plz answer the 12,13 and 14th question
Answers
Answer:
Step-by-step explanation:
12. Area of trapezium= 1/2*d*(a+b) sq. Units.
Where d= perpendicular distance between parallel sides which are a & b respectively, 25 &13 cm.
25–13=12, 12/2= 6 cm is base of RT. Angled ∆.
Non parallel sides=10cm, this is hypotenuese of this ∆.
Find ‘d ‘ to solve.
d=√{10^2–6^2}=8 cm.
Area of trapezium= 1/2*8*(25+13)=152 cm^2
13. Consider a Triangle ACD
it is right angled triangle therefore by pythagoras theorem
AC^2 = AD ^2 +DC^2
41^2 = AD^2 +40^2
1681 = AD^2 +1600
AD^2 = 81
AD = 9cm
Therefore the perpendicular height of trapezium is 9 cm
Now by formula
Area of trapezium = 1/2 × sum of lengths of parallel sides × height
=1/2 × (15 +40) × 9
=1/2 ×55 ×9
=495 /2
=247.5 sq. cm
14. Solution :-
Area of the Δ DEI = 1/2 × Base × Height
⇒ 1/2 × 60 × 40
= 1200 sq m
Area of the Trapezium EFGI = 1/2 × (Sum of the two parallel sides) × Height
⇒ 1/2 × (60 + 50) × 70
⇒ 1/2 × 110 × 70
= 3850 sq m
Area of the Δ AFG = 1/2 × Base × Height
⇒ 1/2 × 50 × 50
= 1250 sq m
Area of the Trapezium ABCH = 1/2 ×(Sum of the two parallel sides) × Height
⇒ 1/2 × (40 + 30) × 80
⇒ 1/2 × 70 × 80
= 2800 sq m
Area of the Δ CDH = 1/2 × Base × Height
The height of this figure is not clear in the image. I think it is 80 m and I have assumed it 80 m.
⇒ 1/2 × 40 × 80
= 1600 sq m
Total area = Area of Δ DEI + Area of the Trapezium EFGI + Area of Δ AFG + Area of Trapezium ABCH + Area of Δ CDH
⇒ 1200 + 3850 + 1250 + 2800 + 1600
= 10700 sq m
So, the area of the figure ABCDEF is 10700 sq m
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let us assume person buy 7 orange
his cost of 7 orange = 1
as he wants 40% profit ,
his selling price for 7 orange will be = 1+ 40% of 1= 1.4
selling price of one orange = 1.4/7 = 0.2
in 1 rupee number of orange he will sell = 1/0.2=5