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Maths 4 Marks Questions

Heron's Formula [NEW] · Gujarat Board (GSEB) STD 9 (GSEB - English Medium). 31 questions.

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Question 14 Marks
Two parallel sides of a trapezium are 60m and 77m and the other sides are 25m and 26m. Find the area of the trapezium.
Answer
Given,
Two parallel sides of trapezium are AB = 77m and CD = 60m
The other two parallel sides of trapezium are BC = 26m, AD = 25m

Join AE and CF
DE is perpendicular to AB and also, CF is perpendicular to AB
Therefore, DC = EF = 60m
Let AE = x
So, BF = 77 - 60 - x
BF = 17 - x
In $\triangle\text{ADE},$
By using Pythagoras theorem,
DE2 = AD2 - AE2
DE2 = 252 - x2
In $\triangle\text{BCF},$
By using Pythagoras theorem,
CF2 = BC2 - BF2
CF2 = 262 - (17 − x)2
Here, DE = CF
So, DE2 = CF2
252 - x2 = 262 - (17 - x)2
252 - x2 = 262 - (172 - 34x + x2)
252 - x2 = 262 - 172 + 34x + x2
252 = 262 - 172 + 34x
x = 7
DE2 = 252 - x2
$\text{DE}=\sqrt{625-49}$
DE = 24m
Area of trapezium $=\frac12\times(60+77)\times24$
Area of trapezium = 1644m2.
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Question 24 Marks
The sides of a quadrilateral, taken in order as 5m, 12m, 14m, 15m respectively. The angle contained by first two sides is a right angle. Find its area.

 

Answer
Given that the sides of the quadrilateral are
AB = 5m, BC = 12m, CD = 14m, DA = 15m
Join AC
Now, in $\triangle\text{ABC}=\frac12\times\text{AB}\times\text{BC}$
$=\frac{1}{2}\times5\times12$
= 30m2
In $\triangle\text{ABC},$ By applying Pythagoras theorem
AC2 = AB2 + BC2
$\text{AC}=\sqrt{5^2+12^2}$
AC = 13m
Now area of $\triangle\text{ADC},$
Perimeter = 2s = AD + DC + AC
2s = 15m + 14m + 13m
s = 21m
By using Heron's Formula
The area of a triangle PSR $=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
$=\sqrt{21\times(21-15)\times(21-14)\times(21-13)}$
= 84m2
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ADC
= (30 + 84) m2
= 114m2
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Question 34 Marks
The sides of a quadrilateral field, taken in order are 26m, 27m, 7m, 24m respectively. The angle contained by the last two sides is a right angle. Find its area.
Answer
Here the length of the sides of the quadrilateral is given as:

