Question

14. ABCD is a rhombus with side equal to 8 cm. CB is produced to R ,such that BR = 4 cm. DR cuts AC and AB at P and Q respectively.
Find
(a) ar(∆ADP)/ar(∆PCR)
(b) ar(∆DCP)/ar(∆APQ)
(c) ar(∆ADP)/ar(∆DCP)
(d) ar(∆ADP)/ar(∆APQ)​

Answer 1

Answer:

Step-by-step explanation:

As AD || CR, angle DAB = angle ABR

angle APD = angle BPR (Vertically opposite angles)

By similarity of triangles criteria AA, ∆APD~∆BPR

=> AD/BR = AP/BP = PD/PR = 2

=> AP = 2 BP

Also, AB = AP + BP

=> 8 cm = 3BP

=> BP = 8/3 cm, AP = 16/3 cm

Now, let H lie on CD such that AH is perpendicular to AB

=> AH is perpendicular to DC

Ar(∆ADP) = AH × AP / 2 = (8 AH / 3) cm

Ar(∆DCP) = AH × CD / 2 = 4 AH cm

Ar(∆ADP) : Ar(∆DCP) = 8 AH / 3 cm : 4 AH cm = 2 : 3

Answer 2

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