Question 14 Marks
Find the area of the of the blades of the magnetic compass shown in Fig. below$\big($ Take $\sqrt{11}=3.32\big).$
AnswerArea 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 $\triangle\text{DEF}=\sqrt{\text{s}(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
$=\sqrt{5.5\times(5.5)\times(4.5)\times(0.5)}$
$= 2.49cm^2$
Also, area of $\triangle\text{ADB}$ = Area of $\triangle\text{CDB}$
Therefore area of the blades of the magnetic compass = $2 \times $ area of $\triangle\text{ADB}$
Area of the blades of the magnetic compass $= 2 \times 2.49$
Area of the blades of the magnetic compass $= 4.98cm^2$
View full question & answer→Question 24 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.
AnswerThe 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$\text{s} = 72\text{m}$
$\text{A}=\sqrt{72(72-36)(72-48)(72-60)}$
$\text{A}=\sqrt{72(36)(24)(12)}$
$\text{A}=\sqrt{746496}$
$\text{A} = 864\text{m}^2.$
View full question & answer→Question 34 Marks
The lengths of the sides of a triangle are in a ratio of $3: 4: 5$ and its perimeter is 144 cm . Find the area of the triangle and the height corresponding to the longest side.
AnswerGiven the perimeter of a triangle is 160 m and the sides are in a ratio of $3: 4: 5$ Let the sides $a, b, c$ of a triangle be $3 x , 4 x , 5 x$ respectively
So, the perimeter $=2 s=a+b+c 144=a+b+c 144=3 x+4 x+5 x$ Therefore, $x=12 cm$
So, the respective sides are: $a=36 cm b =48 cm c =60 cm$
Now, semi perimeters $=\frac{ a + b + c }{2}$
$=\frac{36+45+60}{2}$
$=72 cm$
By using Heron's Formula, The area of a triangle $=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}$
$=\sqrt{72 \times(72-36) \times(72-48) \times(72-60)}$
$=864 cm^2$
Thus, the area of a triangle is $864 cm^2$ The longest side $=60 cm$ Area of the triangle $=\frac{1}{2} \times h \times 60$
$\frac{1}{2} \times h \times 60=864 cm^2$
$h=28.8 cm$
Hence the length of the smallest altitude is 28.8 cm .
View full question & answer→Question 44 Marks
A park in the shape of a quadrilateral $A B C D$, has angle $\angle C=90^{\circ}, A B=9 m, B C=12 m, C D=5 m, A D=8 m$. How much area does it occupy.
AnswerGiven that the sides of the quadrilateral are $A B=9 m, B C=12 m, C D=5 m, DA =8 m$
In $\triangle B C D$, apply Pythagoras theorem $B D^2=B C^2+C D^2 B D^2=12^2+5^2 B D=13 m$
Area of $\triangle BCD =\frac{1}{2} \times BC \times CD$
$=\frac{1}{2} \times 12 \times 5 \\
=30 m^2$
Now, in $\triangle ABD$,
Perimeter $=2 s=9 m+8 m+13 m s=15 m$
By using Heron's Formula The area of a triangle PSR $=\sqrt{ s \times( s - a ) \times( s - b ) \times( s - c )}$
$=\sqrt{15 \times(15-9) \times(15-8) \times(15-13)} \\
=35.49 m^2$
Area of quadrilateral $A B C D=$ Area of triangle $A B D+$ Area of triangle $B C D$
$=(35.496+30) m^2 \\
=65.5 m^2$ View full question & answer→Question 54 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.
AnswerGiven that, Perimeter of a rhombus $=32 m$
We know that, Perimeter of a rhombus $=4 \times$ side $4 \times$ side $=32 m 4 \times a =32 m$ a $=8 m$
Let $AC =10 m O A=12 \times ACOA =12 \times 10 OA =5 m$
By using Pythagoras theorem $O B^2=A B^2-O A^2 O B^2$
$=8^2-5^2$
$OB=\sqrt{39} m$
$BD=2 \times OBBD=2 \sqrt{39} m$
Area of the sheet $=\frac{1}{2} \times BD \times AC$
Area of the sheet $=\frac{1}{2} \times 2 \sqrt{39} \times 10$
Therefore, cost of printing on both sides at the rate of ₹ 5 per $m ^2=₹ 2 \times 10 \sqrt{39} \times 5$
$=₹ 625$. View full question & answer→Question 64 Marks
If each side of a triangle is doubled, the find percentage increase in its area.
