Exercise 11.2

Question 1: The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.

Answer: Area of Trapezium

`=1/2xx` sum of parallel sides `xx` perpendicular distance

Parallel sides = 1 and 1.2 m

And perpendicular distance = 0.8 m

So, Area `=1/2xx(1+1.2)xx0.8`

`=1.1xx0.8=0.88` sq m

Question 2: The area of a trapezium is 34 cm² and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.

Answer: Area of Trapezium

`=1/2xx` sum of parallel sides `xx` perpendicular distance

So, `34=1/2xx(10+x)xx4`

Or, `34=2(10+x)`

Or, `17=10+x`

Or, `x=17-10=7` cm

Question 3: Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

mensuration 3

Answer: If a perpendicular from D to BC is drawn then it will be equal to AB,
then `EC = 48-40 = 8` m.
In ΔDEC `DE^2 = DC^2-EC^2`
`DE^2 = 17^2 - 8^2 = 289 - 64 = 225`
Or, `DE = 15` m

Area of Trapezium

`=1/2xx` sum of parallel sides `xx` perpendicular distance


`=44xx15=660` sq m

Question 4: The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.

mensuration 5

Answer: Area of upper triangle = `1/2xx` height `xx` base

`=1/2xx13xx24=156` sq m

Area of lower triangle `=1/2xx` height `xx` base

`=1/2xx8xx24=96` sq m

Area of the field `= 156 + 96 = 252` square metre

Question 5: The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

Answer: Area of rhombus `=1/2xx` Diagonal 1 `xx` Diagonal 2

`=1/2xx7.5xx12=45` sq cm

Question 6: Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.

Answer: If one of the Diagonals is 8 cms then other diagonal can be derived as follows using Pythagorean Theorem:


Or, `text(Altitude)_1^2=text(Side)^2-text(Altitude)_2^2`


Or, Altitude `=4sqrt2`

So, diagonal 1 `=8sqrt2`

Area `=1/2xx4xx8sqrt2=16sqrt2` sq cm

Question 7: The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m² is Rs 4.

Answer: Area of rhombus `=1/2xx45xx30=675`

Cost of polishing =Total Area `xx` Rate

`=3000xx675xx4=8100000` Rupees

Question 8: Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m2 and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

Answer: Area of Trapezium

`=1/2xx` Sum of parallel sides `xx` Perpendicular distance

Or, `10500=1/2xx(text(Side)_1+text(Side)_2)xx100`

Or, `(text(Side)_1+text(Side)_2)=10500xx(2)/(100)=210`

Or, `3xx\text(Side)_1=210`

(As one of the parallel sides is double of the other side)
Side 1 = 70
Side 2 = 140 This is the side near the river

Question 9: Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.

mensuration 11

Answer: Upper part is in the shape of a trapezium. This is mirrored in the lower part as well. So area of two trapeziums added to the area of middle rectangular portion will be equal to the area of the octagon.
The Base of the triangle runs from the base of the perpendicular to the edge of the octagon. This can be calculated using Pythagorean Theorem:
Base2 = Hypotenuse2 - Perpendicular2
= 52 - 42 = 25 - 16 = 9
So, Base = 3
Hence The Upper Parallel Side of the Trapezium = 11- 6 = 5 cms (As there will be two bases of two triangles adding up to 6 cms)
Second point to note is, as it is a regular octagon so the upper parallel side = side of the octagon = 5 cms

Area of Trapezium

`=1/2xx` Sum of parallel sides `xx` Perpendicular distance


So, Area of 2 trapeziums = 55 sq cms

Now, Area of Rectangle = Length `xx` Breadth
`= 11 xx 5 = 55`
Hence, Area of the platform `= 55 + 55 = 110` sq cms

Question 10: There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways. Find the area of this park using both ways. Can you suggest some other way of finding its area?

mensuration 13

Answer: Jyoti has divided the park in to congruent trapeziums. The parallel sides of the trapezium are measuring 15 metres and 30 metres and the perpendicular distance is 7.5 metres

Area of Trapezium

`=1/2xx` Sum of parallel sides `xx` Perpendicular distance

So, Area of two trapeziums `=(30+15)xx7.5=337.5` sq m

Kavita divided the park into upper triangular portion and lower rectangular portion. Sides of rectangle are 15 metres each. Base and height of triangle are 15 metres each.

Area of square `=15xx15=225`

Area of triangle `=1/2xx15xx15=112/5`

Total Area `=225+112.5=337.5` sq cm

Question 11: Diagram of the adjacent picture frame has outer dimensions =24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is same.

mensuration 16

Answer: Let us draw a perpendicular from any side of the inner rectangle to the side of the outer rectangle as shown above. This will give us 4 congruent triangles.
As `24-16 = 8`  and `28-20 = 8`

So, height and base of each triangle `=8/2=4` cm

Sides of both horizontal rectangles = 4 cm and 16 cm

Sides of both vertical rectangles = 4 cm and 20 cm

Now, area of one triangle `=1/2 xx` base `xx` height

`=1/2xx4xx4=8` sq cm

Area of the horizontal rectangle = 4 x 16 = 64
Hence, Area of lower horizontal section of the frame = 64 + 8 + 8 = 80 sqcms
Area of the vertical rectangle = 4 x 20 = 80

Hence, Area of One Vertical Section of the frame `= 80 + 8 + 8 = 96`

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