Question

ABCD is a rectangle , E is mid-point of AD , F is mid-point of EC , area of Rectangle ABCD = 120cm² and area of triangle BDF = ab cm²
(Where ab is two digit number ) then find the value of (a+b)

Note:- Answer can be single digit integer from 0 to 9

Answer 1

(refer the attachment for figure)

Given :- E is the midpoint of AD and F is the mid point of EC
Construction :- Join BE

Solution:-

Consider ∆EBC and the rectangle. They both lie on the same base (BC) and same parallel (BC || AD), hence,

ar(EBC) = 1/2 ar(ABCD)

=> ar(EBC) = 120/2 = 60 cm²

Now, consider ∆ABE and ∆EDC

ar(ABE) = 1/2 × AE × AB

=> ar(∆ABE) = 1/2 × ED × DC

(since, E is the midpoint and opposite sides are equal in a rectangle)

=> ar(ABE) = ar(EDC)

Now, ar(ABE + EDC + EBC) = ar(ABCD)

=> ar(EDC + EDC) + 1/2ar(ABCD) = ar (ABCD)

=> 2ar(EDC) = ar(ABCD) - 1/2(ABCD)

=> ar(EDC) = 1/4(ABCd)

=> ar(EDC) = 120/4

=> ar(EDC) = 30 cm²

Now, in ∆EDC, DF is the median and thus

ar(DFC) = 1/2 ar(EDC)

(Medians divides the triangle into two triangles of equal area)

=> ar(DFC) = 30/2 = 15 cm²

Now for ∆BEC, BF is the median. In the same way as done above, ar(BFC) = 60/2 = 30cm²

ar(BFDC) = ar(DFC) + ar(BFC)

= ar(BFDC) = 15 + 30 = 45 cm²


Now ar(BDC) = 1/2(ABCD)

(Diagonal divides the rectangle into two triangles of equal area)

=> ar(BDC) = 120/2 = 60 cm²

So, ar(BDF) = ar(BDC) - ar(BFDC)

=> ar(BDF) = 60 - 45

=> ar(BDF) = 15 cm²


Now ar(BDF) = ab cm²

=> ab = 15

Now, a can be 3 or 5 and b can be 5 or 3


So, in both cases, (a + b) = (5 + 3) = 8


Hence, your answer is 8



Answer :- 8

Answer 2

Question :

ABCD is a rectangle , E is mid-point of AD , F is mid-point of EC , area of Rectangle ABCD = 120cm² and area of triangle BDF = ab cm²
(Where ab is two digit number ) then find the value of (a+b)

Note:- Answer can be single digit integer from 0 to 9

Solution:

Refer to the above attachments !!

The required answer is

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