In the parallelogram ABCD- P is mid-point of AB- CP and BD intersect each other at point O- If the area of - P O B-40 cm 2- find the areas of - BOC and PBC -A- 80 cm 2- 120 cm 2B- 60 cm 2- 120 cm 2C- 40 cm 2- 80 cm 2D- 80 cm 2- 100 cm 2

The correct option is A 80 Given: DC = 2 PB Δ POB and Δ DOC are similar triangles by AAA. Hence, Area of Δ POB /Area of Δ DOC = PB^2/DC^2 40/Area of Δ DOC = ...

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Byju's Answer

Standard VIII

Mathematics

Rectangle

In the parall...

Question

In the parallelogram ABCD, P is mid-point of AB. CP and BD intersect each other at point O. If the area of $Δ$ POB = 40 $c m^{2}$ , find the areas of $Δ$ BOC and $Δ$ PBC.

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Solution

The correct option is A
80

Given: DC = 2 PB

$Δ$ POB and $Δ$ DOC are similar triangles by AAA.

Hence,

$\frac{A r e a o f Δ P O B}{A r e a o f Δ D O C} = \frac{P B^{2}}{D C^{2}} \frac{40}{A r e a o f Δ D O C} = \frac{P B^{2}}{(2 P B)^{2}}$

Area of $Δ$ DOC = 40 $\times$ 4

= 160 $c m^{2}$

Let the area of the $Δ$ APC be y $c m^{2}$

Area of $Δ$ ABC = Area of $Δ$ BDC, Base of the triangles are same and they are between same parallel lines.

y + 40 + $Δ$ BOC = $Δ$ BOC + 160

y = 160 -40

= 120 $c m^{2}$

Now, Area of $Δ$ APC = $Δ$ BPC (PC is the median of the triangle ABC)

Area of $Δ$ PBC = 120 $c m^{2}$

Area of $Δ$ PBC = Area of $Δ$ POB + Area of $Δ$ BOC

120 = 40 + Area of $Δ$ BOC

Area of $Δ$ BOC = 120 – 40

= 80 $c m^{2}$

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In the parallelogram ABCD, P is mid-point of AB. CP and BD intersect each other at point O. If the area of Δ POB = 40 cm2, find the areas of Δ BOC and Δ PBC.

In the parallelogram ABCD, P is mid-point of AB. CP and BD intersect each other at point O. If the area of $Δ$ POB = 40 $c m^{2}$ , find the areas of $Δ$ BOC and $Δ$ PBC.