AnswerThe 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 $s_1$ 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 = $A_1- A = 4A - A = 3A$
Percentage increase in area $=\frac{3\text{A}}{\text{A}}\times100$
$= 300\%.$
View full question & answer→Question 74 Marks
The perimeter of a triangular field is 540m and its sides are in the ratio $25 : 17 : 12$. Find the area of triangle.
AnswerLet the sides of the given triangle be $a=25 x, b=17 x, c=12 x$
respectively, So, $a=25 x cm b =17 x cm c =12 x cm$
Given Perimeter $=540 cm 2 s= a + b + c a + b + c =540 cm 25 x +17 x +12 x =540 cm a =250 cm b =170 cm c =$ 120 cm
Now, Semi perimeter $s =\frac{( a + b + c )}{2}=\frac{540}{2}$
$=270 cm$
By using Heron's Formula The area of the triangle
$=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}=\sqrt{270 \times(270-250) \times(270-170) \times(270-120)} \\
=9000 cm^2$
Therefore, the area of the triangle is $9000 cm^2$
View full question & answer→Question 84 Marks
The adjacent sides of a parallelogram $ABCD$ measure $34cm$ and $20cm$, and the diagonal $AC$ measures $42cm$. Find the area of parallelogram.
AnswerThe 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 = $2 \times$ $\big($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
$\triangle\text{ABC}=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
$=\sqrt{48\times(14)\times(28)\times(6)}$
$= 336cm^2$
Therefore, area of parallelogram ABCD $= 2 \times$ $\big($Area of $\triangle\text{ABC}\big)$
Area of parallelogram $= 2 \times 336cm^2$
Area of parallelogram ABCD $= 672cm^2$ View full question & answer→Question 94 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.
AnswerLet '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 $70.42cm^2$ 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.
View full question & answer→Question 104 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.
AnswerHere the length of the sides of the quadrilateral is given as:
$AB=26 m, BC=27 m, CD=7 m, DA=24 m$
Diagonal $A C$ is joined.
Now, in $\triangle ADC$
By applying Pythagoras theorem $AC ^2$
$=A D^2+C D^2 A C^2$
$=14^2+7^2 A C=25 m$
Now area of $\triangle ABC$ Perimeter $=2 s= AB + BC + CA 2 s=26 m+27 m+25 ms=39 m$
By using Heron's Formula The area of a triangle $=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}$
$=\sqrt{39 \times(39-26) \times(39-27) \times(39-25)}$
$=291.84 m^2$
Thus, the area of a triangle is $291.84 m^2$
Now for area of $\triangle ADC$
Perimeter $=2 S= AD + CD + AC =25 m+24 m+7 mS =28 m$ By using Heron's Formula The area of triangle
$=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}=\sqrt{28 \times(28-24) \times(28-7) \times(28-25)}$
$=84 m^2 \text { Thus, the area of a triangle is } 84 m^2$
Therefore, Area of rectangular field $A B C D=$ Area of triangle $A B C+$ Area of triangle $A D C$
$=291.84+84=375.8 m^2$ View full question & answer→Question 114 Marks
A triangle has sides 35cm, 54cm, 61cm long. Find its area. Also, find the smallest of its altitudes.
AnswerGiven, 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.
