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?
8√6m2
10√6m2
12√6m2
16√6m2
SOLUTION
Solution : B
S=Perimeter2 =(2+5+5)2=6 m
Area of the triangle= √s(s−a)(s−b)(s−c)
=√6(6−2)(6−5)(6−5)
=√6×4×1×1
=2√6 m2
Since there are 20 rocks of 4 colours, number of rocks of each colour =204=5
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Therefore the area covered by each type of rock will be =2√6×5 =10√6 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 .
16√3 cm2
64 cm2
√1463 cm2
24√3 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(s−a)(s−b)(s−c)=√19(19−12)(19−8)(19−18)=√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.
True
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.
12 cm
10 cm
9 cm
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?
16√3 cm2
16 cm2
√3 cm2
32 cm2
SOLUTION
Solution : A
Semi perimeter of a triangle (s) = a+b+c2
⇒8+8+82=12cm
Area of a triangle = √s(s−a)(s−b)(s−c)
=√12(12−8)(12−8)(12−8)
=√12×4×4×4
=√3×4×4×4×4
= 16√3 cm2
Question 6
What is the area of a right angled triangle if its sides are 6cm, 8cm and 10cm?
26 cm2
24 cm2
36 cm2
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 A√2. Find the value of A
SOLUTION
Solution : Using the Heron's formula, Area (A)= √s(s−a)(s−b)(s−c).
Semi-Perimeter (s) = a+b+c2 =11+15+62
=16 m
∴A=√16(16−11)(16−15)(16−6) =20√2 m2
3 and a half years = 12 x 3 + 6 = 42 months.
∴ Earning in 3 and a half years =20√2×1000×42 =Rs.840000√2
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?
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=AC2−BC2
Therefore, AB2=132–52⇒AB=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(s−a)(s−b)(s−c)
A=√12(12−6)(12−8)(12−10)
=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.
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(s−a)(s−b)(s−c)
=√18(18−10)(18−12)(18−14)
=√18×8×6×4
=24√6 cm2
∴ area of quadrilateral
=18+24√6 cm2