AB = 26m, BC = 27m, CD = 7m, DA = 24m
Diagonal AC is joined.
Now, in $\triangle\text{ADC}$
By applying Pythagoras theorem
AC2 = AD2 + CD2
AC2 = 142 + 72
AC = 25m
Now area of $\triangle\text{ABC}$
Perimeter = 2s = AB + BC + CA
2s = 26m + 27m + 25m
s = 39m
By using Heron's Formula
The area of a triangle $=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
$=\sqrt{39\times(39-26)\times(39-27)\times(39-25)}$
= 291.84m2
Thus, the area of a triangle is 291.84m2
Now for area of $\triangle\text{ADC}$
Perimeter = 2S = AD + CD + AC
= 25m + 24m + 7m
S = 28m
By using Heron's Formula 
The area of triangle $=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
$=\sqrt{28\times(28-24)\times(28-7)\times(28-25)}$
= 84m2
Thus, the area of a triangle is 84m2
Therefore, Area of rectangular field ABCD
= Area of triangle ABC + Area of triangle ADC
= 291.84 + 84
= 375.8m2
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Question 44 Marks
The perimeter of a triangullar field is 144m and the ratio of the sides is 3 : 4 : 5. Find the area of the field.
Answer
The area of a triangle having sides a, b, c and s as semi-perimeter is given by,
$\text{A}=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})},$ where,
$\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}$
It is given the sides of a triangular field are in the ratio 3 : 4 : 5 and perimeter = 144m
Therefore, a : b : c = 3 : 4 : 5
We will assume the sides of triangular field as
a = 3x; b = 4x; c = 5x
$2\text{s} = 144$
$\text{s}=\frac{144}{2}$
$\text{s} = 72$
$72=\frac{3\text{x}+4\text{x}+5\text{x}}{2}$
$72 \times 2 = 12\text{x}$
$\text{x}=\frac{144}{12}$
$\text{x} = 12$
Substituting the value of x in, we get sides of the triangle as
a = 3x = 3 × 12
a = 36m
b = 4x = 4 × 12
b = 48m
c = 5x = 5 × 12
c = 60m
Area of a triangular field, say A having sides a, b, c and s as semi-perimeter is given by:
a = 36m; b = 48m; c = 60m
$\text{s} = 72\text{m}$
$\text{A}=\sqrt{72(72-36)(72-48)(72-60)}$
$\text{A}=\sqrt{72(36)(24)(12)}$
$\text{A}=\sqrt{746496}$
$\text{A} = 864\text{m}^2.$
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Question 54 Marks
The perimeter of a triangular field is 540m and its sides are in the ratio 25 : 17 : 12. Find the area of triangle.
Answer
Let the sides of the given triangle be a = 25x, b = 17x, c = 12x respectively,
So,
a = 25x cm
b = 17x cm
c = 12x cm
Given Perimeter = 540cm
2s = a + b + c
a + b + c = 540cm
25x + 17x + 12x = 540cm
a = 250cm
b = 170cm
c = 120cm
Now, Semi perimeter $\text{s}=\frac{(\text{a}+\text{b}+\text{c})}{2}$
$=\frac{540}{2}$
$= 270\text{cm}$
By using Heron's Formula
The area of the triangle $=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
$=\sqrt{270\times(270-250)\times(270-170)\times(270-120)}$
$= 9000\text{cm}^2$
Therefore, the area of the triangle is 9000cm2
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Question 64 Marks
The perimeter of a triangular field is 240dm. If two of its sides are 78dm and 50dm, find the length of the perpendicular on the side of length 50dm from the opposite vertex.
Answer
Given,
In a triangle ABC, a = 78dm = AB, b = 50dm = BC
Now, Perimeter = 240dm
Then, AB + BC + AC = 240dm
78 + 50 + AC = 240
AC = 240 - (78 + 50)
AC = 112dm = c
Now, 2s = a + b + c
2s = 78 + 50 + 112
s = 120dm
Area of the triangle ABC $=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
$=\sqrt{120\times(120-78)\times(120-50)\times(120-112)}$
$= 1680\text{dm}^2$
Let AD be a perpendicular on BC
Area of the triangle ABC $=\frac12\times\text{AD}\times\text{BC}$
$\frac{1}{2}\times\text{AD}\times\text{BC}=1680\text{dm}^2$
$\text{AD} = 67.2\text{dm}.$
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Question 74 Marks
The perimeter of a triangle is 300m. If its sides are in the ratio of 3 : 5 : 7. Find the area of the triangle.
Answer
Given the perimeter of a triangle is 300 m and the sides are in a ratio of 3 : 5 : 7
Let the sides a, b, c of a triangle be 3x, 5x, 7x respectively
So, the perimeter = 2s = a + b + c
200 = a + b + c
300 = 3x + 5x + 7x
300 = 15x
Therefore, x = 20m
So, the respective sides are
a = 60m
b = 100m
c = 140m
Now, semi perimeter
$\text{s}=\frac{(\text{a}+\text{b}+\text{c})}{2}$
$\text{s}=\frac{(60+100+140)}{2}$
$= 150\text{m}$
By using Heron's Formula
The area of the triangle $=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
$=\sqrt{150\times(150-60)\times(150-100)\times(150-140)}$
$=1500\sqrt{3}\text{m}^2$
Thus, the area of a triangle is $1500\sqrt{3}\text{m}^2$
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Question 84 Marks
The perimeter of an isosceles triangle is 42cm and its base is $\Big(\frac32\Big)$ times each of the equal side. Find the length of each of the triangle, area of the triangle and the height of the triangle.
Answer
Let 'x' be the length of two equal sides,
Therefore the base $=\frac12\times\text{x}$
Let the sides a, b, c of a triangle be $\frac12\times\text{x},$ x and x respectively
So, the perimeter = 2s = a + b + c
42 = a + b + c
$42=\frac32\times\text{x}+\text{x}+\text{x}$
Therefore, x = 12cm
So, the respective sides are:
a = 12cm
b = 12cm
c = 18cm
Now, semi perimeter
$\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}$
$=\frac{12+12+18}{2}$
$= 21\text{cm}$
By using Heron's Formula,
The area of a triangle $=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
$=\sqrt{21\times(21-12)\times(21-12)\times(21-18)}$
$= 71.42\text{cm}^2$
Thus, the area of a triangle is 70.42cm2
The altitude will be smallest provided the side corresponding to this altitude is longest.
The longest side = 18cm
Area of the triangle $=\frac12\times\text{h}\times18$
$\frac12\times\text{h}\times18=71.42\text{cm}^2$
$\text{h} = 7.94\text{cm}$
Hence the length of the smallest altitude is 7.93cm.
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Question 94 Marks
The lengths of the sides of a triangle are in a ratio of 3 : 4 : 5 and its perimeter is 144cm. Find the area of the triangle and the height corresponding to the longest side.
Answer
Given the perimeter of a triangle is 160m and the sides are in a ratio of 3 : 4 : 5
Let the sides a, b, c of a triangle be 3x, 4x, 5x respectively
So, the perimeter = 2s = a + b + c
144 = a + b + c
144 = 3x + 4x + 5x
Therefore, x = 12cm
So, the respective sides are:
a = 36cm
b = 48cm
c = 60cm
Now, semi perimeter
$\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}$
$=\frac{36+45+60}{2}$
$= 72\text{cm}$
By using Heron's Formula,
The area of a triangle $=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
$=\sqrt{72\times(72-36)\times(72-48)\times(72-60)}$
$= 864\text{cm}^2$
Thus, the area of a triangle is 864cm2
The longest side = 60cm
Area of the triangle $=\frac12\times\text{h}\times60$
$\frac12\times\text{h}\times60=864\text{cm}^2$
$\text{h} = 28.8\text{cm}$
Hence the length of the smallest altitude is 28.8cm.
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Question 104 Marks
The adjacent sides of a parallelogram ABCD measure 34cm and 20cm, and the diagonal AC measures 42cm. Find the area of parallelogram.
Answer
The adjacent sides of a parallelogram ABCD measures 34cm and 20cm, and the diagonal AC measures 42cm.

