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Heron`s Formula Test - 7

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Heron`s Formula Test - 7

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Weekly Quiz Competition

Question 1
1 / -0

In the given figure, if the perimeter of the trapezoid is 34 cm, what will be the area of triangle ADB?

A
cm²

B
cm²

C
cm²

D
cm²

Solution

Perimeter = Sum of all sides
34 = 9 + 8 + 10 + AB
AB = 34 - 27
AB = 7 cm

In triangle ADB,
a = AB =7 cm,
b = AD = 9 cm,
c = BD = 12 cm

S = = = 14 cm

Area of triangle ADB =

=

=

= cm²
Question 2
1 / -0

In a triangle, the lengths of two of the sides are 10 cm and 14 cm. If the perimeter of the triangle is 30 cm, what will be its area?

A
cm²

B
cm²

C
cm²

D
cm²

Solution

Perimeter = Sum of all sides
Let the missing side be x.
x + 14 + 10 = 30
x = 30 - 24
x = 6 cm

Semiperimeter (S) = Perimeter ÷ 2 = 30 ÷ 2 = 15 cm

Area of triangle =

=

=

=

= cm²
Question 3
1 / -0

In the given figure, area of the isosceles triangle = cm², length (BD) of the triangle = 8 cm. Find the measures of the other sides of the isosceles triangle.

A
AB = AD = 10 cm

B
AB = AD = 6 cm

C
AB = AD = 5 cm

D
AB = AD = 7 cm

Solution

Let the other side of the isosceles triangle (AB) = x cm
So, AD is also equal to x cm because in an isosceles triangle, two sides are equal to each other (AB = AD)
By using Heron's Formula:
S =
S =
S = = x + 4 cm
So,
a = 8 cm, b = x cm, c = x cm

Area of an isosceles triangle =
=

Squaring both sides we get,
320 = (x + 4)(x - 4)16
= x² - 16
20 + 16 = x²
36 = x²
So, x = ± 6 cm
Length can not be negative.
So, x = 6 cm
Length of AB and AD = 6 cm
Question 4
1 / -0

In a scalene triangle, the three sides are a, b and c. The ratio of a to b is 2 : 3 and that of b to c is 2 : 3. If the semiperimeter of the triangle is 19 cm, find the area of the triangle (round to the nearest whole number).

A
35 cm²

B
36 cm²

C
37 cm²

D
38 cm²

Solution

According to the question, the triangle is as follows:

Ratio of a to b = 2 : 3
= 4 : 6 ... (i)
Ratio of b to c = 2 : 3
= 6 : 9 ... (ii)
From (i) and (ii), ratio of a to b to c is 4 : 6 : 9.

Let a, b and c be 4x, 6x and 9x, respectively.
Now, the semiperimeter of the triangle (s) = 19

Semiperimeter =
19 =
19 × 2 = 4x + 6x + 9x
38 = 19x
x = 38 ÷ 19
x = 2

Therefore, a = 4x
= 4(2)
= 8 cm

b = 6x
= 6(2)
= 12 cm

c = 9x
= 9(2)
= 18 cm

Area of the triangle =

=

=

=

= 38.24 cm²
= 38 cm²(After rounding to the nearest whole number)
Question 5
1 / -0

A rhombus shaped park has a perimeter of 80 m and 24 m long diagonal. If grass is laid on the park and the total amount spent on it is Rs. 28,800, what is the per m²cost of laying the grass?

A
Rs. 85

B
Rs. 75

C
Rs. 70

D
Rs. 60

Solution

Perimeter of the park = 80 m
Each side = 80 ÷ 4 = 20 m
Now according to the question,
Dimensions of the triangle formed, i.e. MNO = 24 m, 20 m, 20 m
Semiperimeter = = 32 m

Area of Rhombus = 192 × 2 = 384 m²Per m² cost = 28,800 ÷ 384 = Rs. 75
Question 6
1 / -0

What will be cost of painting a triangle at the rate of 25 paise per cm² if the sides of the triangle are (x - 1), (x + 1) and (x – 3), and the perimeter is 84 cm?

