Free Heron's Formula 01 Practice Test - 9th Grade 

Question 1

Areas of triangles with which given sides can be found using Heron's formula?

A. 7 cm, 6 cm, 5 cm
B. 4 cm, 4 cm, 8 cm
C. 12 cm, 18 cm, 7 cm
D. 2 cm, 3 cm, 5 cm

SOLUTION

Solution : A and C

In a triangle, the sum of the lengths of any two sides is greater than the length of the third side. This criterion is met by only options 7 cm, 6 cm, 5 cm and 12 cm, 18 cm, 7 cm.

Question 2

The sides of a triangle are 35 cm, 54 cm and 61 cm. The length of its longest altitude is

A.

245 cm

B.

485 cm

C.

565 cm

D.

108 cm

SOLUTION

Solution : A

Given, a=35 cm; b=54 cm; c=61 cm.

Semi-perimeter
s=a+b+c2=35+54+612=75 cm

A()=s(sa)(sb)(sc)   =75(7535)(7554)(7561)   =75×40×21×14
A()=4205 cm2.

Now, A()=12×base×altitude

The altitude corresponding to shortest base is longest.
The shortest side = 35 cm.

4205=12×35×altitudealtitude=2×420535=245 cm.

Question 3

Heron's formula cannot be used in finding the area of quadrilateral.

A.

True

B.

False

SOLUTION

Solution : B

The given statement is false. Heron's formula can be used in finding the area of quadrilateral as quadrilateral is formed by two triangles.

Question 4

If an isosceles triangle has a  perimeter of 18 cm and base 8 cm, then its area is ___ .

A.

45 cm2

B.

68 cm2

C.

12 cm2

D.

20 cm2

SOLUTION

Solution : C

Let each of its equal sides be x.
Then, 2x+8=18 x=5cm
Hence, the sides are 5 cm, 5 cm and 8 cm.
Using the Heron's formula,
Area(A)=s(sa)(sb)(sc)
where s = semi- perimeter and a,b,c are the three sides of the triangle.
SemiPerimeter(s)=perimeter2
=182 =9 cm
A=9 (95)(95)(98)
=9×4×4×1
=144 cm2
=12 cm2

Question 5

What is the area(in cm2) of this triangle?
___

SOLUTION

Solution : (Area of triangle) =12×base×height

The three sides are (10cm, 6cm and 8cm).

102=62+82

The sides satisfy Pythagoras theorem. Hence, this is a right-angled triangle, with the sides 6cm and 8cm being the base and the height, respectively.

Hence, Area=12×6×8 cm2
=24 cm2

Question 6

The area of a square is 36 cm2. If the side of the square is equal to the side of an equilateral triangle, then find the area of the triangle.

A.

93 cm

B.

63 cm

C.

73 cm

D.

43 cm

SOLUTION

Solution : A

Area of square = Side2
 36 = Side2
Side = 6 cm

Side of triangle = 6 cm

Semi perimeter of the triangle s =6+6+62cm=9 cm

Area of the triangle =s(sa)(sb)(sc)
=9(96)3
=243
=93 cm

Question 7

If the sides of a triangle are 4 cm, 6 cm and 6 cm, then find the height corresponding to the smallest side.

A.

82

B.

42

C.

162

D.

72

SOLUTION

Solution : B

S=Perimeter2=(a+b+c)2=(4+6+6)2=8 cm

Area of the triangle =s(sa)(sb)(sc) = 8(84)(86)(86) = 128=82cm2

Area of a triangle is also find using 12×base×height

 82=12×4×height

height=42cm.

Question 8

If the area of an equilateral triangle is 93 cm2. What is the length of the side of the triangle.

A.

8 cm

B.

9 cm

C.

18 cm

D.

6 cm

SOLUTION

Solution : D

Let the side of the equilateral triangle = a cm.

Semi Perimeter, s = 3a2

sa=3a2a=3a2a2=a2

Now by heron’s formula

Area of the triangle= s(sa)(sb)(sc) 

93=3a2×a2×a2×a2

243=3a416

Squaring both sides,
243=3a416

a4=81×16

a=6 cm

Question 9

Find the area of a parallelogram ABCD in which AB = 3 cm and BC = 4 cm and AC = 5 cm using Heron's formula.

A.

6 cm2

B.

12 cm2

C.

18 cm2

D.

24 cm2

SOLUTION

Solution : B

Since ABCD is a parallelogram, its opposite sides will be equal.

AB = 3 cm, BC = 4 cm and AC = 5 cm

Now in triangle ABC,

Area of the triangle using heron’s formula,

s=Perimeter2=3+4+52=6 cm

Area = s(sa)(sb)(sc)
6(63)(64)(65)
= 36
= 6 cm2

Area of the parallelogram = 2 (Area of triangle ABC)

The area of the parallelogram=2×6=12 cm2

Question 10

12×base×height is the formula for which type of triangles? 

A.

Right angled triangles

B.

Isosceles triangles

C.

Equilateral triangles

D.

Scalene triangles

SOLUTION

Solution : A, B, C, and D

Using formula 12×base×height we can find the area of any triangle irrespective of its type if we know the base and the height of the given triangle.