A rectangle is removed from the middle of a larger rectangular shaped board. What is the area of the remaining board?

Mathematics

1 answer:

jeyben [28]2 years ago

8 0

Answer:

The area of the remaining board is [(L × B) - (l × b)].

Step-by-step explanation:

Suppose the bigger rectangle is labelled as ABCD and the smaller rectangle is labelled as PQRS.

Consider that the length and breadth of the bigger rectangle are L and B respectively. And the length and breadth of the bigger rectangle are l and b respectively.

The area of any rectangle is:

Area = Length × Breadth

The area of the bigger rectangle is:

Area of ABCD = L × B

The area of the smaller rectangle is:

Area of PQRS = l × b

Then the area of the remaining board will be:

Area of remaining board = Area of ABCD - Area of PQRS

= (L × B) - (l × b)

Thus, the area of the remaining board is [(L × B) - (l × b)].

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Licemer1 [7]

Answer:

4 cm

Step-by-step explanation:

Draw the line MD that is parallel to AQ. Consider triangles BMD and BAQ. These triangles are similar, because they have three pairs of congruent angles (angles BMD and BAQ are corresponding, angles BDM and BQA are corresponding and angle B is common). The coefficient of similarity is $\dfrac{1}{2},$ because point M is midpoint of AB. Then

$\dfrac{BD}{BQ}=\dfrac{1}{2}\Rightarrow BQ=2BD.$

Since $BQ=BD+DQ,$ then $DQ=2BD-BD=BD.$

Consider triangles CPQ and CMD. These triangles are similar, because they have three pairs of congruent angles (angles CPQ and CMD are corresponding, angles CQP and CDM are corresponding and angle C is common). The coefficient of similarity is $\dfrac{1}{2},$ because point P is midpoint of CM. Then

$\dfrac{CQ}{CD}=\dfrac{1}{2}\Rightarrow CD=2CQ.$