A
Rs. 64

B
Rs. 72

C
Rs. 84

D
Rs. 90

Solution

Perimeter = (x - 1) + (x + 1) + (x – 3)
84 cm = x – 1 + x + 1 + x – 3
84 = 3x - 3
87 = 3x
x = 29 cm

Sides: a = x – 1 = 28
b = x + 1 = 30
c = x – 3 = 26
Semiperimeter = 84 ÷ 2 = 42 cm

= 336 cm²
Cost of painting a triangle at the rate of 25 paise per cm²_^.

The total cost of painting at the rate of 25 paise per cm²^₌336 × 25 = 8,400 paise or Rs. 84
Question 7
1 / -0

A parallelogram having a base of 25 m is made up of two triangles such that the dimensions of each small triangle are 17 m, 12 m and 25 m. What will be the height of the parallelogram?

A
7.2 m

B
8.4 m

C
12.4 m

D
15.6 m

Solution

Dimensions of one triangle = 17 m, 12 m, 25 m
S = = 27 m
Area =
=
=
= 90 m²
Area of parallelogram = 2 × Area of one triangle
Area of parallelogram = 2 × 90 = 180 m²
Area of parallelogram = Base × Height
180 = 25 × Height
Height = 180 ÷ 25
Height = 7.2 m
Question 8
1 / -0

Two of the sides of a triangle are 12 cm and 16 cm. What will be the value of r if the perimeter of the triangle is 50 cm, and the area of the triangle is cm²?

A
10

B
12

C
14

D
15

Solution

Let the missing side of the triangle be x.
Now,
x + 12 + 16 = 50
x = 50 – 28
x = 22
Semiperimeter = 50 ÷ 2 = 25 cm
Area of the triangle =
=
=
=
r = 15
Question 9
1 / -0

The perimeter of a triangular wall shelf is 126 cm. What will be the area of the shelf if the largest side of the shelf is 3 times the smallest side of the shelf, and the remaining side is 50 cm long?

A
cm²

B
cm²

C
cm²

D
cm²

Solution

Let the smallest side be x cm.
Largest side = 3x cm
Perimeter = 126 cm
x + 50 + 3x = 126
4x = 76
x = 76 ÷ 4
x = 19
Sides = 19 cm, 50 cm, 57 cm
S = = 63 cm
Area of triangle =
Area = = cm²
Question 10
1 / -0

Triangle ABC has a perimeter of 34 cm. If (s - a) = 7 cm and (s - b) = 6 cm, what will be the area of the triangle, given that a and b are two sides of the triangle, and s is the semiperimeter?

A
cm²

B
cm²

C
cm²

D
cm²

Solution

Perimeter = 34 cm
Semiperimeter = 34 ÷ 2 = 17 cm
According to the question,
(s - a) = 7
17 - a = 7
a = 10 cm
Similarly,
(s - b) = 6
17 - b = 6
b = 11 cm
Now,
a + b + c = 34
10 + 11 + c = 34
c = 34 - 21 = 13 cm

Area =

=

=

=

= cm²
Question 11
1 / -0

There are two isosceles triangles. The two equal sides of the bigger isosceles triangle are equal to (x + 2) units and the remaining side of the bigger isosceles triangle is four more than double the unequal side of the smaller isosceles triangle. In the smaller isosceles triangle, an angle bisector, which measures 12 units, is drawn from its vertex to the side which measures different from other sides. The unequal side of the smaller isosceles triangle is of 10 units, and the two equal sides of the smaller triangle are of x units length. What is the difference between the areas of both the triangles?

A
76 sq. units

B
48 sq. units

C
24 sq. units

D
56 sq. units

Solution

In an isosceles triangle, an angle bisector is perpendicular to the unequal side and is the same as median.

