Question

ABCD is a parallelogram of area 162 sq cm. P is a point on Ab such that AP : PB = 1 : 2. Calculate:
i) The area of Triangle APD
ii) The ratio PQ : PD

Answer 1

area=BH=162
P is the point perpendicular to point D
3xh=162
xh=54
X=6;H=9
base=18
area of triangle APD= 1/2*6*9=27

Answer 2

Area of Triangle is 27 $cm^2$

Step-by-step explanation:

We have,

PB = 2AP

AP = $\frac {AB}{3}$

Altitude of the parallelogram be CL

Now,  

ar(ABCD) = 162 $cm^2$

AB(CL) = $162\ cm^2$

So,  

ar(ΔAPD) = $\frac {1}{2} (AP)(CL)$

= $\frac {1}{2} \frac {AB}{3} (CL)$

= $\frac {AB(CL)}{6}$

= $\frac {162}{6}$

= 27 $cm^2$