SSC Exam > SSC Questions > ABCD is parallelogram and E is a point on AD,...
Start Learning for Free
ABCD is parallelogram and E is a point on AD, which divides AD in the ratio 3 : 1. If the area of the triangle EBC is x cm2, then the area of the parallelogram ABCD is equal to
- a)3x cm2
- b)3x/2 cm2
- c)2x cm2
- d)4x/2 cm2
Correct answer is option 'C'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
ABCD is parallelogram and E is a point on AD, which divides AD in the ...
Most Upvoted Answer
ABCD is parallelogram and E is a point on AD, which divides AD in the ...
Given, ABCD is a parallelogram and E is a point on AD dividing AD in the ratio 3:1.
To find: Area of parallelogram ABCD
Approach:
- Draw a perpendicular from B to AD and let it meet AD at F.
- As per the properties of parallelogram, BF is parallel to AC and AD.
- As EF is a line passing through point E, BF and AC, and AD, ∠EBC and ∠FBC are alternate angles and therefore equal.
- Similarly, ∠AFC and ∠BCF are alternate angles and therefore equal.
- Also, ∠ABC and ∠FBE are vertically opposite angles and therefore equal.
- Therefore, ΔEBC and ΔFBM (where M is the midpoint of BC) are similar triangles, as they have two angles equal.
- As E divides AD in the ratio 3:1, we can say that AE:ED = 3:1. Therefore, AM:MD = 3:1.
- As M is the midpoint of BC, BM = MC. Therefore, we can say that BF:FC = 3:1.
Calculation:
- Let the area of triangle EBC be x cm^2.
- As ΔEBC and ΔFBM are similar triangles, we can say that their corresponding sides are in the same ratio.
- Therefore, as BM:MC = 1:1, we can say that EB:BF = 3:1.
- Let the height of parallelogram ABCD be h cm.
- As BF:FC = 3:1, we can say that the height of triangle FBC is h/4.
- Therefore, the area of triangle FBC is (1/2) * BF * (h/4) = (1/8) * BF * h.
- As ΔABC and ΔFBE are similar triangles, we can say that their corresponding sides are in the same ratio.
- Therefore, as AB:BF = 3:1, we can say that the height of triangle FBE is (3/4)h.
- Therefore, the area of triangle FBE is (1/2) * BF * (3/4)h = (3/8) * BF * h.
- As ∠ABC and ∠FBE are equal, we can say that the area of parallelogram ABCD is twice the area of ΔFBE.
- Therefore, the area of parallelogram ABCD = 2 * (3/8) * BF * h = (3/4) * BF * h.
- As E divides AD in the ratio 3:1, we can say that AE:ED = 3:1. Therefore, AE = (3/4)AD and ED = (1/4)AD.
- As BF is parallel to AD, we can say that BF:AD = FC:AD = BF+FC:AD = 1:4.
- Therefore, BF = (1/5)AD and FC = (4/5)AD.
- As the height of parallelogram ABCD is h, we can say that the area of parallelogram ABCD = base * height = AD * h.
- Therefore, the area of parallelogram ABCD = AD * h = (AD
To find: Area of parallelogram ABCD
Approach:
- Draw a perpendicular from B to AD and let it meet AD at F.
- As per the properties of parallelogram, BF is parallel to AC and AD.
- As EF is a line passing through point E, BF and AC, and AD, ∠EBC and ∠FBC are alternate angles and therefore equal.
- Similarly, ∠AFC and ∠BCF are alternate angles and therefore equal.
- Also, ∠ABC and ∠FBE are vertically opposite angles and therefore equal.
- Therefore, ΔEBC and ΔFBM (where M is the midpoint of BC) are similar triangles, as they have two angles equal.
- As E divides AD in the ratio 3:1, we can say that AE:ED = 3:1. Therefore, AM:MD = 3:1.
- As M is the midpoint of BC, BM = MC. Therefore, we can say that BF:FC = 3:1.
Calculation:
- Let the area of triangle EBC be x cm^2.
- As ΔEBC and ΔFBM are similar triangles, we can say that their corresponding sides are in the same ratio.
- Therefore, as BM:MC = 1:1, we can say that EB:BF = 3:1.
- Let the height of parallelogram ABCD be h cm.
- As BF:FC = 3:1, we can say that the height of triangle FBC is h/4.
- Therefore, the area of triangle FBC is (1/2) * BF * (h/4) = (1/8) * BF * h.
- As ΔABC and ΔFBE are similar triangles, we can say that their corresponding sides are in the same ratio.
- Therefore, as AB:BF = 3:1, we can say that the height of triangle FBE is (3/4)h.
- Therefore, the area of triangle FBE is (1/2) * BF * (3/4)h = (3/8) * BF * h.
