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Heron`s Formula Test - 6

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Heron`s Formula Test - 6

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Question 1
1 / -0

The ratio of the unequal side to the perimeter of an isosceles triangle is 1 : 5. Find the area of the triangle if its semi-perimeter is 7.5 cm.

A
cm²

B
cm²

C
cm²

D
cm²

Solution

Let the perimeter of the isosceles triangle be 5x and the length of unequal side be x cm.
Let the length of equal sides of the triangle be 2x each.
Perimeter of the triangle = 7.5 × 2 = 15 cm
Therefore, 5x = 15
x = 3 cm
The sides of the triangle are 3 cm, 6 cm and 6 cm.

Area of the triangle =

=

=

=

=

=

= cm²
Question 2
1 / -0

In a circle with centre O and chord AB = 8 cm, the area of the shaded part is 66 cm². Find the area of triangle AOB. (Use = 22/7)

A
cm²

B
cm²

C
cm²

D
cm²

Solution

Reflex ∠AOB = 360° - 150° = 210°
Area of the shaded part =
66 =

r² =
r²= 36
r = 6 cm
Therefore, OA = OB = 6 cm (Radius of the circle)
The sides of the triangle are 6 cm, 6 cm and 8 cm.
s =
s =
s = 10 cm
Area of the triangle =

=

=
= cm²
Question 3
1 / -0

In the given triangle, AB + AC = 8 cm and AB + BC = 10 cm. Find the area of the triangle.

A
cm²

B
cm²

C
10.5 cm²

D
21 cm²

Solution

∠B = ∠C
This implies that the given triangle is isosceles.
Therefore, AB or AC = 8 ÷ 2 = 4 cm
And BC = 10 – 4 = 6 cm
s = = = 7 cm
Area of the triangle =
=

= cm²
Question 4
1 / -0

The circumference of the given circle with centre O is 44 cm. What will be the approximate area of triangle AOB if M and N are the mid-points of OA and OB, respectively? (Use = 22/7)

A
56.57 cm²

B
67.43 cm²

C
75.65 cm²

D
97.98 cm²

Solution

Circumference of the circle = 2r
44 = 2 × × r
r = 7 cm
OM = ON = 7 cm
OA = OB = 7 × 2 = 14 cm (As M and N are the mid-points of OA and OB, respectively)
Therefore, x = 14 cm

AB = = = 21 - 1 = 20 cm

Sides of the triangle = 14 cm, 14 cm and 20 cm

s =
s = = 24 cm

Area of the triangle =
=
=
=
= 97.98 cm²
Question 5
1 / -0

The perimeter of a triangle whose sides measure (3x + 4) cm, (2x + 1) cm and (4x - 5) cm is 36 cm. Find the area of the triangle.

A
cm²

B
cm²

C
cm²

D
cm²

Solution

Perimeter of the triangle = 3x + 4 + 2 x + 1 + 4x - 5
36 = 9x
x = 4 cm
Sides of the triangle:
3x + 4 = 3 × 4 + 4 = 16 cm
2x + 1 = 2 × 4 + 1 = 9 cm
4x - 5 = 4 × 4 - 5 = 11 cm
Semi-perimeter (s) =
=
= 36 ÷ 2 = 18 cm
Area of the triangle =

=

=

=

= cm²
Question 6
1 / -0

The area of an isosceles triangle, whose base is 8 cm, is 12 cm². What is the length of each equal side?

A
4 cm

B
5 cm

C
6 cm

D
7 cm

Solution

Let the length of each equal side be a cm.
That is,
a = a, b = a, c = 8

s =
=
= a + 4

Area of the triangle =
12 =
12 =
12 = (Identity: (a + b)(a - b) = a² - b²)
9 = a² - 4²9 + 16 = a²a =
a = 5 cm
Question 7
1 / -0

What is the area of an isosceles triangle in which the product of an equal side and the unequal side is 35 and the sum of equal sides is greater than 10 but less than 15?

A
cm²

B
cm²

C
cm²

D
cm²

Solution

Factors of 35: 1, 5, 7 and 35
There are two possibilities, i.e 5 × 7 and 1 × 35.
Ignore 1 × 35 because the sum of equal sides is greater than 10 but less than 15.
So, the only possibility left is 5 × 7 and as we know that the sum of equal sides is greater than 10; the sides of the triangle will be 7 cm, 7 cm and 5 cm.

