Class 9 Maths Exercise 12.2 Solution

The figure is given below:

The dimensions of the park as follows:
Angle \( \mathrm{C}=90^{\circ} \),
\(\mathrm{AB}=9 \mathrm{~m}\)
\(\mathrm{BC}=12 \mathrm{~m}\)
\(\mathrm{CD}=5 \mathrm{~m}\)
And, \( \mathrm{AD}=8 \mathrm{~m} \)
Now, BD is joined.
Thus quadrilateral ABCD can now be divided into two triangles ABD and BCD
Area of quadrilateral \( \mathrm{ABCD}= \) Area of triangle \( \mathrm{ABD}+ \) Area of triangle BCD
Pythagoras Theorem: It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In \( \triangle \mathrm{BCD} \) by Pythagoras theorem,
\(\mathrm{BD}^{2}=\mathrm{BC}^{2}+\mathrm{CD}^{2}\)
\(\mathrm{BD}^{2}=122+5^{2}\)
\(\mathrm{BD}^{2}=169\)
\(\mathrm{BD}=13 \mathrm{~m}\)
As triangle BCD is right angled triangle,
Area of \( \triangle B C D=\frac{1}{2} \times \) base \( \times \) height
Area of \( \triangle B C D=\frac{1}{2} \times 12 \times 5=30 \mathrm{~m}^{2} \)
Now, Semi perimeter of triangle \( (\mathrm{ABD})=\frac{(8+9+13)}{2} \)
\(=\frac{30}{2}=15 \mathrm{~m}\)
Using Heron's formula,
\(\operatorname{ar}(\mathrm{ABD})=\sqrt{s(s-a)(s-b)(s-c)}\)
where, \( \mathrm{s}= \) semi-perimeter of the triangle and \( \mathrm{a}, \mathrm{b} \), and c are the sides of the triangle.
\(=\sqrt{15(15-13)(15-8)(15-9)}\)
\(=\sqrt{15 \times 2 \times 7 \times 6}\)
\(=\sqrt{6 \times 6 \times 5 \times 7}\)
\(=6 \sqrt{35 }\mathrm{m}^{2}\)

\(=6 \times 5.91=35.46 \mathrm{~m}^{2}\)
Area of quadrilateral \( \mathrm{ABCD}=35.46+30 \)
Area of quadrilateral \( \mathrm{ABCD}=65.46 \mathrm{~m}^{2} \)
Thus, the park acquires an area of \( 65.5 \mathrm{~m}^{2} \) (approx).

Given: \( \mathrm{AB}=3 \mathrm{~cm}, \mathrm{BC}=4 \mathrm{~cm}, \mathrm{CD}=4 \mathrm{~cm}, \mathrm{DA}=5 \mathrm{~cm} \) and \( \mathrm{AC}=5 \mathrm{~cm} \)
To find: Area of quadrilateral ABCD.
Now we can divide the quadrilateral into two separate triangles and then calculate their area. Now, in \( \triangle \mathrm{ABC} \),
Using Pythagoras theorem,
\(\mathrm{AC}^{2}=\mathrm{BC}^{2}+\mathrm{AB}^{2}\)
\(5^{2}=3^{2}+4^{2}\)
\(25=16+9\)
\(25=25\)
Thus,
Triangle \( \triangle A B C \) is right angled triangle and since AB and BC are smaller sides so they work as legs.
And we know that,
Area of right angles triangle \( =\frac{1}{2} \times \) base \( \times \) height
\(\operatorname{ar}(\mathrm{ABC})=\frac{1}{2} \times 3 \times 4=6 \mathrm{~cm}^{2}\)
Now,
Perimeter of \( =5+5+4 \)
Semi perimeter of \( \triangle \mathrm{ACD}=\frac{5+5+4}{2} \)
\(=\frac{14}{2}\)
\(=7 \mathrm{~cm}\)
Using Heron's formula,
Area of \( \triangle \mathrm{ACD}=\sqrt{s(s-a)(s-b)(s-c)} \)
where, \( s= \) semi perimeter of the triangle and \( a, b \), and \( c \) are the sides of the triangle
\(=\sqrt{7(7-5)(7-5)(7-4)}\)
\(=\sqrt{2 \times 2 \times 7 \times 3}\)
\(=2 \sqrt{21} \mathrm{~cm}^{2}\)
Take \( =\sqrt{21}=4.58 \)
\(=9.16 \mathrm{~cm}^{2} \text { (approx) }\)
Area of quadrilateral \( \mathrm{ABCD}= \) Area of \( \triangle A B C+ \) Area of \( \triangle \mathrm{ACD} \)
\(=6+9.16\)
\(=15.16 \mathrm{~cm}^{2}\)
By rounding off \( =15.2 \mathrm{~cm}^{2} \) (approx)

