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

9th Grade > Mathematics

HERON S FORMULA MCQs

Total Questions : 51 | Page 3 of 6 pages

Question 21. A triangular wall having sides 3 m, 5 m and 6 m has to be painted. If the cost of painting

1 m^{2}

of a wall is ₹ 5, what is the cost of painting the whole wall?

₹ 16√13
₹ 10√14
₹ 14√10
₹14

Discuss Question

Answer: Option B. -> ₹ 10√14
:
B
Semi perimeter of a triangle =

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

\Rightarrow

\frac{(3 + 5 + 6)}{2} = 7 m

Area of a triangle =

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

\Rightarrow

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

\Rightarrow

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

\Rightarrow

\sqrt{14 \times 4}

\Rightarrow \sqrt{56} = 2 \sqrt{14} m^{2}

Since,cost of painting

1 m^{2}

of wall is ₹5,
The cost of painting the whole wall

= 5 \times 2 \sqrt{14} = ₹ 10 \sqrt{14} .

Question 22.

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

7 cm, 6 cm, 5 cm
4 cm, 4 cm, 8 cm
12 cm, 18 cm, 7 cm
2 cm, 3 cm, 5 cm

Discuss Question

Answer: Option A. -> 7 cm, 6 cm, 5 cm
:
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 23.

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

$24 \sqrt{5} c m$
$48 \sqrt{5} c m$
$56 \sqrt{5} c m$
$108 c m$

Discuss Question

Answer: Option A. ->

24 \sqrt{5} c m

:
A

Given, $a = 35 cm; b = 54 cm; c = 61 cm .$
Semi-perimeter
$s = \frac{a + b + c}{2} = \frac{35 + 54 + 61}{2} = 75 cm$
$A (△) = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{75 (75 - 35) (75 - 54) (75 - 61)} = \sqrt{75 \times 40 \times 21 \times 14}$
$A (△) = 420 \sqrt{5} {cm}^{2} .$
Now, $A (△) = \frac{1}{2} \times base \times altitude$
The altitude corresponding to shortest base is longest.
The shortest side = 35 cm.
$∴ 420 \sqrt{5} = \frac{1}{2} \times 35 \times altitude altitude = \frac{2 \times 420 \sqrt{5}}{35} = 24 \sqrt{5} cm .$

Question 24.

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

$45 c m^{2}$
$68 c m^{2}$
$12 c m^{2}$
$20 c m^{2}$

Discuss Question

Answer: Option C. ->

12 c m^{2}

:
C

Let each of its equal sides be x.
Then, $2 x + 8 = 18$ $\Rightarrow x = 5 c m$
Hence, the sides are $5 cm, 5 cm and 8 cm$ .
Using the Heron's formula,
$A r e a (A) = \sqrt{s (s - a) (s - b) (s - c)}$
where s = semi- perimeter and a,b,c are the three sides of the triangle.
$S e m i - P e r i m e t e r (s) = \frac{perimeter}{2}$
$= \frac{18}{2}$ $= 9 c m$
$A = \sqrt{9 (9 - 5) (9 - 5) (9 - 8)}$
$= \sqrt{9 \times 4 \times 4 \times 1}$
$= \sqrt{144} c m^{2}$
$= 12 c m^{2}$

Question 25.

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

True
False
$56 \sqrt{5} c m$
$108 c m$

Discuss Question

Answer: Option B. -> False
:
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 26.

What is the area(in ${c m}^{2}$ ) of this triangle?
___

Discuss Question

Answer: Option B. -> False
:

(Area of triangle)

= \frac{1}{2} \times base \times height

The three sides are

(10cm, 6cm and 8cm)

10^{2} = 6^{2} + 8^{2}

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,

A r e a = \frac{1}{2} \times 6 \times 8

{c m}^{2}

= 24 c m^{2}

Question 27.

The area of a square is $36 c m^{2}$ . If the side of the square is equal to the side of an equilateral triangle, then find the area of the triangle.

$9 \sqrt{3} c m$
$6 \sqrt{3} c m$
$7 \sqrt{3} c m$
$4 \sqrt{3} c m$

Discuss Question

Answer: Option A. ->

9 \sqrt{3} c m

:
A

Area of square = $S i d e^{2}$
$\Rightarrow$ 36 = $S i d e^{2}$
$\Rightarrow$ Side = 6 cm

Side of triangle = 6 cm

Semi perimeter of the triangle s $= \frac{6 + 6 + 6}{2} c m = 9 c m$
Area of the triangle $= \sqrt{s (s - a) (s - b) (s - c)}$
$= \sqrt{9 (9 - 6)^{3}}$
$= \sqrt{243}$
$= 9 \sqrt{3} c m$

Question 28.

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

8 $\sqrt{2}$
4 $\sqrt{2}$
16 $\sqrt{2}$
7 $\sqrt{2}$

Discuss Question

Answer: Option B. -> 4

\sqrt{2}

:
B

$S = \frac{P e r i m e t e r}{2} = \frac{(a + b + c)}{2} = \frac{(4 + 6 + 6)}{2} = 8 c m$

Area of the triangle = $\sqrt{s (s - a) (s - b) (s - c)}$ = $\sqrt{8 (8 - 4) (8 - 6) (8 - 6)}$ = $\sqrt{128} = 8 \sqrt{2} {c m}^{2}$

Area of a triangle is also find using $\frac{1}{2} \times b a s e \times h e i g h t$

$\Rightarrow 8 \sqrt{2} = \frac{1}{2} \times 4 \times h e i g h t$

$\Rightarrow h e i g h t = 4 \sqrt{2} c m$ .

Question 29.

If the area of an equilateral triangle is $9 \sqrt{3} {c m}^{2}$ . What is the length of the side of the triangle.

8 cm
9 cm
18 cm
6 cm

Discuss Question

Answer: Option D. -> 6 cm
:
D

Let the side of the equilateral triangle = a cm.

Semi Perimeter, s = $\frac{3 a}{2}$
$s - a = \frac{3 a}{2} - a = \frac{3 a - 2 a}{2} = \frac{a}{2}$

Now by heron’s formula

Area of the triangle= $\sqrt{s (s - a) (s - b) (s - c)}$

$\Rightarrow 9 \sqrt{3} = \sqrt{\frac{3 a}{2} \times \frac{a}{2} \times \frac{a}{2} \times \frac{a}{2}}$
$\Rightarrow \sqrt{243} = \sqrt{\frac{3 a^{4}}{16}}$
Squaring both sides,
$\Rightarrow 243 = \frac{3 a^{4}}{16}$
$\Rightarrow a^{4} = 81 \times 16$
$\Rightarrow a = 6 c m$

Question 30.

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

$6 c m^{2}$
$12 c m^{2}$
$18 c m^{2}$
$24 c m^{2}$

Discuss Question

Answer: Option B. ->

12 c m^{2}

:
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,
$s = \frac{P e r i m e t e r}{2} = \frac{3 + 4 + 5}{2} = 6 c m$
Area = $\sqrt{s (s - a) (s - b) (s - c)}$
= $\sqrt{6 (6 - 3) (6 - 4) (6 - 5)}$
= $\sqrt{36}$
= $6 c m^{2}$
Area of the parallelogram = 2 (Area of triangle ABC)
$\Rightarrow The area of the parallelogram = 2 \times 6 = 12 c m^{2}$