Semi-perimeter (s) = = = 9.5 cm

Area of the triangle =
=

=

=

=
= cm²
Question 8
1 / -0

The area of a triangle with sides of lengths 2 cm, 2p cm and p cm is . What is the value of p?

A
cm

B
cm

C
cm

D
cm

Solution

The sides of the triangle measure 2 cm, 2p cm and p cm.

s = = =

Area of the triangle =

=
=
Squaring both sides,

= (Using identity (a + b)(a - b) = a² - b²)

16 = 9p² - 4
16 + 4 = 9p²20 = 9p²p=cm
Question 9
1 / -0

In a regular pentagon PQRST, a triangle is made by drawing diagonals from a vertex. If the perimeter of triangle PTS is 56 cm, what will be the area of the triangle PSR provided that the perimeter of the pentagon is 90 cm?

A
146.25 cm²

B
160.75 cm²

C
185.22 cm²

D
195.23 cm²

Solution

According to the question,
Perimeter of PQRST = 90 cm
Let each side of the pentagon be x.
Now, x + x + x + x + x = 90
5x = 90
x = 90 ÷ 5 = 18 cm
Each side of the pentagon is 18 cm.
In triangle PTS,
PT + TS + PS = 56 cm
18 + 18 + PS = 56 cm
PS = 20 cm
In triangle PSR,
PS = PR = 20 cm
SR = 18 cm
Semi-perimeter of the triangle = = 29
Area of the triangle =

=

=

= 160.75 cm²
Question 10
1 / -0

The sides of a triangle measure x + 2, x + 4 and 2x + 2. If the perimeter is 48 cm, then what will be the area of the triangle?

A
80.77 cm²

B
92.33 cm²

C
75.89 cm²

D
127.33 cm²

Solution

Sides of the triangle: x + 2, x + 4 and 2x + 2
According to the question,
(x + 2) + (x + 4) + (2x + 2) = 48
4x + 8 = 48
4x = 40
x = 10 cm
The sides will be:
x + 2 = 10 + 2 = 12 cm
x + 4 = 10 + 4 = 14 cm
2x + 2 = 20 + 2 = 22 cm
Semi-perimeter (s) = = 24 cm

Area =

=

= = 75.89 cm²
Question 11
1 / -0

A triangular plot is rented for a period of 4 months. The sides of the plot are 20 m, 24 m and 28 m. How much rent will one pay, if the yearly rent is Rs. 4200 per m²?

A
Rs. 5,76,240

B
Rs. 3,29,210

C
Rs. 3,00,450

D
Rs. 2,35,150

Solution

Let the sides of the plot be:
a = 20 m, b = 24 m, c = 28 m
Semi-perimeter = = 36

Area =

=

= = 235.15 m²
Yearly rent = Rs. 4200 per m²
Rent for 4 months = 4200 × = Rs. 1400 per m²Amount to be paid = 1400 × 235.15 = Rs. 3,29,210
Question 12
1 / -0

A triangle has area 9000 cm²and semi-perimeter 270 cm. What will be the difference between the lengths of the two missing sides, if one side measures 120 cm?

A
55

B
70

C
80

D
170

Solution

Let the given side be a and the missing sides be b and c.
s =
270 =
540 = 120 + b + c
So, b + c = 540 – 120 = 420
Let b = x
c = 420 – x
According to Heron's formula,

Area =

9000 =

Squaring both sides,
9000² = {270(270 – 120)(270 – x)(270 – (420 - x)}
9000²= 270(150)(270 – x)(-150 + x)
= (270 – x)(-150 + x)
2000 = -40500 + 270x + 150x - x²
2000 + 40500 = 420x - x²
42500 = 420x – x²
x² – 420x + 42500 = 0
x² – 250x - 170x + 42500 = 0
x(x – 250) - 170(x – 250) = 0
(x – 250)(x – 170) = 0
x = 250 or 170
If x = 250, then b = 250 and c = 420 – 250 = 170
If x = 170, then b = 170 and c = 420 – 170 = 250
Therefore, the difference will be 80.
Question 13
1 / -0

Find the area of an isosceles triangle with base 6 cm and the length of each equal side as 5 cm.

A
12 cm²

B
10 cm²

C
14 cm²

D
30 cm²

Solution

s =
s =
s = = 8
s = 8 cm
Now,
Area =
Area =
Area =
Area =
Area = 2 × 2 × 3 = 4 × 3
Hence, area = 12 cm²
Question 14
1 / -0

A man was making a quadrilateral shape; he took a piece of orange paper with sides 6 cm, 5 cm and 9 cm. He pasted another white paper with sides 6 cm, 8 cm and 10 cm adjacent to the previous paper. He drew a line dividing the paper into two parts from the mid-point of the longest side to the opposite vertex and coloured the central part as blue and rest of the quadrilateral shape white. How much area of the quadrilateral shape is coloured orange, blue and white?

