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ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. If area ∆DFB = 3 cm2, area of parallelogram ABCD is:
- a)6 cm2
- b)12 cm2
- c)18 cm2
- d)9 cm2
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
ABCD is a parallelogram in which BC is produced to E such that CE = BC...
ar(∆AFB) = 1/2 ar (parallelogram ABCD)
{areas between same parallels and same base}
⇒ ar (parallelogram ABCD) = 2 × ar(∆AFB) = 2 × ar(∆DCB) …(i)
In ∆s ADF and ECF.
∠AFD = ∠EFC {vertically opposite angles}
∠ACF = ∠ADF {alternate angles}
BC = CE = AD
∴ ∆ADF ≅ ∆ECF {by AAS congruency}
∴ DF = CF
ar(parallelogram ABCD) = 2 × ar(∆DCB)
= 2 × 2 × (ar(∆DFB))
= 2 × 2 × 3
= 12 cm2
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ABCD is a parallelogram in which BC is produced to E such that CE = BC...
Given information:
- ABCD is a parallelogram
- BC is produced to E such that CE = BC
- AE intersects CD at F
- Area of triangle DFB = 3 cm²
To find: Area of parallelogram ABCD
Proof:
1) Let's consider the area of triangle DFB and parallelogram ABCD.
2) Since DFB is a triangle, its area can be calculated using the formula: Area = (1/2) × base × height.
3) In triangle DFB, the base is DF and the height is FB.
4) Therefore, Area of triangle DFB = (1/2) × DF × FB.
5) Now, let's consider the area of parallelogram ABCD.
6) The area of a parallelogram can be calculated using the formula: Area = base × height.
7) In parallelogram ABCD, the base is AB and the height is the perpendicular distance between AB and CD.
8) Since ABCD is a parallelogram, AB is parallel to CD.
9) AE intersects CD at F, so the perpendicular distance between AB and CD is equal to the length of the line segment FB.
10) Therefore, Area of parallelogram ABCD = AB × FB.
11) From the given information, we know that the area of triangle DFB is 3 cm².
12) Using the formula mentioned in step 4, we can write: (1/2) × DF × FB = 3.
13) Simplifying the equation, we get: DF × FB = 6. (Equation 1)
14) From the given information, we also know that CE = BC.
15) Since ABCD is a parallelogram, opposite sides are equal in length.
16) Therefore, AB = CD.
17) Also, since BC is produced to E such that CE = BC, we can write: BC = CE = BE.
18) Since AE intersects CD at F, we can write: AF = FC.
19) Now, let's consider the triangle AEF.
20) In triangle AEF, AF = FC, so the triangle is isosceles.
21) The base of the isosceles triangle is EF, and the height is the perpendicular distance between EF and AF.
22) Since EF is parallel to AB and AF is parallel to CD, the perpendicular distance between EF and AF is equal to the length of the line segment FB.
23) Therefore, the area of triangle AEF is (1/2) × EF × FB.
24) From equations 11 and 23, we can write: (1/2) × EF × FB = 3.
25) Simplifying the equation, we get: EF × FB = 6. (Equation 2)
26) Comparing equations 1 and 2, we can see that DF × FB = EF × FB.
27) Dividing both sides of the equation by FB, we get: DF = EF.
28) Since AB is parallel to CD, triangles ADF and CEF are similar.
29) Therefore, the corresponding sides are in proportion: AD/CE = DF/EF = AF/FC.
30) Since AD = BC (opposite sides of a parallelogram are equal), we can write: BC/CE = DF/EF.
31)
- ABCD is a parallelogram
- BC is produced to E such that CE = BC
- AE intersects CD at F
- Area of triangle DFB = 3 cm²
To find: Area of parallelogram ABCD
Proof:
1) Let's consider the area of triangle DFB and parallelogram ABCD.
2) Since DFB is a triangle, its area can be calculated using the formula: Area = (1/2) × base × height.
3) In triangle DFB, the base is DF and the height is FB.
4) Therefore, Area of triangle DFB = (1/2) × DF × FB.
5) Now, let's consider the area of parallelogram ABCD.
6) The area of a parallelogram can be calculated using the formula: Area = base × height.
7) In parallelogram ABCD, the base is AB and the height is the perpendicular distance between AB and CD.
8) Since ABCD is a parallelogram, AB is parallel to CD.
9) AE intersects CD at F, so the perpendicular distance between AB and CD is equal to the length of the line segment FB.
10) Therefore, Area of parallelogram ABCD = AB × FB.
11) From the given information, we know that the area of triangle DFB is 3 cm².
12) Using the formula mentioned in step 4, we can write: (1/2) × DF × FB = 3.
13) Simplifying the equation, we get: DF × FB = 6. (Equation 1)
14) From the given information, we also know that CE = BC.
15) Since ABCD is a parallelogram, opposite sides are equal in length.
16) Therefore, AB = CD.
17) Also, since BC is produced to E such that CE = BC, we can write: BC = CE = BE.
18) Since AE intersects CD at F, we can write: AF = FC.
19) Now, let's consider the triangle AEF.
20) In triangle AEF, AF = FC, so the triangle is isosceles.
21) The base of the isosceles triangle is EF, and the height is the perpendicular distance between EF and AF.
22) Since EF is parallel to AB and AF is parallel to CD, the perpendicular distance between EF and AF is equal to the length of the line segment FB.
23) Therefore, the area of triangle AEF is (1/2) × EF × FB.
24) From equations 11 and 23, we can write: (1/2) × EF × FB = 3.
25) Simplifying the equation, we get: EF × FB = 6. (Equation 2)
26) Comparing equations 1 and 2, we can see that DF × FB = EF × FB.
27) Dividing both sides of the equation by FB, we get: DF = EF.
28) Since AB is parallel to CD, triangles ADF and CEF are similar.
29) Therefore, the corresponding sides are in proportion: AD/CE = DF/EF = AF/FC.
30) Since AD = BC (opposite sides of a parallelogram are equal), we can write: BC/CE = DF/EF.
31)
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ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. If area DFB = 3 cm2, area of parallelogram ABCD is:a)6 cm2b)12 cm2c)18 cm2d)9 cm2Correct answer is option 'B'. Can you explain this answer?
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ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. If area DFB = 3 cm2, area of parallelogram ABCD is:a)6 cm2b)12 cm2c)18 cm2d)9 cm2Correct answer is option 'B'. Can you explain this answer? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. If area DFB = 3 cm2, area of parallelogram ABCD is:a)6 cm2b)12 cm2c)18 cm2d)9 cm2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. If area DFB = 3 cm2, area of parallelogram ABCD is:a)6 cm2b)12 cm2c)18 cm2d)9 cm2Correct answer is option 'B'. Can you explain this answer?.
ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. If area DFB = 3 cm2, area of parallelogram ABCD is:a)6 cm2b)12 cm2c)18 cm2d)9 cm2Correct answer is option 'B'. Can you explain this answer? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. If area DFB = 3 cm2, area of parallelogram ABCD is:a)6 cm2b)12 cm2c)18 cm2d)9 cm2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. If area DFB = 3 cm2, area of parallelogram ABCD is:a)6 cm2b)12 cm2c)18 cm2d)9 cm2Correct answer is option 'B'. Can you explain this answer?.
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