View full question & answer→Question 124 Marks
Find the area of a rhombus whose perimeter is 80m and one of whose diagonal is $24m.$
AnswerPerimeter of a rhombus = $80m$
As we know,
Perimeter of a rhombus $=4 \times$ side $=4 \times$ a
$4 \times a=80 m$
$a=20 m$
Let $A C=24 m$
Therefore, $OA =\frac{1}{2} \times AC$
$O A=12 m$
$\text { In } \triangle AOB$
$O B^2=A B^2-O A^2$
$O B^2=20^2-12^2$
$O B=16 m$
Also, $O B=O D$ because diagonal of rhombus bisect each other at $90^{\circ}$
Therefore, $BD =2 OB =2 \times 16=32 m$
Area of rhombus $=\frac{1}{2} \times BD \times AC$
Area of rhombus $=\frac{1}{2} \times 32 \times 24$
Area of rhombus $=384 m^2$ View full question & answer→Question 134 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.
AnswerWe 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 $\triangle ACB$ is right angled at C ,
we have $A C=15 cm ; A B=17 cm A B 2$
$=AC 2+BC 2(17)^2=(15)^2+BC^2 289$
$=225+BC^2 BC^2$
$=289-225$
$BC^2=\sqrt{64}$
$BC =8 cm$ Area of right angled $\triangle ABC$,
say $A_1$ is given by $A_1=\frac{1}{2}$ (Base $\times$ Height),
where, Base $=B C=8 cm$;
Height $=A C=15 cm$
$P=9+12+8+17=46 cm P=46 cm$ View full question & answer→Question 144 Marks
Find the area of a triangle whose sides are respectively $150cm, 120cm$ and $200cm.$
AnswerLet the sides of the given triangle be a, b, c respectively.
So given, $a=150 cm b=120 cm c =200 cm$ By using Heron's Formula The Area of the triangle
$=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}$
Semi perimeter of a triangle
$=s 2 s=a+b+cs=\frac{(a+b+c)}{2}$
$s=\frac{(150+200+120)}{2}$
$s=235 cm$
Therefore, Area of the triangle $=\sqrt{ s \times( s - a ) \times( s - b ) \times( s - c )}$
$=\sqrt{235 \times(235-150) \times(235-200) \times(235-120)}$
$=8966.56 cm^2$
View full question & answer→Question 154 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.
AnswerGiven that the sides of the quadrilateral are $A B=5 m, B C=12 m, C D=14 m, DA =15 m$ Join AC Now, in $\triangle ABC =\frac{1}{2} \times AB \times BC$
$=\frac{1}{2} \times 5 \times 12$
$=30 m^2 \ln \triangle ABC$
By applying Pythagoras theorem
$A C^2=A B^2+B C^2 A C=\sqrt{5^2+12^2}$
$A C=13 $
Now area of $\triangle ADC$,
$\text { Perimeter }=2 s$
$=A D+D C+A C 2 s$
$=15 m+14 m+13 m s$
$=21 m$
By using Heron's Formula The area of a triangle PSR
$=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}$
$=\sqrt{21 \times(21-15) \times(21-14) \times(21-13)}$
$=84 m^2$
Area of quadrilateral $A B C D=$ Area of triangle $A B C+$ Area of triangle $A D C$
$=(30+84) m ^2=114 m^2$
View full question & answer→Question 164 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.
AnswerGiven 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})}$$=\sqrt{150\times(150-60)\times(150-100)\times(150-140)}$
$=1500\sqrt{3}\text{m}^2$
Thus, the area of a triangle is $1500\sqrt{3}\text{m}^2$
View full question & answer→Question 174 Marks
Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.
AnswerWhenever we are given the measurements of all sides of a triangle, we basically look for Heron's formula to find out the area of the triangle. If we denote area of the triangle by A, then the area of a triangle having sides a, b, c and s as semi-perimeter is given by:$\text{A}=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
Where, $\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}$ We are given: a = 18cm b = 10cm, and perimeter = 42cm We know that perimeter = 2s, So, 2s = 42 Therefore, s = 21cm We know that, $\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}$$21=\frac{18+10+\text{c}}{2}$
42 = 28 + c c = 14cm So the area of the triangle is:$\text{A}=\sqrt{21\times(21-18)\times(21-10)\times(21-14)}$
$\text{A}=\sqrt{21\times(3)\times(11)\times(7)}$
$\text{A}=21\sqrt{11}\text{cm}^2$
View full question & answer→Question 184 Marks
Find the area of a triangle whose sides are respectively 9cm, 12cm and 15cm.