Area of the parallelogram = Area of $\triangle\text{ADC}$ + Area of $\triangle\text{ABC}$

Note: Diagonal of a parallelogram divides into two congruent triangles

Therefore,

Area of the parallelogram = 2 × $\big($Area of $\triangle\text{ABC}\big)$

Now, for area of $\triangle\text{ABC}$

Perimeter = 2s = AB + BC + CA

2s = 34cm + 20cm + 42cm

s = 48cm

By using Heron's Formula,

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

$=\sqrt{48\times(14)\times(28)\times(6)}$

= 336cm2

Therefore, area of parallelogram ABCD = 2 × $\big($Area of $\triangle\text{ABC}\big)$

Area of parallelogram = 2 × 336cm2

Area of parallelogram ABCD = 672cm2

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Question 114 Marks

Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.

Answer
Whenever we are given the measurements of all sides of a triangle, we basically look for Heron's formula to find out the area of the triangle.
If we denote area of the triangle by A, then the area of a triangle having sides a, b, c and s as semi-perimeter is given by:
$\text{A}=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
Where, $\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}$
We are given:
a = 18cm
b = 10cm, and perimeter = 42cm
We know that perimeter = 2s,
So, 2s = 42
Therefore, s = 21cm
We know that, $\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}$
$21=\frac{18+10+\text{c}}{2}$
42 = 28 + c
c = 14cm
So the area of the triangle is:
$\text{A}=\sqrt{21\times(21-18)\times(21-10)\times(21-14)}$
$\text{A}=\sqrt{21\times(3)\times(11)\times(7)}$
$\text{A}=21\sqrt{11}\text{cm}^2$
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Question 124 Marks

Find the area of a triangle whose sides are respectively 9cm, 12cm and 15cm.