We know that,
BD = 12 units
AC = 10 units
As BD is the median, it would divide AC in two equal parts.
Therefore,
AD = DC = 1/2 × AC
= 1/2 × 10
= 5 units

So, ΔABD is a right-angled triangle.
Now, using Pythagoras theorem we get:
(AB)²= (AD)²+ (BD)²
(AB)²= (5)²+ (12)²(AB)²= 25 + 144
(AB)²= 169
AB = 13 units
AB = BC = x (The two equal sides of the smaller triangle are x units.)
Therefore, x = 13 units

Area of the smaller triangle = 1/2 × AC × BD = 1/2 × 10 × 12 = 60 sq. units
Sides of the larger triangle :
a = x + 2 = 13 + 2 = 15 units
b = a = 15 units
c = 2(10) + 4 = 20 + 4 = 24 units (The remaining side of the bigger isosceles triangle is four more than double the unequal side of the smaller isosceles triangle.)
The semiperimeter 's' of the bigger isosceles triangle can be calculated as:

= = 27 units

Now, using Heron's formula:
Area of the bigger isosceles triangle =

=

=

=
= 108 sq. units
Hence, 108 sq. units is the area of the bigger isosceles triangle.
Required difference = 108 - 60 = 48 sq. units
Question 12
1 / -0

In the given figure, DCF is an isosceles triangle with two of its sides equalling 12.5 cm each. What would be the ratio of the area of rhombus ABCD to the area of the shaded region if the diagonals of rhombus ABCD are 32 cm and 24 cm long?

A
89 : 120

B
120 : 99

C
128 : 89

D
99 : 128

Solution

Given,

ABCD is a rhombus.
Diagonals of this rhombus are 32 cm and 24 cm long.

Therefore, area of rhombus ABCD =

=

=
= 384 cm²(Diagonals of a rhombus bisect each other at right angles.)

Now,
Side of the rhombus can be calculated by using Pythagoras theorem.
(Side of the rhombus)²= (Half of one diagonal)² + (Half of other diagonal)²
(Side of the rhombus)² = (12)² + (16)²
(Diagonals of a rhombus divide each other in two equal parts.)
(Side of the rhombus)² = 144 + 256
(Side of the rhombus)²= 400
Side of the rhombus = 20 cm
Semiperimeter of triangle BOC =
= = 24 cm

Now, the area of triangle BOC =

=

=

=

= 96 cm²
As there are two shaded triangles BOC and AOD, the area of the two triangles would be: 2 × 96
= 192 cm²

Now, the side of the rhombus is 20 cm and DCF is an isosceles triangle with two of the sides equalling 12.5 cm each.

Therefore, the semiperimeter of the isosceles triangle DCF =
= 22.5 cm
Now, area of the isosceles triangle DCF =

=

=

=

= 75 cm²Hence, the area of the isosceles triangle DCF = 75 cm²The area of the shaded region = (192 + 75) cm²= 267 cm²
The ratio of the area of the rhombus to that of the shaded region = 384 : 267
= 128 : 89
Hence, option 3 is the answer.
Question 13
1 / -0

In a right-angled parallelogram, the smaller side is 5 units and a diagonal is 13 units long. The angle of the parallelogram is changed now but the measure of each side is identical. A point on the larger side of the parallelogram is such that it divides the larger side of the parallelogram in two equal parts. From that point, a line which measures 5 units is drawn to any of the opposite vertices.

What would be the difference between the area of the right-angled triangle and that of the new isosceles triangle?