- As ∠ABC and ∠FBE are equal, we can say that the area of parallelogram ABCD is twice the area of ΔFBE.
- Therefore, the area of parallelogram ABCD = 2 * (3/8) * BF * h = (3/4) * BF * h.
- As E divides AD in the ratio 3:1, we can say that AE:ED = 3:1. Therefore, AE = (3/4)AD and ED = (1/4)AD.
- As BF is parallel to AD, we can say that BF:AD = FC:AD = BF+FC:AD = 1:4.
- Therefore, BF = (1/5)AD and FC = (4/5)AD.
- As the height of parallelogram ABCD is h, we can say that the area of parallelogram ABCD = base * height = AD * h.
- Therefore, the area of parallelogram ABCD = AD * h = (AD
|
Explore Courses for SSC exam
|
|
Similar SSC Doubts
ABCD is parallelogram and E is a point on AD, which divides AD in the ratio 3 : 1. If the area of the triangle EBC is x cm2, then the area of the parallelogram ABCD is equal toa)3x cm2b)3x/2 cm2c)2x cm2d)4x/2 cm2Correct answer is option 'C'. Can you explain this answer?
Question Description
ABCD is parallelogram and E is a point on AD, which divides AD in the ratio 3 : 1. If the area of the triangle EBC is x cm2, then the area of the parallelogram ABCD is equal toa)3x cm2b)3x/2 cm2c)2x cm2d)4x/2 cm2Correct answer is option 'C'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about ABCD is parallelogram and E is a point on AD, which divides AD in the ratio 3 : 1. If the area of the triangle EBC is x cm2, then the area of the parallelogram ABCD is equal toa)3x cm2b)3x/2 cm2c)2x cm2d)4x/2 cm2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ABCD is parallelogram and E is a point on AD, which divides AD in the ratio 3 : 1. If the area of the triangle EBC is x cm2, then the area of the parallelogram ABCD is equal toa)3x cm2b)3x/2 cm2c)2x cm2d)4x/2 cm2Correct answer is option 'C'. Can you explain this answer?.
ABCD is parallelogram and E is a point on AD, which divides AD in the ratio 3 : 1. If the area of the triangle EBC is x cm2, then the area of the parallelogram ABCD is equal toa)3x cm2b)3x/2 cm2c)2x cm2d)4x/2 cm2Correct answer is option 'C'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about ABCD is parallelogram and E is a point on AD, which divides AD in the ratio 3 : 1. If the area of the triangle EBC is x cm2, then the area of the parallelogram ABCD is equal toa)3x cm2b)3x/2 cm2c)2x cm2d)4x/2 cm2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ABCD is parallelogram and E is a point on AD, which divides AD in the ratio 3 : 1. If the area of the triangle EBC is x cm2, then the area of the parallelogram ABCD is equal toa)3x cm2b)3x/2 cm2c)2x cm2d)4x/2 cm2Correct answer is option 'C'. Can you explain this answer?.
Solutions for ABCD is parallelogram and E is a point on AD, which divides AD in the ratio 3 : 1. If the area of the triangle EBC is x cm2, then the area of the parallelogram ABCD is equal toa)3x cm2b)3x/2 cm2c)2x cm2d)4x/2 cm2Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for SSC.
Download more important topics, notes, lectures and mock test series for SSC Exam by signing up for free.
Here you can find the meaning of ABCD is parallelogram and E is a point on AD, which divides AD in the ratio 3 : 1. If the area of the triangle EBC is x cm2, then the area of the parallelogram ABCD is equal toa)3x cm2b)3x/2 cm2c)2x cm2d)4x/2 cm2Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
ABCD is parallelogram and E is a point on AD, which divides AD in the ratio 3 : 1. If the area of the triangle EBC is x cm2, then the area of the parallelogram ABCD is equal toa)3x cm2b)3x/2 cm2c)2x cm2d)4x/2 cm2Correct answer is option 'C'. Can you explain this answer?, a detailed solution for ABCD is parallelogram and E is a point on AD, which divides AD in the ratio 3 : 1. If the area of the triangle EBC is x cm2, then the area of the parallelogram ABCD is equal toa)3x cm2b)3x/2 cm2c)2x cm2d)4x/2 cm2Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of ABCD is parallelogram and E is a point on AD, which divides AD in the ratio 3 : 1. If the area of the triangle EBC is x cm2, then the area of the parallelogram ABCD is equal toa)3x cm2b)3x/2 cm2c)2x cm2d)4x/2 cm2Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice ABCD is parallelogram and E is a point on AD, which divides AD in the ratio 3 : 1. If the area of the triangle EBC is x cm2, then the area of the parallelogram ABCD is equal toa)3x cm2b)3x/2 cm2c)2x cm2d)4x/2 cm2Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice SSC tests.
|
|
Explore Courses for SSC exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup