Q.1Â Â Â Â Â A park, in the shape of a quadrilateral ABCD, has 90Âº, AB = Â 9 m , BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
Sol.
Area of
Using Pythagoras theorem, we have
For ::
Let a = 13 m, b = 8 m and c = 9 m
Now,
and
Therefore
Therefore Required area =
Q.2Â Â Â Â Â 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.
Since
Therefore 90Âº
Area of rt
For :
Let a = 5 cm, b = 4cm and c = 5 cm. Then ,
Now,
and
Area of
Area of quadrilateral ABCD
Q.3Â Â Â Â Â Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used.
Sol. Â Â Â Â Area of region I:
Â Â Â Â Â Â Â Region I is Enclosed by a triangle of sides a = 5 cm, b = 5 cm and c = 1 cm
Let s be the perimeter of the triangle. Then,
Now,
Area of region II:
Region II is a rectangle of length 6.5 cm and breadth 1 cm.
Area of region III:
Region III is an isosceles trapezium.
InÂ Î”ABE, We have
Â Â Â Â Â Â Â Â Â Â Â Â Â AB^{2}Â = AE^{2} + BE^{2}
â‡’ Â Â Â Â Â Â Â Â Â Â Â 1 Â Â = Â 0.25 +Â BE^{2}
â‡’ Â Â Â Â Â Â Â Â Â Â Â BE Â =Â
Area of region IV:
Region IV forms a right triangle whose two sides are of lengths 6 cm and 1.5 cm.
Area of region V:
Region IV and V are congruent.
Area of region V = 4.5 cm^{2}
Hence, total area of the paper used = (2.49 + 6.5 + 1.3 + 4.5 + 4.5) cm^{2}
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 19.29 cm^{2}
Q.4Â Â Â 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.
For the triangle :
Let its sides be a, b and c such that
a = 26 cm, b = 28 cm and c = 30. Then,
Now,
and
Area of the triangle
For the parallelogram :
Area = base Ã— height
Therefore
[Since Area of parallelogram = Area of (given)
Therefore Area of parallelogram = 336 cm^{2} and its base = 28 cm]
Q.5Â Â Â Â A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?
Sol.
We know that the diagonals of the rhombus bisect each other at right angles. Using Pythagoras theorem, we have
Area of one
Therefore Area of rhombus
Grass area for 18 cows
Grass area for 1 cow
Q.6Â Â Â Â Â 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?
In one triangular piece, let a = 20 cm, b = 50 cm and c = 50 cm
Now,
and
Therefore Area of one triangular piece
Therefore Area of 5 red triangles
and , area of 5 green triangles
Q.7Â Â Â Â A kite 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.
ABCD is a square such that AC = BC = 32 cm and AEF is an isosceles triangle in which AE = AF = 6 cm
For the area of shades I and II
Clearly from the figure
Area of shade I = Area of shade II = Area of CDB
[Since ABCD is in square shapes]
sq cm
For the area of shade III (i.e., AEF)
and
Therefore
Area of shade III
Q.8Â Â Â Â A floral design on floor is made up of 16 tiles which areÂ triangularÂ , the sides of the triangle being 9 cm, 28 cm and 35Â cm (see figure). Find the cost of polishing the tiles at the rate of 50 p per .
For one triangular tile :
Let a = 9 cm, b = 28 cm and c = 35 cm
Now ,
and
Area of one tile
Therefore Area of 16 such tiles
Cost of polishing @ 50 P per
Q.9Â Â Â Â A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non- parallel sides are 14 m and 13 m. Find the area of the field.
Sol.Â Â Â Â Â Â Â Â Â Various length are as marked in the figure.
Clearly DM = CN
[Distance between parallel sides are always equal]
Now
Area of the trapezium
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