The figure of the questions is:

Now to find area:

Sides of triangular section \( \mathrm{I}=5 \mathrm{~cm}, 1 \mathrm{~cm} \) and 5 cm
Perimeter of the triangular section \( \mathrm{I}=5+5+1=11 \mathrm{~cm} \)
Semi perimeter,
\( \mathrm{s}=\frac{\text { perimeter of triangular section } I}{2}=\frac{11}{2}=5.5 \mathrm{~cm} \)
We use Heron's formula to find the area of the section I,
Area of Section \( \mathrm{I}=\sqrt{s(s-a)(s-b)(s-c)} \)
\(=\sqrt{5.5(5.5-5)(5.5-5)(5.5-1)}\)
\(=\sqrt{5.5 \times 0.5 \times 0.5 \times 4.5}\)
\(=\sqrt{6.1875} \mathrm{~cm}^{2}\)
Area of Section \( \mathrm{I}=2.48 \mathrm{~cm}^{2} \) (approx.) \( \sim 2.5 \mathrm{~cm}^{2} \)

Length of the sides of the rectangle of section \(\mathrm{II}=6.5 \mathrm{~cm} \) and 1 cm Area of section \( \mathrm{II}=6.5 \times 1=6.5 \mathrm{~cm}^{2} \)

Section III is an isosceles trapezium which is divided into two right triangles and one rectangle.
Now, Area of trapezium
\(=\text { Area (i) }+ \text { Area (ii) + Area (iii) }\)
\(=\frac{ 1 }{ 2 } \times 0.5 \times 1+1 \times 1+\frac{ 1 }{ 2 } \times 0.5 \times 1=1.5 \mathrm{~cm}^{2}\)

Section IV and V are congruent right-angled triangles with height 1.5 cm and base 6 cm .
now area of triangle,
\(=\frac{1}{2} \times \text { base } \times \text { height }=\frac{1}{2} \times 1.5 \times 6\)
\(=4.5 \mathrm{~cm}^{2}\)
Area of region IV and \( \mathrm{V}=2 \times 4.5 \mathrm{~cm}^{2}=9 \mathrm{~cm}^{2} \)
Thus, total area is \( 2.5+6.5+9+1.5=19.5 \mathrm{~sq} \mathrm{~cm} \)

Given, Area of the parallelogram and triangle are equal

Sides of triangle are \( 26 \mathrm{~cm}, 28 \mathrm{~cm} \) and 30 cm .
Semi perimeter of triangle, \( \mathrm{s}=\frac{26+28+30}{2} \)
\(\mathrm{S}=\frac{84}{2}\)
\(=42 \mathrm{~cm}\)
Using Heron's formula,
Area of triangle \( =\sqrt{s(s-a)(s-b)(s-c)} \)
\(=\sqrt{42(42-26)(42-28)(42-30)}\)
\(=\sqrt{42 \times 16 \times 14 \times 12}\)
\(=336 \mathrm{~cm}^{2}\)
\( = \) Let the height of parallelogram be h ,
As parallelogram and triangle having same base and same height have equal areas.
\( \therefore \) Area of parallelogram \( = \) Area of triangle
\(28 \times \mathrm{h}=336\)
\(\mathrm{h}=\frac{336}{28}\)
\(\mathrm{~h}=12 \mathrm{~cm}\)
Hence, the height of the parallelogram is 12 cm .

