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 [1] the area of the sheet required fo r making the box . [2] 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


area(ΔBPQ)=area(ΔBDC)=1/2 area(ΔABC)

[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.





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