Free Heron's Formula 02 Practice Test - 9th Grade 

Question 1

A roof is made by using 20 triangular rocks which are equally divided in 4 different colours.The sides of each piece are 2 m, 5 m and 5 m. What is the area covered by each colour?

A.

86m2

B.

106m2

C.

126m2

D.

166m2

SOLUTION

Solution : B

S=Perimeter2 =(2+5+5)2=6 m

Area of the triangle= s(sa)(sb)(sc)
 =6(62)(65)(65)
 =6×4×1×1
 =26 m2

Since there are 20 rocks of 4 colours, number of rocks of each colour =204=5
`
Therefore the area covered by each type of rock will be =26×5 =106 m2

Question 2

Find the area of a triangle, two sides of which are 8 cm and 12 cm and the perimeter is 38 cm .

A.

163 cm2

B.

64 cm2

C.

1463 cm2

D.

243 cm2

SOLUTION

Solution : C

Third side = perimeter - (sum of other two sides) = 38 - 20 = 18 cm
Semi perimeter of a triangle = Perimeter2=382=19 cm

According to heron's formula,
Area of a triangle =s(sa)(sb)(sc)=19(1912)(198)(1918)=19×7×11×1=1463 cm2

Question 3

If the height of a triangle is 10 cm and its area is 40 cm2, then the base of the triangle is 6 cm.

A.

True

B.

False

SOLUTION

Solution : B

Area of a triangle=12×base×height=40

base=40×2height

base=8010=8 cm
 

Question 4

If two sides of a triangle are 8 cm and 6 cm and its perimeter is 26 cm. Find the third side of the triangle.

A.

12 cm

B.

10 cm

C.

9 cm

D.

11 cm

SOLUTION

Solution : A

Perimeter = a +b +c (i.e. the sum of length of all the sides)
Third side = 26 - 8 - 6
                                          = 26 - (8+6)
                                          = 26 - 14
                                          =12

Third side = 12 cm

Question 5

What is the area of an equilateral triangle whose side is 8 cm?

A.

163  cm2

B.

16 cm2

C.

3  cm2

D.

32 cm2

SOLUTION

Solution : A

 Semi perimeter of a triangle (s) = a+b+c2
8+8+82=12cm
Area of a triangle = s(sa)(sb)(sc)
=12(128)(128)(128)
=12×4×4×4
=3×4×4×4×4
= 163 cm2

Question 6

What is the area of a right angled triangle if its sides are 6cm, 8cm and 10cm?

A.

26 cm2

B.

24 cm2

C.

36 cm2

D.

40 cm2

SOLUTION

Solution : B

Area of triangle = 12×base×height


The hypotenuse is the longest side of a right-angled triangle. Hence, the base and the height are 6 cm and 8 cm.
area=12×6×8
=24 cm2

Question 7

A triangular advertisement board has sides 11m, 15m and 6m. If the advertisement yields an earning of Rs1000/m2per month. In 3 and half years, the company will earn A2. Find the value of A ___.

SOLUTION

Solution : Using the Heron's formula, Area (A)= s(sa)(sb)(sc).
Semi-Perimeter (s) = a+b+c2 =11+15+62
=16 m
A=16(1611)(1615)(166) =202 m2
3 and a half years = 12 x 3 + 6 = 42 months.
Earning in 3 and a half years =202×1000×42 =Rs.8400002
The value of A is 840000.

Question 8

Which of these cannot be the area of a right-angled triangle if the hypotenuse of the  triangle is 13 m and one other side is 5 m?

A. 32.5 m2
B. 78 m2
C. 30 m2
D. 60 m2

SOLUTION

Solution : A, B, and D


Let the triangle be ABC where AC = 13 m and BC = 5 m

Using Pythagoras theorem we can write,

AB2=AC2BC2

              
Therefore, AB2=13252AB=12 m
Area of triangle = 12×base×height=12×5×12=30 m2

Question 9

The edges of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting it at the rate of 25 paise per cm2 is Rs. ___

SOLUTION

Solution :

a=6 cm,b=8 cm,c=10 cm
s=a+b+c2
s=6+8+102
=12cm
A=s(sa)(sb)(sc)
A=12(126)(128)(1210)
=24 cm2
The cost of painting it =24×25=600 paise
=Rs. 6

Question 10

The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order) respectively and the angle between the first two sides is a right angle. Find its area.

A. 18+242 cm2
B. 18+246 cm2
C. 36+242 cm2
D. 36+246 cm2

SOLUTION

Solution : B


Area of ABC=12×base×height
=12×6×8
=24cm2

Diagonal AC = BC2+AB2=10 cm

For ACD,
s=10+12+142=18
Area =s(sa)(sb)(sc)
=18(1810)(1812)(1814)
=18×8×6×4
=246 cm2

area of quadrilateral 
=18+246 cm2