Diagonals divide the rhombus ABCD into two congruent triangles of equal area.

Semi perimeter of \( \triangle A B C=\frac{30+30+48}{2}=54 \mathrm{~m} \)
Using Heron's formula,
\(\text {Area of } \triangle A B C=\sqrt{s(s-a)(s-b)(s-c)}\)
\( \sqrt{54(54-30)(54-30)(54-48)}\)
\(=\sqrt{54 \times 24 \times 24 \times 6}\)
\(=432 \mathrm{~m}^{2}\)
Area of field \( =2 x \) area of field \( \triangle A B C \)
\(=2 \times 432\)
\(=864 \mathrm{~m}^{2}\)
Thus,
Area of grass field which each cow will be getting \( =\frac{864}{18}=48 \mathrm{~m}^{2} \)

To Find: Total Amount of cloth required
Given: Umbrella is made up of 10 triangular pieces of sides 50, 50 and 20 cm
Perimeter of each triangular piece of cloth \( =50+50+20 \)
Semi perimeter of each triangular piece of cloth \( =\frac{50+50+20}{2} \)
\(=\frac{120}{2}\)
\(=60 \mathrm{~cm}\)
Using Heron's formula,
Area of one triangular piece \( =\sqrt{s(s-a)(s-b)(s-c)} \)
where, as is the semi-perimeter of the triangle and \( \mathrm{a}, \mathrm{b} \) and c are the sides of the triangle.
\(=\sqrt{60(60-50)(60-50)(60-20)}\)
\(=\sqrt{60 \times 10 \times 10 \times 40}\)
\(=200 \sqrt{6} \mathrm{~cm}^{2}\)
Area of piece of cloth of one colour \( =5 \times 200 \sqrt{ 6} \)
\(=1000 \sqrt{6} \mathrm{~cm}^{2}\)

For I and II section:
For calculating the area of the square, Let the length of side \( =x \mathrm{~cm} \)

Each Interior Angle of square \( =90^{\circ} \) Pythagoras Theorem: Square of Hypotenuse equals to the sum of squares of other two sides
Now by Pythagoras theorem, \( x^{2}+x^{2}=3^{2} \times 3^{2} \)
\(2 x^{2}=3^{2} \times 3^{2}\)
\(x^{2}=16 \times 32=512 \mathrm{~cm}^{2}\)
Area of Square \( =(\text {side})^{2} \)
And this will be the area of Square.
Area of square \( =512 \mathrm{~cm}^{2} \)
Area of section \( \mathrm{I}= \) Area of section \( \mathrm{I I}=\frac{1}{2} \) Area of square
Now, we need half of this area,
So half of the area of square \( =256 \mathrm{~cm}^{2} \)
For the III section
Length of the sides of triangle \( =6 \mathrm{~cm}, 6 \mathrm{~cm} \) and 8 cm
Perimeter of triangle \( =6+6+8=20 \mathrm{~cm} \)
Semi-Perimeter of Triangle, \( \mathrm{s}=10 \mathrm{~cm} \)
Using Heron's formula,
Area of the III triangular piece \( =\sqrt{s(s-a)(s-b)(s-c)} \)
where, \( \mathrm{s}= \) semi-perimeter and \( \mathrm{a}, \mathrm{b} \), and c are the sides of the triangle
\(\text {Area of Triangle } =\sqrt{10(10-6)(10-6)(10-8)}\)
\( =\sqrt{4 \times 10 \times 2 \times 4}\)
Area of Triangle \( =8 \sqrt{5} \mathrm{~cm}^{2} \)
\(=17.92 \mathrm{~cm}^{2}\)