AnswerLet 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 $=\sqrt{\text{s}\times(\text{s}-\text{a})\times(\text{s}-\text{b})\times(\text{s}-\text{c})}$
$=\sqrt{18\times(18-9)\times(18-12)\times(18-15)}$
$= 54\text{cm}^2$
View full question & answer→Question 194 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.
AnswerGiven, 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$
$\text{AD} = 67.2\text{dm}.$
View full question & answer→Question 204 Marks
Find the area of the quadrilateral $A B C D$ in which $A B=3 cm, B C=4 cm, C D=4 cm, D A=5 cm$ and $A C=5 cm$.
AnswerFor $\triangle\text{ABC}$
$A C^2=B C^2+A B^2 25=9+16$
So, $\triangle ABC$ is a right angle triangle right angled at point $R$ Area of triangle
$ABC=12 \times AB \times BC=\frac{1}{2} \times 3 \times 4$
$=6 cm^2 \text { From } \triangle CAD$
$\text { Perimeter }=2 s=A C+C D+D A 2 s$
$=5 cm+4 cm+5 cm 2 s$
$=14 cm s$
$=7 cm$
By using Heron's Formula Area of the triangle CAD
$=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}$
$=\sqrt{7 \times(7 \times 5) \times(7-4) \times(7-5)}$
$=9.16 cm^2$
Area of ABCD = Area of ABC + Area of CAD
$=(6+9.16) cm^2$
$=15.16 cm^2$ View full question & answer→Question 214 Marks
Find the area of the shaded region in fig. below
AnswerArea of the shaded region $=$ Area of $\triangle ABC -$ Area of $\triangle ADB$
Now in triangle $A D B A B^2=A D^2+B D^2 \ldots$ (i) Given, $A D=12 cm, B D=16 cm$
Substituting the value of $A D$ and $B D$ in eq (i),
we get $A B^2=12^2+16^2=400 cm^2 A B=20 cm$
Now, area of a triangle $=\frac{1}{2} \times AD \times BD =96 cm^2$ Now in triangle $ABC , s =\frac{1}{2} \times( AB + BC + CA )$
$=\frac{1}{2} \times(52+48+20)$
$=60 cm$
By using Heron's Formula The area of a triangle
$=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}$
$=\sqrt{60 \times(60-20) \times(60-48) \times(60-52)}$
$=480 cm^2$
Thus, the area of a triangle is $480 cm^2$ Area of shaded region $=$ Area of $\triangle ABC -$ Area of $\triangle ADB$
$=(480-96) cm^2$
$=384 cm^2$
Area of shaded region $=384 cm^2$
View full question & answer→Question 224 Marks
Find the area of an isosceles triangle having the base x cm and one side y cm.
AnswerLet 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}}$$\text{A}=\frac12(\text{Base}\times\text{Height})$
$=\frac12\times\text{x}\Big(\sqrt{\text{y}^2-\frac{\text{x}^2}{4}}\Big)$
$=\frac{\text{x}}{2}\Big(\sqrt{\text{y}^2-\frac{\text{x}^2}{4}}\Big)$
View full question & answer→Question 234 Marks
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are $13 cm, 14 cm$ and 15 cm and the parallelogram stands on the base 14 cm , find the height of a parallelogram.
AnswerThe sides of the triangle $D C E$ are: $D C=15 cm, C E=13 cm$,
Let the $h$ be the height of parallelogram $A B C D$
Now, for the area of $\triangle DCE$
Perimeter = DC + CE +
$ED 2 s=15 cm+13 cm+14 cm s=21 cm$
By using Heron's Formula, Area of the
$\triangle AOB=\sqrt{s(s-a) \times(s-b) \times(s-c)}$
$=\sqrt{21 \times(7) \times(8) \times(6)}$
$=84 cm^2$
Also, area of $\triangle DCE =$ Area of parallelogram $A B C D$
$\Rightarrow 84 cm^2 24 \times h$
$=84 cm^2 h$
$=6 cm .$
View full question & answer→Question 244 Marks
If each side of a equilateral triangle is tripled then what is the percentage increase in the area of the triangle?
