Heron S Formula(9th Grade > Mathematics ) Questions and answers for exam Preparation

9th Grade > Mathematics

HERON S FORMULA MCQs

Total Questions : 51 | Page 2 of 6 pages

Question 11. 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

Discuss Question

Answer: Option B. -> 24 cm2
:
B
Area of triangle =

\frac{1}{2} \times b a s e \times h e i g h t

The hypotenuse is the longest side of a right-angled triangle. Hence, the base and the height are 6 cm and 8 cm.

∴ area = \frac{1}{2} \times 6 \times 8

= 24 c m^{2}

Question 12. 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

Discuss Question

Answer: Option C. -> √1463 cm2
:
C
Third side = perimeter - (sum of other two sides) =38 - 20 =18 cm
Semi perimeter of a triangle=

\frac{Perimeter}{2} = \frac{38}{2} = 19 cm

According to heron's formula,
Area of a triangle

= \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{19 (19 - 12) (19 - 8) (19 - 18)} = \sqrt{19 \times 7 \times 11 \times 1} = \sqrt{1463} {cm}^{2}

Question 13. 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

c m^{2}

is Rs. ___

Discuss Question

a = 6 c m, b = 8 c m, c = 10 cm

s = \frac{a + b + c}{2}

s = \frac{6 + 8 + 10}{2}

= 12 c m

A = \sqrt{s (s - a) (s - b) (s - c)}

A = \sqrt{12 (12 - 6) (12 - 8) (12 - 10)}

= 24 {cm}^{2}

The cost of painting it

= 24 \times 25 = 600

paise

= Rs. 6

Question 14. 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

Discuss Question

Answer: Option A. -> 12 cm
:
A
Perimeter = a +b +c (i.e. the sum of length of all the sides)

\Rightarrow

Third side = 26 - 8 - 6
= 26 - (8+6)
= 26 - 14
=12
Third side = 12 cm

Question 15. 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

Discuss Question

Answer: Option B. -> 10√6m2
:
B

S = \frac{P e r i m e t e r}{2}

= \frac{(2 + 5 + 5)}{2} = 6 m

Area of the triangle=

\sqrt{s (s - a) (s - b) (s - c)}

= \sqrt{6 (6 - 2) (6 - 5) (6 - 5)}

= \sqrt{6 \times 4 \times 1 \times 1}

= 2 \sqrt{6}

m^{2}

Since there are 20 rocks of 4 colours, number of rocks of each colour

= \frac{20}{4} = 5

`
Therefore the area covered by each type of rock will be

= 2 \sqrt{6} \times 5

= 10 \sqrt{6}

m^{2}

Question 16. A triangular advertisement board has sides 11m, 15m and 6m. If the advertisement yields an earning of

R s 1000 / m^{2}

per month. In 3 and half years, the company will earn

A \sqrt{2}

. Find the value of A ___.

Discuss Question

:
Using the Heron's formula, Area (A)=

\sqrt{s (s - a) (s - b) (s - c)}

.
Semi-Perimeter (s) =

\frac{a + b + c}{2}

= \frac{11 + 15 + 6}{2}

= 16 m

∴ A = \sqrt{16 (16 - 11) (16 - 15) (16 - 6)}

= 20 \sqrt{2} m^{2}

3 and a half years = 12 x 3 + 6 = 42 months.

∴

Earning in 3 and a half years

= 20 \sqrt{2} \times 1000 \times 42

= R s . 840000 \sqrt{2}

The value of A is 840000.

Question 17. 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.

18+24√2 cm2
18+24√6 cm2
36+24√2 cm2
36+24√6 cm2

Discuss Question

Answer: Option B. -> 18+24√6 cm2
:
B

The Sides Of A Quadrilateral ABCD Are 6 Cm, 8 Cm, 12 Cm And ...

Area of

△ A B C = \frac{1}{2} \times base \times h e i g h t

= \frac{1}{2} \times 6 \times 8

= 24 c m^{2}

Diagonal AC =

\sqrt{B C^{2} + A B^{2}} = 10 c m

For

△ A C D,

s = \frac{10 + 12 + 14}{2} = 18

Area

= \sqrt{s (s - a) (s - b) (s - c)}

= \sqrt{18 (18 - 10) (18 - 12) (18 - 14)}

= \sqrt{18 \times 8 \times 6 \times 4}

= 24 \sqrt{6} c m^{2}

∴

area of quadrilateral

= 18 + 24 \sqrt{6} c m^{2}

Question 18. The sides of a triangle are 5 cm, 6 cm and 3 cm long. The area of the triangle is

\sqrt{A} c m^{2}

. Then the value of A is ___.

Discuss Question

s = \frac{5 + 6 + 3}{2}

=7 cm
Area

A = \sqrt{s (s - a) (s - b) (s - c)}

= \sqrt{7 (7 - 5) (7 - 6) (7 - 3)}

= \sqrt{7 \times 2 \times 1 \times 4}

= \sqrt{56} c m^{2}

The value of A is 56.

Question 19. The perimeter of a triangular field is 420 m and its sides are in the ratio 6:7:8. Find the area of the triangular field.

420√15 m2
1050√7 m2
1050√15 m2
2100√15 m2

Discuss Question

Answer: Option D. -> 2100√15 m2
:
D
Let the sides be 6x, 7x, 8x.
Then,

6 x + 7 x + 8 x = 420

\Rightarrow 21 x = 420

x = 20

∴

the sides are:
a =120 m, b =140 m, c =160 m
Semi-perimeter s =

\frac{420}{2} = 210 m

A = \sqrt{s (s - a) (s - b) (s - c)}

= \sqrt{210 (210 - 120) (210 - 140) (210 - 160)}

= \sqrt{210 \times 90 \times 70 \times 50}

= 2100 \sqrt{15} m^{2}

Question 20. The area of a rectangle is

96 c m^{2}

and its length is

12 c m

. The perimeter of the rectangle is same as the perimeter of a triangle. Find the semi perimeter of the triangle.

10 cm
40 cm
20 cm
80 cm

Discuss Question

Answer: Option C. -> 20 cm
:
C
Area of a rectangle = length

\times

breadth

\Rightarrow

96 = 12

\times

breadth
Breadth = 8 cm
Perimeter of the rectangle = 2(length + breadth) = 2(12+8) cm = 40 cm
Perimeter of the rectangle = Perimeter of the triangle
Therefore, semi perimeter of the triangle