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Exercise 12.2

1. A park, in the shape of a quadrilateral ABCD, has ÐC = 90°, AB .= 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

Sol. Triangle BCD is right angled at C.

∴           $BD = \sqrt {{{(5)}^2} + {{(12)}^2}} m = \sqrt {25 + 144} m$
$= \sqrt {169} m = 13m$
ar(ABCD) = ar(ABD) + ar(BCD)                   …(i)

$or(BCD) = {1 \over 2} \times 12 \times 5$
= 30 m²                                                                …(ii)
For triangle ABD ,

$S = {{8 + 9 + 13} \over 2} = 15m$

$ar(ABD) = \sqrt {15(15 - 9)(15 - 8)(15 - 13)} {m^2}$

$= \sqrt {15 \times 6 \times 7 \times 2} {m^2} = 6\sqrt {35} {m^2}$
=  35.5 m² (approx. )

2. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

Sol. ar(ABCD) = ar(ABC) + ar(ACD)                                      …(i)

For ΔABC,
$S = {{3 + 4 + 5} \over 2} = 6m$
$ar(ABC) = \sqrt {6(6 - 3)(6 - 4)(6 - 5)} {cm^2}$
$= \sqrt {6 \times 3 \times 2 \times 1} = 6c{cm^2}$                          …(ii)
For ΔACD,
$S = {{5 + 5 + 4} \over 2} = 7cm$
$ar(ACD) = \sqrt {7(7 - 5)(7 - 5)(7 - 4)} c{m^2}$
$= \sqrt {7 \times 2 \times 2 \times 3} = 2\sqrt {21} c{m^2} = 9.17c{m^2}$   ….(iii)
Substituting the values of areas from (ii) and (iii) in (i), we get
ar(ABCD) = (6 + 9.17) cm² = 15.17 cm².

3. Radha made a picture of an aeroplane with coloured paper as shown in figure given below. Find the total area of the paper used.

Sol. Region I: There is a triangle.
Sides of the triangle are 5 cm, 5 cm and 1 cm.
$S = {{5 + 5 + 1} \over 2} = 5.5cm$
$Area(I) = \sqrt {5.5(5.5 - 5)(5.5 - 5)(5.5 - 1)} c{m^2}$
$= \sqrt {5.5 \times 0.5 \times 0.5 \times 4.5} c{m^2} = \sqrt {6.19} c{m^2}$
= 2.5 cm² (approx.)
Region II: There is a rectangle.
Area (II) = 6.5 × 1 cm² = 6.5 cm².
Region III: There is a trapezium.
Parallel sides are 2 cm, 1 cm and other sides are 1 cm and 1 cm.
Look the trapezium in the adjacent figure.
AL = 0.5 cm

$DL = \sqrt {{{(1)}^2} - {{(0.5)}^2}} cm$
$= \sqrt {1 - {{(0.5)}^2}} cm$
$= \sqrt {0.75} cm$
= 0.9 cm (approx ).
∴  Area (III) $= {1 \over 2}(2 + 1) \times 0.9 = 3 \times 0.45 = 1.35c{m^2}$
Regions IV and V: There are two equal right angled triangles.
For each, two perpendicular sides are 6 cm and 1.5 cm.
Area (IV and V)  $= 2 \times {1 \over 2} \times 6 \times 1.5 = 9c{m^2}$
∴   Total area of the paper used
= area (I + II + III + IV + V)
= (2.5 + 6.5 + 1.35 + 9.0) cm²
= 19.35 cm².
Hence, the required area is approximately 19.4 cm².

4. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.

Sol.     Area of parallelogram =   area of triangle.                                             …(i)
Let h be the height in cm of the parallelogram.
Sides of the triangle are 26 cm, 28 cm and 30 cm.
$S = {{26 + 28 + 30} \over 2} = 42$
Now, area of the triangle
$= \sqrt {42(42 - 26)(42 - 28)(42 - 30)} c{m^2}$
$= \sqrt {42 \times 16 \times 14 \times 12} c{m^2}$
= 336 cm²
Area of the parallelogram = base × height = 28 × h
From (i), we have
Substituting the values of areas from (ii) an.d (iii) in (i), we have
28 × h = 336 ⇒  h = 12 cm.
Hence, height of the parallelogram = 12 cm.

5. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 in and its longer diagonal is 48 m, how much area of grass field will each cow be getting?

Sol. In right angled ΔAOD, AD = 30 m, AO =  ${1 \over 2}$ m = 24 m.

$DO = \sqrt {{{(30)}^2} - {{(24)}^2}}$
$m = \sqrt {900 - 576} m$
$= \sqrt {324} m = 18m$
∴ DB = 2 × 18 m = 36 m.

∴  Area of the rhombus (field) $= {1 \over 2} \times 48 \times 36 = 864{m^2}$
Area of grass field for each cow $= {{864} \over {18}}{m^2} = 48{m^2}$

6. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm, and 50 cm. How much cloth of each colour is required for the umbrella?

Sol. Each piece is triangular piece
(5 pieces of one colour + 5 pieces of other colour).
There is a total of 10 pieces.
Dimensions of a piece are 50 cm, 50 cm, 20 cm.
$S = {{50 + 50 + 20} \over 2} = 60cm$
∴ Area or one piece  $= \sqrt {60(60 - 50)(60 - 50)(60 - 20)} c{m^2}$
$= \sqrt {60 \times 10 \times 10 \times 40}$
$= 200\sqrt 6 c{m^2}$
∴   Area of cloth of each colour = 5 × 200 √6 cm²
=  1000 √6  cm²

7. A late in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. How much paper of each shade has been used in it?

Sol. Square have diagonals 32 cm each.
∴  Area of square =   $= {1 \over 2} \times 32 \times 32c{m^2} = 512c{m^2}$
∴   Paper used for each of the shades I and II
$= {{512} \over 2} = 256c{m^2}$
Dimensions of triangle are 8 cm, 6 cm and 6 cm.
∴                       $S = {{8 + 6 + 6} \over 2} = 10$
Area of shade     III  $= \sqrt {10(10 - 8)(10 - 6)(10 - 6)} c{m^2}$
$= \sqrt {10 \times 2 \times 4 \times 4} = 8\sqrt 5 c{m^2}$
= 17.89 cm2

8. A floral design on a floor is made up of 16 tiles which are triangular; the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). Find the cost of polishing the ties at the rate of 50 p per cm².

Sol. Dimensions of one triangular side are 28 cm, 35 cm and 9 cm.
$S = {{28 + 35 + 9} \over 2} = {{72} \over 2} = 36$
∴    $Area = \sqrt {36(36 - 28)(36 - 35)(36 - 9)} c{m^2}$
$= \sqrt {36 \times 8 \times 1 \times 27} c{m^2} = 88.20c{m^2}$
∴   Cost of polishing 16 tiles $= {{50} \over {100}} \times 16 \times 88.20$
= Rs. 705.60.

9. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 in and 13 m. Find the area of the field.

Sol. Draw $CE\parallel AD$.
AE = 10 m;
BE = (25 – 10) m = 15 m

Dimensions of triangle CEB are
CE = 14 m, BE = 15 m,
BC = 13 m.
$S = {{14 + 1513} \over 2} = 21m$
Area of triangle BCE $= \sqrt {21(21 - 14)(21 - 15)(21 - 13)} {m^2}$
$= \sqrt {21 \times 7 \times 6 \times 8} {m^2} = 84{m^2}$
Let height of the triangle = h m.
$84 = {1 \over 2} \times 15 \times h \Rightarrow h = {{84 \times 2} \over {15}}m = {{56} \over 5}m$
∴      Area of the field (trapezium) = $= {1 \over 2}(10 + 25) \times {{56} \over 5}m.$
$= {1 \over 2} \times 35 \times {{56} \over 5}{m^2} = 196{m^2}$