AnswerArea of an equilateral triangle having each side a cm is given by:$\text{A}=\frac{\sqrt{3}\text{a}^2}{4}$
Now, Area of an equilateral triangle, say $A_1$ if each side is tripled is given by; a = 3a$\text{A}_1=\frac{\sqrt{3}}{4}\text{a}^2$
$\text{A}_1=\frac{\sqrt{3}}{4}(3\text{a})^2$
$\text{A}_1=\frac{9\sqrt{3}\text{a}^2}{4}\text{cm}^2$
Therefore, increase in area of triangle = $A_1 - A$$=\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$=\frac{\frac{8\sqrt{3}\text{a}^2}{4}}{\frac{\sqrt{3}\text{a}^2}{4}}\times100$
$= 800\%.$
View full question & answer→Question 254 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.
AnswerGiven, Two parallel sides of trapezium are $A B=77 m$ and $C D=60 m$ The other two parallel sides of trapezium are $B C$ $=26 m, AD =25 m$
Join $A E$ and $C F D E$ is perpendicular to $A B$ and also, $C F$ is perpendicular to $A B$
Therefore, $DC = EF =60 m$ Let $AE = xSo , BF =77-60- x BF =17- x \ln$
$\triangle ADE$, By using Pythagoras theorem, $DE ^2$
$=A D^2-A E^2 D E^2=25^2-x^2 \ln$
$\triangle BCF$, By using Pythagoras theorem, $CF ^2$
$=B C^2-B F^2 C F^2=26^2-(17-x)^2$
Here, $D E=C F$ So, $DE ^2$
$=C F^2 25^2-x^2$
$=26^2-(17-x)^2 25^2-x^2$
$=26^2-\left(17^2-34 x+x^2\right) 25^2-x^2$
$=26^2-17^2+34 x+x^2 25^2$
$=26^2-17^2+34 x x$
$=7 DE^2$
$=25^2-x^2$
$DE=\sqrt{625-49}$
$DE =24 m$ Area of trapezium
$=\frac{1}{2} \times(60+77) \times 24$
Area of trapezium $=1644 m^2$. View full question & answer→Question 264 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 $\sqrt{3}=1.73\big)$
AnswerGiven that, in a quadrilateral ABCD in which $AD =24 cm, \angle BAD =90^{\circ}$
$B C D$ is an equilateral triangle and the sides $B C=C D=B D=26 cm \operatorname{In} \triangle B A D$, by applying Pythagoras theorem,
$B A^2=B D^2-A D^2 B A^2=26^2+24^2 BA =\sqrt{ 1 0 0 }$
$BA =10 cm$ Area of the $\triangle BAD =\frac{1}{2} \times BA \times AD$
Area of the $\triangle BAD =\frac{1}{2} \times 10 \times 24$ Area of the triangle $B A D=120 cm^2$
Area of the equilateral triangle $=\sqrt{\frac{3}{4}} \times$ side
Area of the equilateral $\triangle QRS =\sqrt{\frac{3}{4}} \times 36$
Area of the equilateral triangle $B C D=292.37 cm^2$
Therefore, the area of quadrilateral $A B C D=$ Area of $\triangle B A D+$ Area of the $\triangle B C D$
The area of quadrilateral $A B C D=120+292.37=412.37 cm^2$ View full question & answer→Question 274 Marks
Find the area of quadrilateral ABCD in which AB = 42cm, BC = 21cm, CD = 29cm, DA = 34cm and the diagonal BD = 20cm.