Answer
Let the sides of the given triangle be a, b, c respectively.
So given,
a = 9cm
b = 12cm
c = 15cm
By using Heron's Formula
The area of the triangle $=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
Semi perimeter of a triangle = s
2s = a + b + c
$\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}$
$\text{s}=\frac{(9+12+15)}{2}$
$\text{s} = 18\text{cm}$
$\therefore$ Area of the triangle $=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
$=\sqrt{18\times(18-9)\times(18-12)\times(18-15)}$
$= 54\text{cm}^2$
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Question 134 Marks
Let $\triangle$ be the area of a triangle. Find the area of a triangle whose each side is twice the side of the given triangle.
Answer
We are given assumed value $\triangle$ is the area of a given $\triangle\text{ABC}$
We assume the sides of the given triangle ABC be a, b, c
The area of a triangle having sides a, b, c and s as semi-perimeter is given by,
$\text{A}=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$\triangle=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
Where,
$\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}$
$2\text{s} = \text{a} + \text{b} + \text{c}$
We take the sides of a new triangle as 2a, 2b, 2c that is twice the sides of previous one
Now, the area of a triangle having sides 2a, 2b, and 2c and s1 as semi-perimeter is given by,
$\text{A}_1=\sqrt{\text{s}_1(\text{s}_1-2\text{a})(\text{s}_1-2\text{c})(\text{s}_1-2\text{c})},$ where
$\text{s}_1=\frac{2\text{a}+2\text{b}+2\text{c}}{2}$
$\text{s}_1=\frac{2(\text{a}+\text{b}+\text{c})}{2}$
$\text{s}_1 = \text{a} + \text{b} + \text{c}$
$\text{s}_1 = 2\text{s}$
Now,
$\text{A}_1=\sqrt{2\text{s}(2\text{s}-2\text{a})(2\text{s}-2\text{b})(2\text{s}-2\text{c})}$
$\text{A}_1=\sqrt{2\text{s}\times2(\text{s}-\text{a})\times2(\text{s}-\text{b})\times2(\text{s}-\text{c})}$
$\text{A}_1=4\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$\text{A}_1=4\triangle$
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Question 144 Marks
In a triangle $\triangle\text{ABC},$ AB = 15cm, BC = 13cm and AC = 14cm. Find the area of triangle ABC and hence its altitude on AC.
Answer
Let the sides of the given triangle be AB = a, BC = b, AC = c respectively.
So given,
a = 15cm
b = 13cm
c = 14cm
By using Heron's Formula
The Area of the triangle $=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
Semi perimeter of a triangle = 2s
2s = a + b + c
$\text{s}=\frac{(\text{a}+\text{b}+\text{c})}{2}$
$\text{s}=\frac{(15+13+14)}{2}$
$\text{s} = 21\text{cm}$
$\therefore$ Area of the triangle $=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
$=\sqrt{21\times(21-15)\times(21-13)\times(21-14)}$
$= 84\text{cm}^2$
BE is a perpendicular on AC
Now, area of triangle = 84cm2
$\frac12\times\text{BE}\times\text{AC}=84\text{cm}^2$
$\text{BE} = 12\text{cm}$
The length of BE is 12cm.
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Question 154 Marks
If each side of a triangle is doubled, the find percentage increase in its area.
Answer
The area of a triangle having sides a, b, c and s as semi-perimeter is given by,
$\text{A}=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
where,
$\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}$
2s = a + b + c
We take the sides of a new triangle as 2a, 2b, 2c that is twice the sides of previous one
Now, the area of a triangle having sides 2a, 2b, and 2c and s1 as semi-perimeter is given by,
$\text{A}_1=\sqrt{\text{s}_1(\text{s}_1-2\text{a})(\text{s}_1-2\text{c})(\text{s}_1-2\text{c})},$
Where,
$\text{s}_1=\frac{2\text{a}+2\text{b}+2\text{c}}{2}$
$\text{s}_1=\frac{2(\text{a}+\text{b}+\text{c})}{2}$
$\text{s}_1 = \text{a} + \text{b} + \text{c}$
$\text{s}_1 = 2\text{s}$
Now,
$\text{A}_1=\sqrt{2\text{s}(2\text{s}-2\text{a})(2\text{s}-2\text{b})(2\text{s}-2\text{c})}$
$\text{A}_1=\sqrt{2\text{s}\times2(\text{s}-\text{a})\times2(\text{s}-\text{b})\times2(\text{s}-\text{c})}$
$\text{A}_1=4\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$\text{A}_1=4\triangle$
Therefore, increase in the area of the triangle
= A1 - A
= 4A - A
= 3A
Percentage increase in area 
$=\frac{3\text{A}}{\text{A}}\times100$
$= 300\%.$
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Question 164 Marks
If each side of a equilateral triangle is tripled then what is the percentage increase in the area of the triangle?
Answer
Area of an equilateral triangle having each side a cm is given by:
$\text{A}=\frac{\sqrt{3}\text{a}^2}{4}$
Now, Area of an equilateral triangle, say A1 if each side is tripled is given by;
a = 3a
$\text{A}_1=\frac{\sqrt{3}}{4}\text{a}^2$
$\text{A}_1=\frac{\sqrt{3}}{4}(3\text{a})^2$
$\text{A}_1=\frac{9\sqrt{3}\text{a}^2}{4}\text{cm}^2$
Therefore, increase in area of triangle
= A1 - A
$=\frac{9\sqrt{3}\text{a}^2}{4}-\frac{\sqrt{3}\text{a}^2}{4}$
$=\frac{8\sqrt{3}\text{a}^2}{4}$
Percentage increase in area
$=\frac{\frac{8\sqrt{3}\text{a}^2}{4}}{\frac{\sqrt{3}\text{a}^2}{4}}\times100$
$= 800\%.$
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Question 174 Marks
Find the perimeter and area of the quadrilateral ABCD in which AB = 17cm, AD = 9cm, CD = 12cm, $\angle\text{ACB}=90^\circ$ and AC = 15cm.
Answer
We assume ABCD be the quadrilateral having sides AB, BC, CD, DA and $\angle\text{ACB}=90^\circ.$
We take a diagonal AC, where AC divides ABCD into two triangle $\triangle\text{ACB}$ and $\triangle\text{ADC}$

