# Free Heron's Formula 03 Practice Test - 9th Grade

### Question 1

A triangular wall having sides 3 m, 5 m and 6 m has to be painted. If the cost of painting 1 m2 of a wall is ₹ 5, what is the cost of painting the whole wall?

₹ 16√13

₹ 10√14

₹ 14√10

₹14

#### SOLUTION

Solution :B

Semi perimeter of a triangle = (a+b+c)2 ⇒ (3+5+6)2=7m

Area of a triangle

=√s(s−a)(s−b)(s−c) ⇒√7(7−3)(7−5)(7−6)

⇒ √7×4×2×1

⇒ √14×4⇒√56=2√14 m2

Since,cost of painting 1 m2 of wall is ₹5,

The cost of painting the whole wall =5×2√14=₹ 10√14.

### Question 2

If the area of a rhombus-shaped box is 6√7cm2 and its one diagonal is 6cm. Find the side of the rhombus.

8 cm

3 cm

4 cm

5 cm

#### SOLUTION

Solution :C

Let the side of the rhombus = a

Area of the triangle containing the diagonal = 12 times area of the rhombusOne of the diagonal = 6 cm

Semi perimeter = 6+a+a2=a+3

Now the area of the triangle containing the diagonal =√s(s−a)(s−b)(s−c)

3√7 = √(a+3)(a+3−a)(a+3−a)(a+3−6)

= √(a+3)(a−3)(9)

= √(a2−9)(9)

We take square on both sides and get,

63 = (a2−9)×9

⇒(a2−9) = 7

⇒a2 = 7+9 = 16⇒ a= 4cm

### Question 3

If the side of a rhombus is 6 cm and its one diagonal is 8 cm. Find the area of the rhombus in cm2.

9√3

3√9

16√5

8√5

#### SOLUTION

Solution :C

Side of the rhombus = 6 cm

One of the diagonal = 8 cm

Semi perimeter of the triangle containing the diagonal = (6+6+8)2 = 10 cm

Area of the triangle containing the diagonal = √s(s−a)(s−b)(s−c)

= √10(10−6)(10−6)(10−8)

= √10×4×4×2

= √2×5×4×4×2

⇒8√5cm2

Area of rhombus = 2 (Area of the triangle containing the diagonal)Area of the rhombus = 2×8√5 = 16√5 cm2

### Question 4

The area of a rectangle is 96 cm2 and its length is 12 cm. 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

#### SOLUTION

Solution :C

Area of a rectangle = length × breadth

⇒ 96 = 12 × breadthBreadth = 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 =402=20 cm

### Question 5

State whether true or false:

If in a right angled triangle only two sides are given, then its area can be calculated.

True

False

#### SOLUTION

Solution :A

In a right angled triangle, if the base and the height are given, the area can be calculated directly by using 12 × base × height. If the hypotenuse and either the base or the height are given, then the third side can be found using Pythagoras' theorem. Then the area can be calculated using the same formula.

### Question 6

The sides of a triangular plot are in the ratio of 16: 8: 10 and its perimeter is 340 m. Its area in m2 is 100√A. The value of A is

#### SOLUTION

Solution :Let the sides of the triangle be 16x, 8x and 10x.

Then, 16x + 8x + 10x = 340

⇒ 34x = 340

⇒ x = 34034

⇒ x =10.Hence the sides are 160 cm, 80 cm and 100 cm.

Using the Heron's formula, Area A = √s(s−a)(s−b)(s−c)Semi-Perimeter s = perimeter2

=3402

=170cm

∴A=√170(170−160)(170−100)(170−80)

=√170×10×70×90

=100√1071cm2

The value of A is 1071

### Question 7

The area of an equilateral triangle whose side is 2 cm is √A. The integral value of A is

#### SOLUTION

Solution :The area of an equilateral triangle is √34a2 , where a = 2 cm.

Hence the area is √34×22

=√3 cm2

Thus the value of A = 3.

### Question 8

The sides of a triangle are 5 cm, 6 cm and 3 cm long. The area of the triangle is √A cm2. Then the value of A is

#### SOLUTION

Solution :s=5+6+32

=7 cm

Area A=√s(s−a)(s−b)(s−c)

=√7(7−5)(7−6)(7−3)

=√7×2×1×4

=√56 cm2

The value of A is 56.

### Question 9

Which of the following are correct regarding the given figure?

The area of a parallelogram is 120 cm2

The area of a parallelogram is 180 cm2

The length of the altitude is 15 cm

The length of the altitude is 30 cm

#### SOLUTION

Solution :B and C

For △BCD,

s=25+17+122=27

Using Heron's formula,

A=√27(27−25)(27−17)(27−12)

=√27×2×10×15

=√3×3×3×2×2×5×3×5

=3×3×2×5

=90 cm2

The parallelogram can also be divided into 2 triangles by diagonal AC, as shown below, the area of each triangle being =90 cm2.

Area of parallelogram =90×2=180 cm2

In, △ACD

Area of △ACD=12×base×Altitude

90=12×12×Altitude

⇒Altitude=15 cm

### Question 10

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.

#### SOLUTION

Solution :D

Let the sides be 6x, 7x, 8x.

Then, 6x+7x+8x=420

⇒21x=420

x=20

∴ the sides are:

a =120 m, b =140 m, c =160 m

Semi-perimeter s = 4202=210 m

A=√s(s−a)(s−b)(s−c)

=√210(210−120)(210−140)(210−160)

=√210×90×70×50

=2100√15 m2