A
18 sq. units

B
12 sq. units

C
10 sq. units

D
8 sq. units

Solution

Given,
In a right-angled parallelogram, the smaller side is 5 units and a diagonal is 13 units long.
A figure can be drawn for the same:

Here, ABCD is a right-angled parallelogram, the smaller side (both AD and BC) is 5 units and a diagonal (AC) is 13 units long.
Using Pythagoras theorem we get,
(AC)² = (AD)² + (DC)²
(13)² = (5)² + (DC)²
169 = 25 + (DC)²
(DC)² = 169 – 25
(DC)² = 144
DC = 12 units
Hence, the area of the right-angled triangle = × Base × Perpendicular height
= × 5 × 12 = 30 sq. units
Now,
As the angle of the parallelogram is changed and a point (on the longer side of the new parallelogram) is such that it divides the longer side of that parallelogram in two equal parts.
It can be shown from this figure:

Here, PQRS is a parallelogram and T is the point on the longer side of the parallelogram.
So, T divides PS in two equal parts and both the sides of the parallelogram are identical.
Therefore, PS = QR = 12 units
TS = 12 ÷ 2 = 6 cm
PQ = SR = 5 units
Also, line TR = 5 units (given)
Hence, △STR is an isosceles triangle.
Now,
Semiperimeter of △STR (s) =
=
= 8 units
Area of △STR = =
=
=
= 12 sq. units
Now, the difference between the area of the right-angled triangle and that of the new isosceles triangle = (30 – 12) sq. units = 18 sq. units
Hence, option 1 is the answer.
Question 14
1 / -0

Find the area of the shaded region if △GEF is an isosceles triangle, where GE = GF, the lengths of its sides are 13 cm, 13 cm and 10 cm, and the sum of the parallel sides of the trapezium ABCD is 38 cm.

A
125 cm²

B
114 cm²

C
168 cm²

D
170 cm²

Solution

Given,

Here,
△GEF is an isosceles triangle with sides as 13 cm, 13 cm and 12 cm.
Therefore, the semiperimeter of the triangle (s) =
= 18 cm
Now, the area of △GEF = Area of the triangle =
=
=
=
= 60 cm²
Now, the area of trapezium ABCD = × (Sum of parallel sides) × Perpendicular distance between the parallel sides … eq 1
After drawing a line perpendicular from point G to side CD of the trapezium, the figure would be like this:

Now, we know that the altitude, the angle bisector and the median are the same for an isosceles triangle.
Therefore,
Median GH divides side EF of an isosceles triangle in two halves.
EH = HF =
== 5 cm
In △GEH,
(GE)² = (GH)² + (HE)²
(13)²= (GH)² + (5)²
169 = (GH)²+ 25
(GH)² = 169 – 25
(GH)² = 144
GH = 12 cm
Area of the trapezium ABCD =× (38) × 12 = 228 cm²
Now, area of the shaded region = Area of trapezium ABCD – Area of △GEF
= (228 – 60) cm²
= 168 cm²
Hence, option 3 is the answer.
Question 15
1 / -0

The sides of a triangle, formed by a piece of wire, are 30 cm, 30 cm and 28 cm. If the same wire is molded to form a circle, then what is the increase in the area covered by these two shapes? (Take π =and )

A
235.57 cm²

B
242.57 cm²

C
228.57 cm²

D
245.28 cm²

Solution

Given,
Sides of triangle are 30 cm, 30 cm and 28 cm.
Therefore, the semiperimeter of the triangle =
=
=
= 44 cm
Area of the triangle =

=

=

= =

= 112 × 3.31

= 370.72 cm²
Now, the same wire is molded to form a circle.
The circumference of the circle would be:
(30 + 30 + 28) cm = 88 cm
We know that,
Circumference of the circle = 2πr
2πr = 88

r =

r =

r = 14 cm

Therefore, the area of the circle = πr²

= = 616 cm²
Hence, the amount of the area increased when the triangular wire is molded into a circular wire = (616 – 370.72) cm² = 245.28 cm²
Hence, option 4 is the answer.
Question 16
1 / -0

The following shape has been cut out from a square shaped piece of origami paper having sides equal to 20 cm each. Find the total area of the left over paper after removal of this shape, provided that the four small triangles are identical and the shape is square.