Since $\triangle\text{ACB}$ is right angled at C, we have
AC = 15cm; AB = 17cm
AB2 = AC2 + BC2
(17)2 = (15)2 + BC2
289 = 225 + BC2
BC2 = 289 - 225
$\text{BC}^2=\sqrt{64}$
BC = 8cm
Area of right angled $\triangle\text{ABC},$ say A1 is given by
$\text{A}_1=\frac12(\text{Base}\times\text{Height}),$ where,
Base = BC = 8cm; Height = AC = 15cm
P = 9 + 12 + 8 + 17
= 46cm
P = 46cm.
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Question 184 Marks
Find the area of the shaded region in fig. below

Answer
Area of the shaded region = Area of $\triangle\text{ABC}-$ Area of $\triangle\text{ADB}$
Now in triangle ADB
AB2 = AD2 + BD2....(i)
Given, AD = 12cm, BD =16cm
Substituting the value of AD and BD in eq (i), we get
AB2 = 122 + 162
= 400cm2
AB = 20cm
Now, area of a triangle $=\frac12\times\text{AD}\times\text{BD}$
= 96cm2
Now in triangle ABC,
$\text{s}=\frac12\times(\text{AB}+\text{BC}+\text{CA})$
$=\frac12\times(52+48+20)$
$= 60\text{cm}$
By using Heron's Formula
The area of a triangle $=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
$=\sqrt{60\times(60-20)\times(60-48)\times(60-52)}$
= 480cm2
Thus, the area of a triangle is 480cm2
Area of shaded region = Area of $\triangle\text{ABC}-$ Area of $\triangle\text{ADB}$
= (480 - 96)cm2
= 384cm2
Area of shaded region = 384cm2
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Question 194 Marks
Find the area of the quadrilateral ABCD in which AD = 24cm, $\angle\text{BAD}=90^\circ$ and BCD forms an equilateral triangle whose each side is equal to 26cm. $\big($ Take $\sqrt{3}=1.73\big)$
Answer
Given that, in a quadrilateral ABCD in which AD = 24cm,
$\angle\text{BAD}=90^\circ$
BCD is an equilateral triangle and the sides BC = CD = BD = 26cm
In $\triangle\text{BAD},$ by applying Pythagoras theorem,

