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?
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
24√5 cm
48√5 cm
56√5 cm
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(s−a)(s−b)(s−c) =√75(75−35)(75−54)(75−61) =√75×40×21×14
A(△)=420√5 cm2.
Now, A(△)=12×base×altitude
The altitude corresponding to shortest base is longest.
The shortest side = 35 cm.
∴420√5=12×35×altitudealtitude=2×420√535=24√5 cm.
Question 3
Heron's formula cannot be used in finding the area of quadrilateral.
True
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 ___ .
45 cm2
68 cm2
12 cm2
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(s−a)(s−b)(s−c)
where s = semi- perimeter and a,b,c are the three sides of the triangle.
Semi−Perimeter(s)=perimeter2
=182 =9 cm
A=√9 (9−5)(9−5)(9−8)
=√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.
9√3 cm
6√3 cm
7√3 cm
4√3 cm
SOLUTION
Solution : A
Area of square = Side2
⇒ 36 = Side2
⇒ Side = 6 cmSide of triangle = 6 cm
Semi perimeter of the triangle s =6+6+62cm=9 cm
Area of the triangle =√s(s−a)(s−b)(s−c)
=√9(9−6)3
=√243
=9√3 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.
8√2
4√2
16√2
7√2
SOLUTION
Solution : B
S=Perimeter2=(a+b+c)2=(4+6+6)2=8 cm
Area of the triangle =√s(s−a)(s−b)(s−c) = √8(8−4)(8−6)(8−6) = √128=8√2cm2
Area of a triangle is also find using 12×base×height
⇒8√2=12×4×height
⇒height=4√2cm.
Question 8
If the area of an equilateral triangle is 9√3 cm2. What is the length of the side of the triangle.
8 cm
9 cm
18 cm
6 cm
SOLUTION
Solution : D
Let the side of the equilateral triangle = a cm.
Semi Perimeter, s = 3a2
s−a=3a2−a=3a−2a2=a2Now by heron’s formula
Area of the triangle= √s(s−a)(s−b)(s−c)
⇒9√3=√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.
6 cm2
12 cm2
18 cm2
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(s−a)(s−b)(s−c)
= √6(6−3)(6−4)(6−5)
= √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?
Right angled triangles
Isosceles triangles
Equilateral triangles
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.