Question

Best answer will be marked brainliest...​

Answer 1

Answer:

Given P is the mid point of AD

so AP = PD

By using BPT in ∆ADB and ∆BCD

i)

So,

DP/AP = CR/RB [DP = AP]

1 = CR/RB

CR = RB

So R is also the mid point of side BC.

Hence proved.

ii)

Given

AB =12cm and CD = 8cm

In ∆ADB, p is the midpoint

By using mid point theorem in triangle.

PQ = 1/2 AB

PQ = 1/2 x 12 = 6

PQ = 6cm

And in ∆BCD

R is the mid point then use mid point theorem.

QR = 1/2 x CD

QR = 1/2 x 8

QR = 4cm

PR = PQ + QR

PR = 6 + 4

PR = 10cm

.........Hope this will help you..........

Answer 2

1. Here p is the midpoint of AB

Therefore , DP=AP

OR, DP/AP = 1 (Transporting AP to RHS)----eq(1)

Now in triangle ABD,

AB is parallel to PQ

{since AB is parallel to PR Therefor AB is also parallel to PQ as W lies on PR}

Therefore,

DP/AP = DQ/QB = 1 ------- eq(2)

{By Basic proportionality theorem (BPT) and by --eq(1)}

Similarly in triangle ADC,

RQ is parallel to DC

{since PR is parallel to DC there RQ is also parallel to DC as Q lies on PR}

Therefore,

DQ/QB = CR/BR

OR, 1 = CR/BR

{from --eq(2)}

Or, BR=CR

{ Transporting BR to LHS}

THEREFORE R DIVIDES BC IN TWO EQUAL PARTS. THUS WE CAN SAY THAT R IS THE MIDPOINT OF BC

THEREFORE R DIVIDES BC IN TWO EQUAL PARTS. THUS WE CAN SAY THAT R IS THE MIDPOINT OF BCHENCE PROVED......

2) HERE,

PQ = 1/2 × AB

Therefore,

PQ = 6cm -----eq(3)

similarly

RQ=1/2 × CD

Therefore,

RQ = 4cm ----- eq(4)

Now,

PR = PQ+RQ

OR, PR = 6cm+4cm

Therefore,

PR = 10CM.......

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