BA2 = BD2 − AD2
BA2 = 262 + 242
$\text{BA}=\sqrt{100}$
BA = 10cm
Area of the $\triangle\text{BAD}=\frac12\times\text{BA}\times\text{AD}$
Area of the $\triangle\text{BAD}=\frac12\times10\times24$
Area of the triangle BAD = 120cm2
Area of the equilateral triangle $=\sqrt{\frac34}\times\text{side}$
Area of the equilateral $\triangle\text{QRS}=\sqrt{\frac34}\times36$
Area of the equilateral triangle BCD = 292.37cm2
Therefore, the area of quadrilateral ABCD = Area of $\triangle\text{BAD}$ + Area of the $\triangle\text{BCD}$
The area of quadrilateral ABCD = 120 + 292.37
= 412.37cm2
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Question 204 Marks
Find the area of the quadrilateral ABCD in which AB = 3cm, BC = 4cm, CD = 4cm, DA = 5cm and AC = 5cm.
Answer
For $\triangle\text{ABC}$

AC2 = BC2 + AB2
25 = 9 + 16
So, $\triangle\text{ABC}$ is a right angle triangle right angled at point R
Area of triangle ABC = 12 × AB × BC
$=\frac12\times3\times4$
= 6cm2
From $\triangle\text{CAD}$
Perimeter = 2s = AC + CD + DA
2s = 5cm + 4cm + 5cm
2s = 14cm
s = 7cm
By using Heron's Formula
Area of the triangle CAD $=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
$=\sqrt{7\times(7\times5)\times(7-4)\times(7-5)}$
= 9.16cm2
Area of ABCD = Area of ABC + Area of CAD
= (6 + 9.16)cm2
= 15.16 cm2
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Question 214 Marks
Find the area of the of the blades of the magnetic compass shown in Fig. below
$\big($ Take $\sqrt{11}=3.32\big).$

Answer
Area of the blades of magnetic compass = Area of $\triangle\text{ADB}$ + Area of $\triangle\text{CDB}$
Now, for the area of $\triangle\text{ADB}$
Perimeter = 2s = AD + DB + BA
2s = 5cm + 1cm + 5cm
s = 5.5cm
By using Heron's Formula,
Area of the $\triangle\text{DEF}=\sqrt{\text{s}(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
$=\sqrt{5.5\times(5.5)\times(4.5)\times(0.5)}$
= 2.49cm2
Also, area of $\triangle\text{ADB}$ = Area of $\triangle\text{CDB}$
Therefore area of the blades of the magnetic compass = 2 × area of $\triangle\text{ADB}$
Area of the blades of the magnetic compass = 2 × 2.49
Area of the blades of the magnetic compass = 4.98cm2
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Question 224 Marks
Find the area of quadrilateral ABCD in which AB = 42cm, BC = 21cm, CD = 29cm, DA = 34cm and the diagonal BD = 20cm.