A
38.14 cm²

B
34.14 cm²

C
32.24 cm²

D
30.04 cm²

Solution

Let ABC be the orange coloured paper. Also, ACD be the paper which has been divided into two parts by joining the mid-point E of AD to point C.
For area of triangle ABC,
a = 5 cm, b = 6 cm, c = 9 cm
s = , where a, b and c are the sides of the triangle.
s =
s = 10 cm
Area of the orange coloured paper =

=

=

= = 10 = 14.14 cm²
Area of triangle ACD:

s = = 12 cm

Area =

=

=

= = 24 cm²
Now, total area of the quadrilateral shape = 14.14 + 24 = 38.14 cm²
Question 15
1 / -0

In the following diagram, BC is the diameter of the circle. AB = 10 cm, AC = 24 cm, BD = 13 cm and CD = 15 cm. What is the area of quadrilateral ABCD?(Take = 3.7)

A
170.5 cm²

B
186.6 cm²

C
190.6 cm²

D
195.5 cm²

Solution

∠BAC = 90° (Angle in a semi-circle)
Thus, BAC is the right-angled triangle.
BC²= AB² + AC²
= 10² + 24²
= 676
BC² = 26²
BC = 26 cm
Area of triangle ABC = × 10 × 24 = 120 cm²
Area of triangle BCD:
a = 13 cm, b = 15 cm, c = 26 cm

s = = = 27 cm

Area =

⁼

=

=

= 66.6 cm²
Total area = 120 + 66.6 = 186.6 cm²
Question 16
1 / -0

The perimeter of the following shape is 40 cm and that of the square in between the shape is 24 cm. If all of the triangles are congruent to each other and of same size, find the area of the shape given that all the triangles are isosceles.

A
60 cm²

B
64 cm²

C
68 cm²

D
72 cm²

Solution

Perimeter of the shape = 40 cm
We know that all the triangles are congruent.
Number of sides of triangle that make up the perimeter of the shape = 8
Since all these sides are similar, each outer side of the shape = Perimeter ÷ Total number of outer sides of the shape
= 40 ÷ 8 = 5 cm

Therefore, two sides of each triangle 'a' and 'b' (Ones that are representing the perimeter of the shape) = 5 cm each

Now, perimeter of the square = 24 cm
Each sides of the square = Perimeter ÷ Number of sides
= 24 ÷ 4 = 6 cm

Therefore, third side of each triangle 'c' = 6 cm

Area of 4 triangles = 4 × 12 = 48 cm²Area of the square = (side)²= (4)²= 16 cm²Total area of the shape = Area of the triangles + Area of the square
= (48 + 16) cm²= 64 cm²
Question 17
1 / -0

In the given parallelogram ABCD, AB = 12 cm and AC is the diagonal. If the semi-perimeter of △ADE is 12 cm and AE = 10 cm, then what is the difference between the areas of △AEC and △ABC, given that E is the mid-point of DC?

A
24 cm²

B
20 cm²

C
16 cm²

D
14 cm²

Solution

AB = 12 cm (Given)
In parallelogram ABCD, AB = DC (Opposite sides of a rectangle are equal)
DC = 12 cm
Since E is the mid-point of DC;
DE =
DE = 6 cm

In △ADE, DE = 6 cm
AE = 10 cm (Given)
Let AD be x cm.
Semi-perimeter (S) of △ADE = 15 cm (Given)
Semi perimeter =
12 =
24 = 6 + 10 + x
24 = 16 + x
x = 24 – 16
x = 8 cm
Therefore, area of △ADE =

=

=

=

= 24 cm²

Now, Since E is the mid-point of side EC(Median EC divides the triangle ADC in two triangles of equal area)

Area of △ADE = Area of △AEC
Therefore, Area of △AEC = 24 cm²Now, we know that the diagonal AC will divide parallelogram ABCD into two triangles ADC and ABC of equal areas.
Area of △ABC = Area of △ADC
And, Area of △ABC = Area of △ADC = 2 (Area of △AEC) = 48 cm²
Area of △ABC = 48 cm²Area of △AEC = 24 cm²
Required difference = (48 – 24) cm²= 24 cm²
Question 18
1 / -0

A farmer is trying to create a fence around his field, which is in the shape of a rhombus. He uses 320 m of the fencing wire. If the length of the diagonal of the field is 100 m, then what will be its area?

