²√3 sq. units

 Suppose that the sides, in metres, are 3x, 5x and 7x 

Then, we know that 3x + 5x + 7x = 300 (perimeter of the triangle) 

Therefore, 15x = 300, which gives x = 20. 

So the sides of the triangle are 3 × 20 m, 5 × 20 m and 7 × 20 m 

i.e., 60 m, 100 m and 140 

= 1500 √(3) = 2598.07621135

By using herons formula 
We get s=13+14+15/2
S=21cm
Now area=√s(s-a) (s-b) (s-c) 
√21(21-13)(21-14)(21-15)
√21(8)(7)(6)
√21(336)
√7056
84 cm² is the area.