 

Answer
Given:
AB = 42cm, BC = 21cm, CD = 29cm, DA = 34cm, and the diagonal

BD = 20cm.
Now, for the area of triangle ABD
Perimeter of triangle ABD 2s = AB + BD + DA
2s = 34cm + 42cm + 20cm
s = 48cm
By using Heron’s Formula,
Area of the $\triangle\text{ABD}=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
$=\sqrt{48\times(48-42)\times(48-20\times(48-34)}$
= 336cm2
Now, for the area of triangle BCD
Perimeter of triangle BCD 2s = BC + CD + BD
2s = 29cm + 21cm + 20cm
s = 35cm
By using Heron's Formula,
Area of the $\triangle\text{BCD}=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
$=\sqrt{35\times(14)\times(6)\times(15)}$
= 210cm2
Therefore, Area of quadrilateral ABCD = Area of $\triangle\text{ABC}$ + Area of $\triangle\text{BCD}$
Area of quadrilateral ABCD = 336 + 210
Area of quadrilateral ABCD = 546cm2
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Question 234 Marks
Find the area of a triangle whose sides are respectively 150cm, 120cm and 200cm.
Answer
Let the sides of the given triangle be a, b, c respectively.
So given,
a = 150cm
b = 120cm
c = 200cm
By using Heron's Formula
The Area of the triangle $=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
Semi perimeter of a triangle = s
2s = a + b + c
$\text{s}=\frac{(\text{a}+\text{b}+\text{c})}{2}$
$\text{s}=\frac{(150+200+120)}{2}$
s = 235cm
Therefore,
Area of the triangle $=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
$=\sqrt{235\times(235-150)\times(235-200)\times(235-120)}$
= 8966.56cm2
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Question 244 Marks
Find the area of a triangle whose sides are 3cm, 4cm and 5cm respectively.
Answer
The area of a triangle having sides a, b, c and s as semi-perimeter is given by,
$\text{A}=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})},$ where
$\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}$
Therefore the area of a triangle, say having sides 3cm, 4cm and 5cm is given by:
a = 3cm; b = 4cm; c = 5cm
$\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}$
$\text{s}=\frac{2+4+5}{2}$
$\text{s}=\frac{12}{2}$
$\text{s} = 6\text{cm}$
Now, area $\text{A}=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{6(6-3)(6-4)(6-5)}$
$=\sqrt{6\times3\times2\times1}$
$=\sqrt{36}$
$= 6\text{cm}^2.$
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Question 254 Marks
Find the area of a rhombus whose perimeter is 80m and one of whose diagonal is 24m.
Answer
Perimeter of a rhombus = 80m

As we know,
Perimeter of a rhombus = 4 × side = 4 × a
4 × a = 80m
a = 20m
Let AC = 24m
Therefore, $\text{OA}=\frac12\times\text{AC}$
OA = 12m
In $\triangle\text{AOB}$
OB2 = AB2 − OA2
OB2 = 202 − 122
OB = 16m
Also, OB = OD because diagonal of rhombus bisect each other at 90°
Therefore, BD = 2 OB = 2 × 16 = 32m
Area of rhombus $=\frac12\times\text{BD}\times\text{AC}$
Area of rhombus $=\frac12\times32\times24$
Area of rhombus $= 384\text{m}^2$
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Question 264 Marks
Find the area of an isosceles triangle having the base x cm and one side y cm.
Answer
Let us assume triangle ABC be the given isosceles triangle having sides AB = AC and base BC. The area of a triangle ABC, say A having given sides AB and AC equals to y cm and given base BC equals to x cm is given by:
$\text{A}=\frac12(\text{Base}\times\text{Height})$
Where,
Base = BC = x cm; Height $=\sqrt{\text{y}^2-\frac{\text{x}^2}{4}}$
$\text{A}=\frac12(\text{Base}\times\text{Height})$
$=\frac12\times\text{x}\Big(\sqrt{\text{y}^2-\frac{\text{x}^2}{4}}\Big)$
$=\frac{\text{x}}{2}\Big(\sqrt{\text{y}^2-\frac{\text{x}^2}{4}}\Big)$
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Question 274 Marks
A triangle has sides 35cm, 54cm, 61cm long. Find its area. Also, find the smallest of its altitudes.
Answer
Given,
The sides of the triangle are
a = 35cm
b = 54cm
c = 61cm
Perimeter 2s = a + b + c
2s = 35 + 54 + 61cm
Semi perimeter s = 75cm
By using Heron's Formula,
$=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
$=\sqrt{75\times(75-35)\times(75-54)\times(75-61)}$
$= 939.14\text{cm}^2$
The altitude will be smallest provided the side corresponding to this altitude is longest.
The longest side = 61cm
Area of the triangle $=\frac12\times\text{h}\times61$
$\frac12\times\text{h}\times61=939.14\text{cm}^2$
$\text{h} = 30.79\text{cm}$
Hence the length of the smallest altitude is 30.79cm.
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Question 284 Marks
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 13cm, 14cm and 15cm and the parallelogram stands on the base 14cm, find the height of a parallelogram.
Answer
The sides of the triangle DCE are:
DC = 15cm,
CE = 13cm,
Let the h be the height of parallelogram ABCD
Now, for the area of $\triangle\text{DCE}$
Perimeter = DC + CE + ED
2s = 15cm + 13cm + 14cm
s = 21cm
By using Heron's Formula,
Area of the $\triangle\text{AOB}=\sqrt{\text{s}(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
$=\sqrt{21\times(7)\times(8)\times(6)}$
= 84cm2
Also, area of $\triangle\text{DCE}$ = Area of parallelogram ABCD ⇒ 84cm2
24 × h = 84cm2
h = 6cm.
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Question 294 Marks
A rhombus sheet, whose perimeter is 32m and whose diagonal is 10m long, is painted on both the sides at the rate of ₹ 5 per meter square. Find the cost of painting.
Answer
Given that,
Perimeter of a rhombus = 32m
We know that,
Perimeter of a rhombus = 4 × side
4 × side = 32m
4 × a = 32m
a = 8m
Let AC = 10m
OA = 12 × AC
OA = 12 × 10
OA = 5m