A
268 cm²

B
250 cm²

C
225 cm²

D
200 cm²

Solution

Area of the given shape = Area of 4 triangles + Area of the square

Area of the given triangle using heron's formula:

s = = = 3 cm

Area of triangle =

=

=

=

= 1.5 cm²
Since there are 4 such triangles, the area of the triangles would be equal to 4 × 1.5 = 6 cm²

Area of the square given in the image = 12 × 12 = 144 cm²

Total area of the given figure = 6 + 144 = 150 cm²

Area of the whole piece of paper = 20 × 20 = 400 cm²

The left over shape will be:

400 – 150 = 250 cm²
Question 17
1 / -0

In a cemented circular playground of a play-way, % of the area is covered by 6 triangles of the same size, which are made for the children for extra-curricular activities. If the sides of each triangle are 3 m, 4 m and 5 m, then find the perimeter of the circular ground. (pi = 22/7)

A
21 m

B
22 m

C
23 m

D
24 m

Solution

According to the question sides of each triangle,
a = 3 m, b = 4 m and c = 5 m

Semiperimeter (s) =
=

= 6 m

According to Heron's formula, area of each triangle =

=

=

=

= 6 m²

Area of all the triangles = 6 × 6 = 36 m²We know that this is % of the total area of the circular playground.
Let the area of the circular playground be x m².

According to the question,
or = 36 × 100

7200x = 36 × 100 × 77
x = (36 × 100 × 77) ÷ 7200
x = (36 × 77) ÷ 72
x = 38.5 m²Area of the circle = πr²38.5 =

38.5 × 7 = 22 × r²269.5 ÷ 22 = r²12.25 = r²r = 3.5 m

Perimeter (circumference) of the circular playground = 2πr

= = 22 m
Question 18
1 / -0

State 'T' for true and 'F' for false for each of following.

1. Altitude to the smallest side of the triangle, having sides 28 cm, 15 cm and 41 cm, is 16.8 cm.
2. If the sides of a triangle are in ratio 7 : 8 : 9, the area will be more than 250 cm², provided that the perimeter is 72 cm.
3. Two triangles are made by drawing a diagonal in a parallelogram. If the semiperimeter of a triangle is given, we can find the area of the parallelogram, provided that the lengths of a diagonal and a base of the parallelogram are given.
4. Mr. Sharma decided to take a triangular piece of land on rent for a period of 6 months. If the sides of the triangle are 40 m, 24 m and 32 m, he will pay Rs. 25,000, provided that the yearly rent is Rs. 125 per m².

A
T, F, T, T

B
T, F, T, F

C
F, F, T, T

D
T, F, F, T

Solution

1. Sides of the triangle = 28 cm, 15 cm and 41 cm

Semiperimeter (s) = = 42 cm

Area of triangle =

=

=

= 126 cm²

Area of a triangle =× b × h
126 = × 15 × h
126 × 2 = 15h
252 ÷ 15 = 16.8 cm
So, 1 is true.

2. Let the sides of triangle be 7x, 8x and 9x.
Perimeter = 72 cm
7x + 8x + 9x = 72
24x = 72
x = 72 ÷ 24
x = 3 cm

Sides will be:
7x = 7 × 3 = 21 cm
8x = 24 cm
9x = 27 cm

s = 72 ÷ 2 = 36 cm

Area =

=

=

= 241.5 cm²So, this is False.

3.

