EXERCISE 25 C
Q1 | Ex-25C | Class 8 | S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 | myhelper
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Question 1
Find the area of each of the following parallelogram whose base=340mm , height=17cm.
Sol :
Area of parallelogram=base×height
=340mm×17cm
[1cm=10mm]
=34cm×17cm
Q2 | Ex-25C | Class 8 | S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 | myhelper
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Question 2
The adjacent sides of a parallelogram are 8m and 5m .The distance between the longer sides is 4m. What is the distance between the shorter sides ?
Sol :
Left side shows distance between the longer sides(base) is 4m , Right side shows distance between the shorter sides(base) is h
Area of parallelogram with longer side as base=8×4
=32m2
Area of parallelogram with shorter side as base=32m2
Q3 | Ex-25C | Class 8 | S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 | myhelper
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Question 3
The area of a parallelogram is 98cm2 and its altitude is twice the corresponding base. Find the altitude of the parallelogram.
Sol :
H=2B
Area of parallelogram=98cm2
B×H=98cm2
B×2B=98cm2
2B2=98cm2
$B^2=\frac{98}{2}$cm
B=√49 cm
B=7 cm
H=2×7=14cm
Q4 | Ex-25C | Class 8 | S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 | myhelper
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Question 4
A parallelogram has a length of 15cm and a height of 20cm. It is divided into two congruent triangles. What is the area of each triangle ?
Sol :
Area of triangle$=\frac{1}{2}\times b\times h$
$=\frac{1}{2}\times 15\times 20$
=150cm2
Q5 | Ex-25C | Class 8 | S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 | myhelper
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Question 5
The two adjacent sides of a parallelogram are 60m and 40m and one of the diagonals is 80m long. What is the area of the parallelogram ?
Sol :
Suppose we divide parallelogram in to two equal triangle
Area of triangle
$s=\frac{a+b+c}{2}$
$=\frac{60+40+80}{2}$
s=90m
Area=√s(s-a)(s-b)(s-c)
=√90(90-60)(90-40)(90-80)
=√90(30)(50)(10)
=√9×3×5×1×10000
=100√3×3×3×5
=300√3×5
=300√15
Area of parallelogram equal to two equal triangle=2×300√15
Q6 | Ex-25C | Class 8 | S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 | myhelper
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Question 6
The area of a parallelogram and square are the same. The perimeter of the square is 84cm and the base of the parallelogram is 7cm, find the height of the parallelogram.
Sol :
ATQ,
Area of the parallelogram = area of the square
Given perimeter of the square = 84cm
We know that,
Side × 4 = perimeter of a square
➡ side × 4 = 84cm
➡ side = $\frac{84}{4}$
➡ side = 21cm
Therefore it's area = side²
= 21 × 21
= 441cm²
Since area of the parallelogram = area of the square.
i.e Area of the parallelogram = 441cm²
Given it's base = 7cm
We know that,
Base × Height = area of a parallelogram
➡ 7 × Height = 441cm²
➡ Height $=\frac{441}{7}$
➡ Height = 63cm
Hence, the height of the parallelogram is 63cm.
Q7 | Ex-25C | Class 8 | S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 | myhelper
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Question 7
Find the area of the rhombus whose base (b)=2.5cm and height (h)=13.2cm
Sol :
Area of the rhombus=base×height
=2.5×13.2 cm²
=33 cm²
Q8 | Ex-25C | Class 8 | S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 | myhelper
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Question 8
Find the area of rhombus ABCD in each case.
(i) AC=8cm , BD=10cm
(ii)
AC=15cm , BD=36cm
Sol :
(i)
Area of rhombus $=\frac{1}{2}\times (\text{first diagonal}) \times (\text{second diagonal})$
$=\frac{1}{2}\times 8\times 10$
=4×10 cm²
=40 cm²
(ii)
Area of rhombus $=\frac{1}{2}\times (\text{first diagonal}) \times (\text{second diagonal})$
$=\frac{1}{2}\times 15 \times 36$
=15×18 cm²
=270 cm²
Q9 | Ex-25C | Class 8 | S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 | myhelper
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Question 9
The area of a rhombus is 72cm2 . If one of its diagonals is 18cm, what is the length of the other diagonal ?
Sol :
Area of rhombus=72cm2
$\frac{1}{2}\times (\text{first diagonal})\times (\text{second diagonal})=72$
$\frac{1}{2}\times 18\times (\text{second diagonal})=72$
$9 \times (\text{second diagonal})=72$
Second diagonal$=\frac{72}{9}$
=8cm
Q10 | Ex-25C | Class 8 | S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 | myhelper
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Question 10
The area of a rhombus is 2016 cm2 and its sides are each equal to 63 cm. Find its height,
Sol :
Area of the rhombus=base×height
$height=\frac{2016}{63}$
Height=32 cm
Q11 | Ex-25C | Class 8 | S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 | myhelper
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Question 11
The perimeter of a rhombus is 40cm and the length of one of its diagonals is 16cm. Find its area.
Sol :
Perimeter of a rhombus =40cm
4×side=40cm
$\text{side}=\frac{40}{4}$cm
Side=10cm
Diagonals of rhombus bisect each other at right angles.
By Pythagoras theorem
OB=√AB2-OA2
=√(10)2-(8)2
=√100-64
=√36
=6
OB=OD=6cm
diagonal BD=6+6=12cm
Area of rhombus$=\frac{1}{2}\times (\text{first diagonal})\times (\text{second diagonal})$
$=\frac{1}{2}\times 16\times 12$
=8×12
=96cm2
Q12 | Ex-25C | Class 8 | S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 | myhelper
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Question 12
Two fields in the shape of parallelogram and rhombus are equal in area. The diagonals of rhombus are 110m and 44m , respectively. If one of the sides of the parallelogram is 88m, find its altitude.
Sol :
ATQ,
Area of rhombus = Area of parallelogram
Given diagonals of the rhombus are 110m and 44m.
Area of the rhombus $=\frac{1}{2}\times (\text{product of diagonals})$
$=\frac{1}{2}\times 110\times 44$
= 55 × 44
= 2420m²
Therefore area of parallelogram is also 2420m²
Base(side) of the parallelogram is given 88m.
Therefore b × h = 2420m²
⇒88 × h = 2420m²
⇒h $= \frac{2420}{88}$
⇒h = 27.5m
Hence, altitude of the parallelogram is 27.5m
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