A

B

C

D

Solution

A rhombus is a quadrilateral with all sides equal.
So, perimeter = 4a
320 = 4a
a = 80 m
So, AB = BC = DC = DA = 80 m and AC = 100 m
Semi-perimeter of triangle ABC = (80 + 80 + 100) ÷ 2 = = 130
Area of triangle ABC =

=

=

= m²

Similarly, area of triangle ACD = m²

Area of field ABCD = m² + m²

=
Question 19
1 / -0

In a triangle ABC with sides 12 cm, 17 cm and 25 cm, find the difference between the altitudes which can be made from the shortest and longest sides of the triangle.

A
7.4 cm

B
7.8 cm

C
6.8 cm

D
6.2 cm

Solution

Sides of the triangle: 12 cm, 17 cm and 25 cm
Area of the triangle:

S =
S = 27 cm
Area =

=

=

= 90 cm²
Let altitude to the smallest side be 'h'.
Then, => h = 15 cm
Let altitude to the longest side be 'p'.
Then, = => p = 7.2 cm

Required difference = 15 - 7.2 = 7.8 cm.
Question 20
1 / -0

If the equal sides of an isosceles triangle ABC are '2m' units each and the third side is 'n' unit , then which of the following will represent the area of 4 such triangles?

A

B

C

D

Solution

AB = AC = 2m and BC = n

Semi-perimeter of the triangle =
=
Area of triangle ABC =

=
=

=
=

=

So, area of 4 triangle = 4
So, area of 4 triangles = 4 ×
=
Question 21
1 / -0

Find the perpendicular height of a parallelogram ABED formed inside a trapezium ABCD with total area of 196 cm². The triangle formed inside the trapezium has a base of 15 cm, with sides BE and BC measuring 14 cm and 13 cm, respectively. The base of the parallelogram so formed is 10 cm.

A
10.0 cm

B
11.2 cm

C
12.4 cm

D
13.6 cm

Solution

Total area = 196 cm²
Area of the triangle:
By Heron's formula
s = = 21
Area =

=
= 84 cm²
Area of the parallelogram = b × h = Area of trapezium ABCD - Area of triangle BEC = 196 – 84 = 112 cm²
10 × h = 112 cm²
h = 11.2 cm
Question 22
1 / -0

In the following figure, triangles are formed by joining the vertices of the square as shown. Triangle BEH is an isosceles triangle with its base as one of the sides of the square. Similarly, triangle BAE is also an isosceles triangle as shown in the figure. Also, triangle BAE is congruent to other three triangles which have one of their vertices as that of the square; and similarly, triangle BEH is congruent to other three triangles which have one of their sides as that of the square. Find the area of the given figure. (Take = 4.58)

A
100.64 cm²

B
95.6 cm²

C
84.64 cm²

D
65.32 cm²

Solution

Given: BE = BH = 5 cm
EH = 4 cm
Area of triangle BEH:Using Heron's formula,

s =

= = 7 cm
Now, area of triangle =

=

= = 9.16 cm²

Now, there are 4 such triangles.
9.16 × 4 = 36.64 cm²Now, for the area of triangle BEA:

s =
= 9 cm
Now, area of the triangle =

=

= 12 cm²

As there are 4 such triangles;
12 × 4 = 48 cm²
Now, area of the square = Side² = 4² = 16 cm²
Total area of the figure = 16 + 48 + 36.64 = 100.64 cm²
Question 23
1 / -0

The area of a rectangle ABCD is 48 cm² and its perimeter is 28 cm with AB as the largest side. E and F are the mid-points of AB and DA, respectively. G is a point on CD such that an isosceles triangle EFG is formed in which FG = GE. Find the area of the ΔFEG, if length GF = 6.5 cm.