So, we need to find the area of all 16 tiles or we can say that Area of 16 triangles.
Perimeter of each triangular shaped tiles \( =28+9+35 \)
Semi perimeter of each triangular shaped tiles, \( \mathrm{s}=\frac{28+9+35}{2} \)
\( \mathrm{s}=\frac{72}{2} \)
\( \mathrm{s}=36 \mathrm{~cm} \)
Now,
Using Heron's formula,
Area of each triangular shape \( =\sqrt{s(s-a)(s-b)(s-c)} \)
where \( \mathrm{s}= \) semi perimeter of triangle and \( \mathrm{a}, \mathrm{b}, \mathrm{c} \) are the sides of the triangle
Area of 1 triangular shape \( =\sqrt{36(36-28)(36-9)(36-35)} \)
\( =\sqrt{36 \times 8 \times 1 \times 27} \)
Area of 1 tile \( =36 \sqrt{6} \mathrm{~cm}^{2} \)
Total area of 16 tiles \( =16 \times 36 \sqrt{6} \mathrm{~cm}^{2}=576 \sqrt{6} \mathrm{~cm}^{2} \)
Total Area of 16 tiles \( =576 \times 2.45 \mathrm{~cm}^{2} \)\( \ \ [\sqrt{6}=2.45] \)
Total Area of 16 tiles \( =1411.2 \mathrm{~cm}^{2} \)
Rate of polishing tiles \( =50 \mathrm{~p} \) per \( \mathrm{cm}^{2} \)
Hence,
The total cost of polishing tiles \( =50 \times 1411.2 \)
\( =70560 \) paise As 1 paisa \( =\frac{1}{100} \mathrm{Rs}=\mathrm{RS} .705 .60 \)

Let ABCD be the given trapezium whose
parallel sides are:
\( \mathrm{AB}=25 \mathrm{~m} \) and \( \mathrm{CD}=10 \mathrm{~m} \)
And,
Non-parallel sides are \( \mathrm{AD}=13 \mathrm{~m} \) and \( \mathrm{BC}=14 \mathrm{~m} \)

Now,
CM perpendicular to AB and \( \mathrm{CE} \| \mathrm{AD} \)
In \( \triangle B C E \)
\( \mathrm{BC}=14 \mathrm{~m}, \mathrm{CE}=\mathrm{AD}=13 \mathrm{~m} \) and
\( \mathrm{BE}=\mathrm{AB}-\mathrm{AE} \)
\( =25-10 \)
\( =15 \mathrm{~m} \)
Semi perimeter of \( \triangle B C E=\frac{15+13+14}{2} \)
\(=\frac{42}{2}\)
\( =21 \mathrm{~cm} \)
Now,
Using Heron's formula,
Area of \( \triangle B C E=\sqrt{s(s-a)(s-b)(s-c)} \)
where, \( \mathrm{s}= \) semi perimeter, \( \mathrm{a}, \mathrm{b} \) and c are the sides of the triangle.
\(=\sqrt{21(21-15)(21-13)(21-14)}\)
\(=\sqrt{21 \times 8 \times 6 \times 7}\)
\(=84 \mathrm{~m}^{2}\)
Now from \( \Delta \) CEB, Area of triangle \( =\frac{ 1 }{ 2 } \times \) base \(\times\) height
For CEB, CM will be the height and 15 cm will be base of the triangle.
And we calculated the area from above. So,
\(\frac{1}{2} \times 15 \times \mathrm{CM}=84\)
\(\mathrm{CM}=\frac{56}{5} \mathrm{~m}\)
Area of parallelogram \( \mathrm{AECD}= \) Base \( \times \) Altitude
\(=\mathrm{AE} \times \mathrm{CM}\)
\(=10 \times \frac{56}{5}\)
\(=112 \mathrm{~m}^{2}\)
Area of trapezium \( \mathrm{ABCD}= \) Area of AECD + Area of triangle BCE
\(=(112+84) \mathrm{m}^{2}\)
\(=196 \mathrm{~m}^{2}\)