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Q.

In the given figure the area of trapezium PQFE is half of the area of trapezium EFRS. If SR॥ EF ॥ PQ and SR and PQ is 7 cm and 10 cm respectively, then EF =

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a

9 cm

b

√83 cm

c

√6 cm

d

√87cm

answer is B.

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Detailed Solution

Given, SR= 7 cm PQ = 10 cm
Let area of PQFE = A
So, area of EFRS = 2A
So, area of PQRS = 2A + A = 3A
Hence, 3A = ½ (10+7) H
Let height of PQRS = h Length of EF = 7+k
Now, Height of trapezium SREF= kh/3
Height of trapezium PQFE = (3-k)/3 h
Area of SREF = 2x area of PQFE
= 1/2 (kh/3) (7+7+k) = 2x1/2 (3-kh/3) (10+7+k)
= k (14+k) = 2 (3-k) (17+k)
= 14k +k2 = 2 (51-14k- k2)
= 3k2 + 42k - 102= 0
= k2 + 14k -34 = 0
Hence a=1
b=14
c=-34
= K= -14±√ 142 -4% (-34)/2.1 (∵ K= -b I√b2 - 40c/2a)
= -14±√ 196+136/2
= K= -14±√332/2 = 7±√83
= K= -7 +√ 83 on K= -7 √83
Ef = 7+ K= √83
Hence, we get that EF = √83.
Type: CBSE
Class: Grade 10

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In the given figure the area of trapezium PQFE is half of the area of trapezium EFRS. If SR॥ EF ॥ PQ and SR and PQ is 7 cm and 10 cm respectively, then EF =