A
16 cm²

B
15 cm²

C
20 cm²

D
24 cm²

Solution

According to the question:

Let the length AB of rectangle ABCD be x cm and the width BC be y cm.
A.T.Q.
Area of rectangle ABCD = length × width
48 = x × y
x = ... (i)
Perimeter of rectangle ABCD = 2(length + width)
28 = 2(x + y)
2x + 2y = 28
(Using x = from (i))

96 + 2y² = 28y
2y² – 28y + 96 = 0
Dividing this by 2,
y² – 14y + 48 = 0
y² – 6y – 8y + 48 = 0
y(y – 6) – 8(y – 6) = 0
(y - 8)(y - 6) = 0
y = 8 or 6
Since AB is the largest side; y = 6 cm
Putting the value of y in (i) we have
x = 48 ÷ 6
x = 8
Therefore, length AB and width BC of rectangle ABCD are 8 cm and 6 cm, respectively.
Since E and F are the mid-points of AB and DA, respectively.
AE and EB = 4 cm
DF and AF = 3 cm
Clearly, △FAE is right angled at A. (Interior angle of a rectangle always equals 90°)

According to Pythagoras theorem, in right angled △FAE,
(Hypotenuse)²= (Base)² + (Perpendicular)²(FE)² = (FA)² + (AE)²
(FE)² = 4² + 3²
(FE)² = 16 + 9
(FE)² = 25
(FE)= 5 cm

We know that FE = 5 cm, GF = EG = 6.5 cm
Using Heron's formula, s =
s =
s =
s = 9 cm

Area of ΔFAE =

=
=
= 15 cm²
Question 24
1 / -0

Take:
State T for true and F for false.

(i) Rohit is making a triangular design piece from a sheet. Two of the sides of the sheet are 16 cm each. The perimeter of the sheet is 50 cm. Thus, he will need 218.8 cm²of the sheet.

(ii) Rajiv's mother is preparing paranthas for him in the shape of triangle. The sides of a parantha are 12 cm, 8 cm and 6 cm. She has ghee to apply on 96 cm². Thus, she can apply ghee on 4 paranthas.

(iii) The sides of a quadrilateral are AB = BC = 20 cm, AD = 30 cm, CD = 18 cm and BD = 30 cm. Its area will be 649 cm².

(iv) We can find the area of a triangle by only knowing its perimeter.

A
(i) = F, (ii) = F, (iii) = F, (iv) = F

B
(i) = T, (ii) = F, (iii) = T, (iv) = F

C
(i) = T, (ii) = T, (iii) = T, (iv) = F

D
(i) = T, (ii) = F, (iii) = F, (iv) = T

Solution

(i)
a = 16 cm, b = 16 cm, c = ? cm
Perimeter = a + b + c
16 + 16 + c = 50
c = 18 cm
s = = = 25 cm
Area =

=

=
=
= 118.8 cm²
Thus, statement (i) is false.

(ii)
a = 12 cm, b = 8 cm, c = 6 cm

s = = =13 cm

Area =

=
=
= cm²Area will be between 21 cm² and 22 cm².
Area of 4 paranthas will be between 84 cm² and 88 cm²
Thus, statement (ii) is false.

(iii)

Area of triangle ABD:
a = 20 cm, b = 30 cm, c = 30 cm

s = = = 40 cm

Area =

=

= 200 cm²
= 282 cm²

Area of triangle BCD:
a = 20 cm, b = 18 cm, c = 30 cm

s = = = 34 cm

Area =

=

=
= cm²Area will be around 187 cm²
Total area will be around 282 + 187 i.e. 469 cm²
Therefore, statement (iii) is false.

(iv) To find the area of a triangle using Heron's formula, we need measures of the three sides.
For example, if a, b and c are the sides of a triangle, then semi-perimeter 's' can be calculated using the formula:

And the area of the triangle can be calculated using the formula:

Thus, statement (iv) is false.
Question 25
1 / -0

In the given figure, the central square is of side 8 cm and the area of triangles ABC and GEF are equal. If AF divides quadrilateral ABCD as well as triangle GEF into two equal triangles, find the total area of the given shape which is to be put on a flag after the square shaped region from the quadrilateral formed by ABCD has been cut out. (Figure not drawn to scale)

A
796 cm²

B
776 cm²

C
724 cm²

D
736 cm²

Solution

Area of triangle CEF:

s = = 42 cm

Area =

=

= 336 cm²
Area of triangle CGF = 336 cm² (It is given that CF divides triangle GEF into two equal triangles)
Area of triangle GEF = 336 + 336 = 672 cm²
Now, area of triangle ABC:

s = = 21 cm

Area = = cm²

Area of triangle ADC = 84 cm² (It is given that CF divides quadrilateral ABCD into two equal triangles)
Area of square = 8 x 8 = 64 cm²
Area left after the removal of square from ABCD = 84 + 84 – 64 = 104 cm²
Total area = 104 + 672 = 776 cm²

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