By using Pythagoras theorem
OB2 = AB2 − OA2
OB2 = 82 − 52
$\text{OB}=\sqrt{39}\text{m}$
BD = 2 × OB
$\text{BD}=2\sqrt{39}\text{m}$
Area of the sheet $=\frac{1}{2}\times\text{BD}\times\text{AC}$
Area of the sheet $=\frac12\times2\sqrt{39}\times10$
Therefore, cost of printing on both sides at the rate of ₹ 5 per m2
$=₹2\times10\sqrt{39}\times5$
$= ₹ 625.$
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Question 304 Marks
A park in the shape of a quadrilateral ABCD, has angle $\angle\text{C}=90^\circ,$ AB = 9m, BC = 12m, CD = 5m, AD = 8m. How much area does it occupy.
Answer
Given that the sides of the quadrilateral are
AB = 9m, BC = 12m, CD = 5m, DA = 8m

In $\triangle\text{BCD},$ apply Pythagoras theorem
BD2 = BC2 + CD2
BD2 = 122 + 52
BD = 13m
Area of $\triangle\text{BCD}=\frac12\times\text{BC}\times\text{CD}$
$=\frac12\times12\times5$
= 30m2
Now, in $\triangle\text{ABD},$
Perimeter = 2s = 9 m + 8m + 13m
s = 15m
By using Heron's Formula
The area of a triangle PSR $=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
$=\sqrt{15\times(15-9)\times(15-8)\times(15-13)}$
= 35.49m2
Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD
= (35.496 + 30)m2
= 65.5m2
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Question 314 Marks
A hand fan is made by sticking 10 equal size triangular strips of two different types of paper as shown in the figure. The dimensions of equal strips are 25cm, 25cm and 14cm. Find the area of each type of paper needed to make the hand fan.

Answer
Given that,
AO = 25cm
OB = 25cm
BA = 14cm
Area of each strip = Area of $\triangle\text{AOB}$
Now, for the area of $\triangle\text{AOB}$
Perimeter = AO + OB + BA
2s = 25cm +25cm + 14cm
s = 32cm
By using Heron's Formula,
Area of the $\triangle\text{AOB}=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
$=\sqrt{32\times(7)\times(4)\times(18)}$
= 168cm2
Also, area of each type of paper needed to make a fan = 5 × Area of $\triangle\text{AOB}$
Area of each type of paper needed to make a fan = 5 × 168cm2
Area of each type of paper needed to make a fan = 840cm2.
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