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ABCD is an isosceles trapezium such that AD || BC, AB = 5 cm, AD = 8 cm and BC = 14 cm. What is the area (in cm2) of trapezium?
- a)36
- b)44
- c)88
- d)144
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
ABCD is an isosceles trapezium such that AD || BC, AB = 5 cm, AD = 8 c...
Let AE and DF are the perpendiculars dropped
As can be observed in figure,
BC can be divided as, BC = BE + EF + FC,
such that, EF = AD = 8 cm
and BE = FC =
= 3 cm
Solving, AE = DF =
= 4 cm
Area trapezium ABCD
As can be observed in figure,
BC can be divided as, BC = BE + EF + FC,
such that, EF = AD = 8 cm
and BE = FC =
= 3 cmSolving, AE = DF =
= 4 cmArea trapezium ABCD
► Area △ABE + Area rectangle AEFD + Area △CDF
► (1/2) x 3 x 4 + 8 x 4 + (1/2) x 3 x 4 = 44 cm2
► (1/2) x 3 x 4 + 8 x 4 + (1/2) x 3 x 4 = 44 cm2
Most Upvoted Answer
ABCD is an isosceles trapezium such that AD || BC, AB = 5 cm, AD = 8 c...
Given: Isosceles trapezium ABCD, AD || BC, AB = 5 cm, AD = 8 cm, BC = 14 cm.
To find: Area of trapezium.
Approach: We can find the height of the trapezium using Pythagoras theorem and then use the formula for the area of trapezium to find the answer.
Solution:
- Draw a rough figure and mark the given values.
- Since AD || BC, we can draw a perpendicular from A to BC.
- Let E be the foot of the perpendicular from A to BC.
- We know that AE = 8 cm, AB = 5 cm, and BC = 14 cm.
- Using Pythagoras theorem, we can find the height of the trapezium.
- In right triangle ABE, BE2 + AE2 = AB2
- BE2 + 82 = 52
- BE2 = 25 - 64 = -39 (which is not possible)
- This means that the foot of the perpendicular E lies outside the line segment BC.
- We can extend BC to meet the perpendicular from A at F.
- Let FC = x cm.
- Then, BF = 14 - x cm.
- In right triangle AEF, AF2 + EF2 = AE2
- AF2 + x2 = 82
- AF2 = 64 - x2
- In right triangle BFC, BF2 + FC2 = BC2
- (14 - x)2 + x2 = 196
- 196 - 28x + 2x2 + x2 = 196
- 3x2 - 28x + 0 = 0
- Solving for x, we get x = 8/3 cm (approx)
- Therefore, BF = 14 - x = 38/3 cm (approx)
- Now, the height of the trapezium is EF = AF - AE = sqrt(64 - x2) - 8
- Plugging in the values, we get EF = 4/3 cm (approx)
- The area of the trapezium is given by (sum of parallel sides) x (height) / 2
- Area = (AB + DC) x EF / 2
- Area = (5 + 14) x (4/3) / 2
- Area = 44/3 cm2 (approx)
Therefore, the area of the trapezium is 44 cm2 (approx). Hence, option B is the correct answer.
To find: Area of trapezium.
Approach: We can find the height of the trapezium using Pythagoras theorem and then use the formula for the area of trapezium to find the answer.
Solution:
- Draw a rough figure and mark the given values.
- Since AD || BC, we can draw a perpendicular from A to BC.
- Let E be the foot of the perpendicular from A to BC.
- We know that AE = 8 cm, AB = 5 cm, and BC = 14 cm.
- Using Pythagoras theorem, we can find the height of the trapezium.
- In right triangle ABE, BE2 + AE2 = AB2
- BE2 + 82 = 52
- BE2 = 25 - 64 = -39 (which is not possible)
- This means that the foot of the perpendicular E lies outside the line segment BC.
- We can extend BC to meet the perpendicular from A at F.
- Let FC = x cm.
- Then, BF = 14 - x cm.
- In right triangle AEF, AF2 + EF2 = AE2
- AF2 + x2 = 82
- AF2 = 64 - x2
- In right triangle BFC, BF2 + FC2 = BC2
- (14 - x)2 + x2 = 196
- 196 - 28x + 2x2 + x2 = 196
- 3x2 - 28x + 0 = 0
- Solving for x, we get x = 8/3 cm (approx)
- Therefore, BF = 14 - x = 38/3 cm (approx)
- Now, the height of the trapezium is EF = AF - AE = sqrt(64 - x2) - 8
- Plugging in the values, we get EF = 4/3 cm (approx)
- The area of the trapezium is given by (sum of parallel sides) x (height) / 2
- Area = (AB + DC) x EF / 2
- Area = (5 + 14) x (4/3) / 2
- Area = 44/3 cm2 (approx)
Therefore, the area of the trapezium is 44 cm2 (approx). Hence, option B is the correct answer.
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ABCD is an isosceles trapezium such that AD || BC, AB = 5 cm, AD = 8 cm and BC = 14 cm. What is the area (in cm2) of trapezium?a)36b)44c)88d)144Correct answer is option 'B'. Can you explain this answer?
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ABCD is an isosceles trapezium such that AD || BC, AB = 5 cm, AD = 8 cm and BC = 14 cm. What is the area (in cm2) of trapezium?a)36b)44c)88d)144Correct answer is option 'B'. Can you explain this answer?, a detailed solution for ABCD is an isosceles trapezium such that AD || BC, AB = 5 cm, AD = 8 cm and BC = 14 cm. What is the area (in cm2) of trapezium?a)36b)44c)88d)144Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of ABCD is an isosceles trapezium such that AD || BC, AB = 5 cm, AD = 8 cm and BC = 14 cm. What is the area (in cm2) of trapezium?a)36b)44c)88d)144Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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