the parallel sides of trapzium are 20m and 30 m and its non parellal side are 6 m and 8m fid the area
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The parallel sides of a trapezium are 20m and 30m, and its non-parallel sides are 6m and 8m. What is the area of the trapezium?
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Consider a trapezium ABCD with AB=20 m, BC = 6 m, CD = 30 m and DA = 8 m.
Drop a perpendicular from A on CD to meet that line at E. We consider the figure ABCD as a combination of a rectangle ABCE and a triangle AED. <B = <C = 90.
Area of ABCE = 20*6 = 120 sq m.
Area of AED can EB found by Heron’s formula where 2s =6+10+8 = 24, or s = 12
Area AED = [12(12–6)(12–10)(12–8)]^0.5
= [12*6*2*4]^0.5
= 24 sq m.
Total area of ABCE + AED = 120+24 = 144 sq m.
3.6k Views ·
Let ABCD be the trapezium in which AB is parallel to DC.
Let AB = 30cm, DC=20 cm,and the non parallel sides AD =6cm and BC = 8cm.
Draw perpendiculars DE and CF from D and C to AB respectively. So DE = CF. DC=EF = 20cm.
Let AE = x cm . Then BF = 10-x.
In right triangle ADE DE^2 = AD^2 - AE^2
In right triangle BCF , CF^2 = BC^2 - BF^2
=> AD^2-AE^2 = BC^2-BF^2
=> 36-x^2 = 64 -(10-x)^2 => 20x = 72 => x=3.6
So DE^2 = 36 - 3.6^2 = 36 - 12.96 =23.04
So DE =√23.04 = 4.8 cm
So area of trapezium = (1/2)h(a+b)
= (1/2)×4.8×(30+20) = 2.4×50 =120
1.9k Views ·
Draw CE II AD such that AECD be a parallelogram.
Also,draw CL is Perpendicular EB
Now, AE = CD =20 cm
CE = AD =8 cm
EB = AB-AE = 24 cm - 14 cm
In triangle EBC , (semi perimeter) = 10+8+6 \ 2 cm =12 cm =
Now , area of triangle EBC = √12(12–10)(12–8)(12–6)
=√12*2*4*6 = √6*2*2*4*6
=√6*6^4*4 = 6^4 = 24 ………….(1)
Also, area of triangle EBC = 1\2^10 cm^ CL = 5 cm^CL ………….(2)
FROM (1) and (2) , we get
5 cm^ CL =24
CL = 4.8 cm
Now , area of the trapezium ABCD
= 1\2 ^ (20 30m) ^ 4.8 cm
1\2^ 50^ 4.8 cm = 120
please mark me as brainliest. I beg you mate. this is perfectly right. believe me. . . . . . . . . . . . . . . . .
How does DuckDuckGo know where I am?
Consider a trapezium ABCD with AB=20 m, BC = 6 m, CD = 30 m and DA = 8 m.
Drop a perpendicular from A on CD to meet that line at E. We consider the figure ABCD as a combination of a rectangle ABCE and a triangle AED. <B = <C = 90.
Area of ABCE = 20*6 = 120 sq m.
Area of AED can EB found by Heron’s formula where 2s =6+10+8 = 24, or s = 12
Area AED = [12(12–6)(12–10)(12–8)]^0.5
= [12*6*2*4]^0.5
= 24 sq m.
Total area of ABCE + AED = 120+24 = 144 sq m.
3.6k Views ·
Let ABCD be the trapezium in which AB is parallel to DC.
Let AB = 30cm, DC=20 cm,and the non parallel sides AD =6cm and BC = 8cm.
Draw perpendiculars DE and CF from D and C to AB respectively. So DE = CF. DC=EF = 20cm.
Let AE = x cm . Then BF = 10-x.
In right triangle ADE DE^2 = AD^2 - AE^2
In right triangle BCF , CF^2 = BC^2 - BF^2
=> AD^2-AE^2 = BC^2-BF^2
=> 36-x^2 = 64 -(10-x)^2 => 20x = 72 => x=3.6
So DE^2 = 36 - 3.6^2 = 36 - 12.96 =23.04
So DE =√23.04 = 4.8 cm
So area of trapezium = (1/2)h(a+b)
= (1/2)×4.8×(30+20) = 2.4×50 =120
1.9k Views ·
Draw CE II AD such that AECD be a parallelogram.
Also,draw CL is Perpendicular EB
Now, AE = CD =20 cm
CE = AD =8 cm
EB = AB-AE = 24 cm - 14 cm
In triangle EBC , (semi perimeter) = 10+8+6 \ 2 cm =12 cm =
Now , area of triangle EBC = √12(12–10)(12–8)(12–6)
=√12*2*4*6 = √6*2*2*4*6
=√6*6^4*4 = 6^4 = 24 ………….(1)
Also, area of triangle EBC = 1\2^10 cm^ CL = 5 cm^CL ………….(2)
FROM (1) and (2) , we get
5 cm^ CL =24
CL = 4.8 cm
Now , area of the trapezium ABCD
= 1\2 ^ (20 30m) ^ 4.8 cm
1\2^ 50^ 4.8 cm = 120
please mark me as brainliest. I beg you mate. this is perfectly right. believe me. . . . . . . . . . . . . . . . .
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