AnswerGiven: AB = 42cm, BC = 21cm, CD = 29cm, DA = 34cm, and the diagonal
$B D=20 cm .$
Now, for the area of triangle $A B D$
Perimeter of triangle $A B D 2 s$
$=A B+B D+D A 2 s$
$=34 cm+42 cm+20 cm s$
$=48 cm$
By using Heron's Formula, Area of the
$\triangle ABD =\sqrt{ s \times( s - a ) \times( s - b ) \times( s - c )}$
$=\sqrt{48 \times(48-42) \times(48-20 \times(48-34)}$
$=336 cm^2$
Now, for the area of triangle BCD Perimeter of triangle BCD 2s
$=B C+C D+B D 2 s$
$=29 cm+21 cm+20 cm s$
$=35 cm$
By using Heron's Formula, Area of the
$\triangle BCD =\sqrt{ s \times( s - a ) \times( s - b ) \times( s - c )}$
$=\sqrt{35 \times(14) \times(6) \times(15)}$
$=210 cm^2$
Therefore, Area of quadrilateral $A B C D=$ Area of $\triangle ABC +$ Area of $\triangle BCD$ Area of quadrilateral $A B C D$
$=336+210$
Area of quadrilateral $A B C D=546 cm^2$ View full question & answer→Question 284 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.
AnswerLet the sides of the given triangle $b e A B=a, B C=b, A C=c$ respectively.
So given, $a =15 cm b =13 cm c =14 cm$
By using Heron's Formula The Area of the triangle
$=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}$
Semi perimeter of a triangle $=2 s 2 s= a + b + c$
$s=\frac{(a+b+c)}{2}$
$s=\frac{(15+13+14)}{2}$
$s=21 cm$
$\therefore \text { Area of the triangle }=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}$
$=\sqrt{21 \times(21-15) \times(21-13) \times(21-14)}$
$=84 cm^2$
BE is a perpendicular on AC Now, area of triangle $=84 cm^2$
$\frac{1}{2} \times BE \times AC=84 cm^2$
$BE=12 cm$
The length of $B E$ is 12 cm .
View full question & answer→Question 294 Marks
Find the area of a triangle whose sides are 3cm, 4cm and 5cm respectively.
AnswerThe 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)}$
$=\sqrt{6\times3\times2\times1}$
$=\sqrt{36}$
$= 6\text{cm}^2.$
View full question & answer→Question 304 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.
AnswerWe are given assumed value $\Delta$ is the area of a given
$\triangle ABC$ We assume the sides of the given triangle $A B C$ be $a, b, c$ The area of a triangle having sides $a, b, c$ and $s$ as semi-perimeter is given by,
$A=\sqrt{s(s-a)(s-b)(s-c)}$
$\Delta=\sqrt{s(s-a)(s-b)(s-c)}$
Where $s =\frac{ a + b + c }{2}$
$2 s=a+b+c$
We take the sides of a new triangle as $2 a , 2 b, 2 c$ that is twice the sides of previous one Now, the area of a triangle having sides $2 a , 2 b$, and 2 c and $s _1$ as semi-perimeter is given by,
$A _1=\sqrt{ s _1\left(s_1-2 a \right)\left( s _1-2 c \right)\left( s _1-2 c \right)}$, where
$s_1=\frac{2 a+2 b+2 c}{2}$
$s_1=\frac{2(a+b+c)}{2}$
$s_1=a+b+c$
$s_1=2 s$
$\text { Now, } A_1=\sqrt{2 s(2 s-2 a)(2 s-2 b)(2 s-2 c)}$
$A_1=\sqrt{2 s \times 2(s-a) \times 2(s-b) \times 2(s-c)}$
$A_1=4 \sqrt{s(s-a)(s-b)(s-c)}$
$A_1=4 \triangle$
View full question & answer→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.
AnswerGiven that, $A O=25 cm O B=25 cm B A=14 cm$
Area of each strip $=$ Area of $\triangle AOB$
Now, for the area of $\triangle AOB$
$\text { Perimeter }=A O+O B+B A 2 s=25 cm+25 cm+14 cm s=32 cm$
By using Heron's Formula, Area of the
$\triangle AOB=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}$
$=\sqrt{32 \times(7) \times(4) \times(18)}$
$=168 cm^2$
Also, area of each type of paper needed to make a fan $=5 \times$ Area of $\triangle AOB$
Area of each type of paper needed to make a fan $=5 \times 168 cm^2$
Area of each type of paper needed to make a fan $=840 cm^2$.
View full question & answer→