ABCD is a parallelogram.
Diagonal AC divides it into two triangles, i.e. ADC and ABC.
In triangle ADC, AC is given and one side (Base of the parallelogram) is given.
As semiperimeter is given, we can work the missing side of the triangle.
Using all the sides of the triangle and the semiperimeter, area of triangle ADC can be worked out.
As the diagonal divides a parallelogram into two equal parts,
Area of triangle ADC = Area of Triangle ABC
So, this is True.
Area of triangle ADC + Area of triangle ABC = Area of parallelogram ABCD

4. Sides of the field = 40 m, 24 m and 32 m

Semiperimeter == 48 m

Area =

=

=

= 384 m²Yearly rent per m² = Rs. 125
Half-yearly rent per m², i.e. rent for 6 months = 125 ÷ 2 = Rs. 62.5
Rent to be paid for 1 year = 124 × 384 = Rs. 48,000
Rent to be paid for 6 months = 48,000/2 = Rs. 24,000
So, this is False.
Question 19
1 / -0

A field is in the shape of a trapezium. The owner of this field occupies this field at a rate of Rs. 650 per sq m per year. The parallel sides of this field are 49 m and 24 m and the non-parallel sides are 12 m and 17 m. A farmer occupies this field for 5 months. How much will he pay after 5 months?

A
Rs. 71,175

B
Rs. 72,456

C
Rs. 74,578

D
Rs. 75,622

Solution

Let ABCD be the trapezium.
Draw a line BE parallel to AD and draw perpendicular h on CD.
This means that ABED is a parallelogram.
BE = AD = 12 m
ED = AB = 24 m
EC = CD – DE
49 – 24 = 25 m
For ΔBEC,
Semiperimeter = s === 27 m
ar(ΔBEC) =

=

=

= 3 × 3 × 5 × 2

= 90 m²
ar(ΔBEC) =× h × EC
90 = × h × 25

h =

h = 7.2 m
ar(ABED) = h × DE
= 7.2 × 24 = 172.8 m²
Total area = ar(ΔBEC) + ar(ABED)
= 90 + 172.8
= 262.8 m²
Cost for 1 year = Rs. 650 per m²
Cost for 5 months =
= Rs. 71,175
Question 20
1 / -0

Pahul is preparing a crown from a cardboard. She has joined three triangles and a rectangle to form this crown. Two identical triangles, which are joined at the ends, have three sides which measure 61 cm, 60 cm and 11 cm. The triangle which is at the middle has three sides which measure 50 cm, 24 cm and 34 cm. The base of this crown is in the shape of a rectangle whose length is 46 cm and width is 13 cm. What will be the area of this crown?

A
1510 cm²

B
1618 cm²

C
1735 cm²

D
2318 cm²

Solution

The sides of the two triangles which are at the sides = 61 cm, 11 cm and 60 cm

Semiperimeter = s === 66 cm

Area =

=

=

= 6 × 5 × 11 = 330 cm²So, the area of two triangles = 330 × 2 = 660 cm² ... (1)
The lengths of the triangle which is in the middle = 24 cm, 34 cm and 50 cm

Semiperimeter = = 54 cm

Area =

=

= 360 cm² ... (2)
Area of the base = l × b = 46 × 13
= 598 cm² ... (3)
Total area of the crown = Area of the rectangular base + 2 × Area of identical triangles + Area of the middle triangle
From eq (1), (2) and (3):
Total area of the crown = 598 + 360 + 660 = 1618 cm²
Question 21
1 / -0

Sameer makes a bird by cutting coloured paper in triangles of different sizes. Each of the length is in cm. If the cost of the coloured paper is Rs. 14 per m², what is the the approximate cost of making this whole bird? ()

A
Rs. 1,705

B
Rs. 1,642

C
Rs. 1,441

D
Rs. 1,253

Solution

There are 5 triangles with different sizes.

Triangle 1: a = 12, b = 13, c = 5

Triangle 2: a = 5, b = 4, c = 3

Triangle 3: a = 12, b = 15, c = 9

Triangle 4: a = 5, b = 4, c = 3

Triangle 5: a = 4, b = 4, c = 4

Triangle 1:
Sides of triangle = 12 cm, b = 13 cm and c = 5 cm

s ==

Area of a triangle =

=

= =

= 30 cm²

Triangles 2 and 4 are the same.
Given sides of the triangles:

a = 5 cm, b = 4 cm and c = 3 cm

s == = 6

Area of a triangle =

=

=

=

= 6 cm²

Triangle 3
Given sides of the triangle:
a = 12 cm, b = 15 cm and c = 9 cm

s == = 18

Area of triangle =

=

=

=

= 54 cm²

Triangle 5 is an equilateral triangle because all its sides are of length 4 cm.
So,
Area = = = 6.93 cm²

