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 respectively.
ar(ADP)
ar(APCR)
ar{ ADP)
(c)
ar(ADCP)apq
(DCP)k
ar(APO)
ar(ADP)
(d)
ar( ZAPO)​

Answer 1

Answer:

DR can cut AB at only one point say P.

Since P lies on AB it follows that PB is || DC. Hence triangle RBP ~triangle RCD. And since BR:CR = 4:8 = 1:2 it follows that PB:DC is also 1:2 Therefore PB = 4 cms. So P is the midpoint of AB. Area of triangle ADP is half the area of triangle ADB. And since area of triangle ADB = area of triangle BCD = area of triangle DCP it follows that area of triangle ADP = one-half the area of triangle DCP.

Step-by-step explanation:

plzz mark.me as brainlist and mainly follow me guys for more of such answers...