I'm unable to solve this problem
Attachments:
TIRTH5828:
i suggest question is incomplete
Answers
Answered by
1
ABCD is a rectangle. P is on AD. Q is on AB.
PR is drawn || AB. QS is drawn || to AD.
As AB ⊥ QS, AB ⊥ AD, we have AD || QS.
As PR ⊥ AD , PR ⊥ BC, we PR || AB.
PR and QS intersect at O.
So we have OQ || PA, and PO || AQ. So APOQ is a parallelogram.
As the angles between adjacent sides are 90 deg., APOQ is a rectangle.
We also have : AP = OQ = BR = 1/4 * BC
AQ = PO = DS = 1/4 * CD
Area of APOQ = AP * PO = 1/16 * BC * CD
Area of ABCD = AD * AB = AD * DC = BC * CD
Hence, the ratio of areas : 1/16 : 1 or 1 : 16
PR is drawn || AB. QS is drawn || to AD.
As AB ⊥ QS, AB ⊥ AD, we have AD || QS.
As PR ⊥ AD , PR ⊥ BC, we PR || AB.
PR and QS intersect at O.
So we have OQ || PA, and PO || AQ. So APOQ is a parallelogram.
As the angles between adjacent sides are 90 deg., APOQ is a rectangle.
We also have : AP = OQ = BR = 1/4 * BC
AQ = PO = DS = 1/4 * CD
Area of APOQ = AP * PO = 1/16 * BC * CD
Area of ABCD = AD * AB = AD * DC = BC * CD
Hence, the ratio of areas : 1/16 : 1 or 1 : 16
Similar questions