Sum = 30 + 6 + 6 + 54 + 6.93 = 102.93

Now,

Cost of paper = 102.93 × 14 = Rs. 1,441.02 = Rs. 1,441 (Approximately)
Question 22
1 / -0

Samya joined some identical triangular and square pieces of cardboard together to make multiple little wall hangings. One of them is given in the figure below (Not to the scale). Find the total area of the wall hanging.

A
213.12 cm²

B
213.02 cm²

C
212.2 cm²

D
203.12 cm²

Solution

We know that the triangle has sides as 3.6 cm, 4.8 cm and 6 cm.

Semiperimeter of each triangle =
=
=

= 7.2 cm

Area of a triangle =

=

= = 8.64 cm²

Now, in the wall hanging there are 8 triangles so,
Area of 8 triangles = 8.64 × 8 = 69.12 cm²
Also there are 4 squares inside, with each side = 6 cm
So, area of one square = 6 × 6 = 36 cm²
Area of four squares = 36 × 4 = 144 cm²
So, total area = 69.12 + 144 = 213.12 cm²
Question 23
1 / -0

Harmeet wants to put a boy scout symbol outside his place which is made up of 4 similar isosceles triangles as shown below. The area shared by the four triangles is 60% of the the total area of the four triangles. The area shared by the four triangles is to be coloured silver and the rest is to be coloured golden. If the cost of silver paint is 30 paise/cm² and that of golden paint is 70 paise/cm², then find the total cost to be incurred in painting the symbol, given that the base of the triangle is 24 cm and the other two sides are 15 cm each.

A
Rs. 170.25

B
Rs. 181.60

C
Rs. 185.25

D
Rs. 198.72

Solution

Since the given triangles are isosceles, the two sides a and b of one triangle = 15 cm
Base is 24 cm, so the side c of one triangle = 24 cm
Now, semiperimeter of one triangle =
=
= 27 cm

Area of one triangle =

=

=

=
= 108 cm²
Now there are four triangles, so total area = 108 × 4 = 432 cm²
Area shared by the 4 triangles = 60% of 432 = 0.6 × 432 = 259.2 cm²
Area of the part that is to be shaded silver = 259.2 cm²
Area of the part that is to be shaded golden = 432 - 259.2 = 172.8 cm²
Cost of colouring the part which is to be painted silver = 259.2 × 30 = 7,776 paise = Rs 77.76
Cost of coloring the part which is to be painted golden = 172.8 × 70 = 12,096 paise = Rs 120.96
Total cost = 77.76 + 120.96 = Rs. 198.72
Question 24
1 / -0

Somya made a paper boat by joining 3 triangles as shown. What is the total area of the boat?

A
212 cm²

B
204 cm²

C
198 cm²

D
186 cm²

Solution

There are three triangles:
Triangle ABC: a = 20, b = 16, c = 12
Triangle CDG: a = 15, b = 12, c = 9
Triangle EFB: a = 15, b = 12, c = 9
Area of triangle ABC:

S = = = 24 cm

Area of triangle ABC =

Area of triangle ABC =

Area of triangle ABC =

Area of triangle ABC =

Area of triangle ABC = 96 cm²Now,
Area of triangle CDG:

S = = = 18 cm

Area of triangle CDG =

Area of triangle CDG =

Area of triangle CDG =

Area of triangle CDG =

Area of triangle CDG = 54 cm²As triangle CDG and triangle EFB have same dimensions,
Area of triangle CDG = Area of Triangle EFB = 54 cm²
Hence, the total area of the boat = 96 + 54 + 54 = 204 cm²

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