EXERCISE 25D
Question 1
Area of trapezium$=\frac{1}{2}\times (\text{sum of parallel sides})\times (\text{height})$
$=\frac{1}{2}\times (6+8)\times 10$
$=\frac{14}{2}\times 10$
=7×10
=70 cm2
(ii) Height=4cm
Parallel sides=6.4cm and 4.6cm
Sol :
$=\frac{1}{2}\times (6.4+4.6)\times 4$
$=\frac{11}{2}\times 4=\frac{44}{2}$
=22 cm2
(iii) Height=8.4cm
Parallel sides=15cm and 9.2cm
Sol :
$=\frac{1}{2}\times (15+9.2)\times 8.4$
$=\frac{24.2}{2}\times 8.4$
=12.1×8.4
=101.64 cm2
(iv) Height=7cm
Parallel sides=12cm and 16cm
Sol :
$=\frac{1}{2}\times (12+16)\times 7$
$=\frac{28}{2}\times 7$
=98 cm2
Question 2
Find the height of a trapezium. whose sum of the lengths of the bases is 31 cm and whose area is 124 cm2
Sol :
Area = 124 cm²
Sum of parallel sides = 31cm
Area of trapezium$=\frac{1}{2}\times (\text{sum of parallel sides})\times (\text{height})$
$124 =\frac{1}{2}\times 31 \times \text{height}$
Height$=\frac{124\times 2}{31}$
Height = 8 cm
Question 3
The ratio of the length of the parallel sides of a trapezium is 3:2 . The shortest distance between them is 15cm . If the area of the trapezium is 450cm2 , what is the sum of the lengths of the parallel sides ?
Sol :
Height is the shortest distance between the parallel sides. So, Height=15cm
Parallel sides =3x and 2x
Area of trapezium=450cm2
$\frac{1}{2}\times (3x+2x)\times 15=450$
$\frac{5x}{2}\times 15=450$
$\frac{5x}{2}=\frac{450}{15}$
5x=30×2
5x=60cm
Sum of the lengths of the parallel sides
=5x=60cm
Question 4
The difference between two parallel sides of a trapezium is 8cm . The perpendicular distance between them is 24m . Find the lengths of the two parallel sides , if the area of the trapezium is 312m2
Sol :
Given: (x-y)=8m..(i)
Height=24m
Area of trapezium is 312m2
$\frac{1}{2}\times (x+y)\times 24=312$
$(x+y)=\frac{312}{12}$
(x+y)=26m..(ii)
On adding (i) and (ii), we get
(x+y)+(x-y)=(26+8)m
x+y+x-y=34
2x=34
$x=\frac{34}{2}$m
x=17m
putting x in (i), we get
(x-y)=8m
17-y=8
17-8=y
y=9m
Lengths of two parallel sides=17m and 9m
Question 5
In the given diagram, CDE is an isosceles triangle with an area of 24cm2. If AB=8cm and AD=12cm , calculate the area of the trapezium ABED
Sol :
Area of rectangle ABFD= 12×8=96cm2
In isosceles triangle the altitude divides the triangle into 2 equal parts so area of triangle DFE= 24÷2=12cm2
Area of ABED=Area of ABFE-Area of DFE
=96-12
=84cm2
Question 6
The following figure shows a parallelogram and a trapezium whose areas are equal. Find the value of 'a'.
Sol :
Area of parallelogram=b×h
=8.8×3cm2
=26.4cm2
Area of parallelogram=Area of trapezium
Area of trapezium=26.4cm2
$\frac{1}{2}\times (\text{sum of parallel sides})\times (\text{height})=26.4$
$\frac{1}{2}\times (a+8.4)\times 4=26.4$
(a+8.4)2=26.4
$(a+8.4)=\frac{26.4}{2}$
(a+8.4)=13.2
a=13.2-8.4
a=4.8cm
Question 7
Two parallel sides of an isosceles trapezium are 16cm and 24cm respectively. If the lengths of each non-parallel side is 5cm, find the area of trapezium.
Sol :
In triangle AED
$AE=\frac{24-16}{2}$
AE=4cm
Using Pythagoras theorem
DA2=DE2+AE2
DE2=DA2-AE2
DE=√(5)2-(4)2
DE=√25-16
DE=√9
DE=3cm=Height
Area of trapezium$=\frac{1}{2}\times (\text{sum of parallel sides})\times (\text{height})$
$=\frac{1}{2}\times (16+24)\times 3$
$=\frac{40}{2}\times 3$
=20×3
=60cm2
Question 8
A grassy area is in the shape of a trapezoid whose bases measures 150m and 100m and the height is 52m . A 3m walkway is constructed inside this grassy area which runs perpendicular from the two bases. Calculate the area of the grassy patches of land after the construction of this walkway.
Sol :
Area of Trapezium= 1/2(sum of parallel sides)× hieght
Area of Trapezium = 1/2(150+100)×52
= 1/2×250×52
=6500 m²
Area of rectangle= l×b
L=52
B=3
Area = 52×3
=156m²
Area of grassy patches = area of Trapezium- area of rectangle
= 6500-156
=6344 m²
25e
ReplyDeleteAt times the units are changes like in Q2 it's 3:2 but while solving it is 3:4 .
ReplyDeleteI think the person should be focused and carefull
Thanks for pointing out mistake.
Delete