q-8]A plastic box 1.5cmlong,1.25mwide and 65cm deep is to be made.It is to be open at the top. Ignoring the thickness if the plastic sheet, determine  the area of the sheet required fo r making the box .  the cost of sheet for it @costs Rs.20/m2.
q-9] In triangle ABC,D are the mid point of AB.P is any on BC .CQIIPD meets AB at Q. show that area [triangle BPQ] =1/2 area [triangle ABC].
q-10]PQRS is a llgm .A is a any point on PS such that PA=1/3PS and B is appoint on QR such that BR=1/2 QR.prove that PBRA is a llgm.
(1) the length of the plastic box l=1.5cm
width of the plastic box b=1.25m=125cm
depth of the plastic box h=65cm
surface area of the 4 sides=2h(l+b)=2*65(1.5+125)=130*(126.5)=16445 cm square
it is open at top. therefore area of the base=lb=1.5*125=187.5 cm square
total area of the sheet =16445+187.5=16632.5 cm square=1.66325 m square
cost of the sheet =1.66325*20=33.27 /-
given: D is the mid-point of BC. and P is any point on AB . and CQ is parallel to PD.
TPT: area [triangle BPQ] =1/2 area [triangle ABC]
Proof: construction: join CD and PQ.
ΔPDQ and ΔPDC are on the same base PD and between same parallel lines PD and CQ.
therefore area(ΔPDQ) = area(ΔPDC)
adding area of ΔBDP on both sides, we get
[since AD is the median of triangle ABC and median divides the triangle into two equal areas]
hence area(ΔBPQ)=1/2 area(ΔABC)
if the question is :PQRS is a llgm .A is a any point on PS such that PA=1/3PS and B is appoint on QR such that BR=1/3 QR.prove that PBRA is a llgm.
given: PQRS is a parallelogram. and PA=1/3PS and BR=1/3QR
TPT: PBRA is a parallelogram
proof: since PQRS is a parallelogram.
hence PABR is a parallelogram.