Answer

Verified

405.3k+ views

**Hint:**Draw a line passing through $P$ and parallel to side $AD$ cutting the sides $AB$ and $CD$ at $E$ and $F$ respectively.

Then, find the area of Triangles $APB$ and $CPD$ by using the formula $area = \dfrac{1}{2} \times B \times H$ where $B$ is the base and $H$ is the height of the triangle. Then, add the area of both triangles. And compare it with the area of the rectangle.

**Complete step by step answer:**

Given, ABCD is a rectangle and area of rectangle $ABCD = AB \times AD$

($\because $ area of rectangle $ = $ length $ \times $ breadth)

Now, In triangle $APB$, $AB$ is the base of the triangle and $EP$ is the height of the triangle because $EP$ is perpendicular to $AB$.

So, area of triangle $APB = \dfrac{1}{2} \times AB \times EP$

Similarly, In triangle $CPD$, $CD$ is the base of the triangle and $FP$ is the height of the triangle.

So, area of triangle $CPD = \dfrac{1}{2} \times CD \times FP$

Now, adding the area of both triangles we get,

$\Rightarrow ar\left( {APB} \right) + ar\left( {CPD} \right) = \left( {\dfrac{1}{2} \times AB \times EP} \right) + \left( {\dfrac{1}{2} \times CD \times FP} \right)$

Taking $\dfrac{1}{2}$ common from both in right hand side we get,

$\Rightarrow ar\left( {APB} \right) + ar\left( {CPD} \right) = \dfrac{1}{2} \times \left( {\left( {AB \times EP} \right) + \left( {CD \times FP} \right)} \right)$

Since $ABCD$ is a rectangle so opposite sides are equal.

$\therefore AB = CD$

$\Rightarrow ar\left( {APB} \right) + ar\left( {CPD} \right) = \dfrac{1}{2} \times \left( {\left( {AB \times EP} \right) + \left( {AB \times FP} \right)} \right)$

Taking $AB$ common from both we get,

$\Rightarrow ar\left( {APB} \right) + ar\left( {CPD} \right) = \dfrac{1}{2} \times AB \times \left( {EP + FP} \right)$

From figure it is obvious that $EF = EP + FP$

$\therefore ar\left( {APB} \right) + ar\left( {CPD} \right) = \dfrac{1}{2} \times AB \times EF$

Now in rectangle $ABCD$, $AD = EF = BC$ because opposite sides are equal and parallel.

Thus, we can write

$\Rightarrow ar\left( {APB} \right) + ar\left( {CPD} \right) = \dfrac{1}{2} \times AB \times AD$

Since $ar\left( {ABCD} \right) = AB \times AD$ proved above.

We can finally write $ar\left( {APB} \right) + ar\left( {CPD} \right) = ar\left( {ABCD} \right)$

**Hence, (1) is proved.**

(2) Draw a line passing through $P$ and parallel to $AB$ then find the area of triangles $APD$ and $BPC$ using formula $area = \dfrac{1}{2} \times B \times H$ where $B$ is the base and $H$ is the height of the triangle. Then, add the area of both triangles.

Given, ABCD is a rectangle and area of rectangle $ABCD = AB \times AD$

($\because $ area of rectangle $ = $ length $ \times $ breadth)

Now, In triangle $APD$, $AD$ is the base of triangle and $GP$ is the height of triangle because $GP$ is perpendicular to $AD$.

So, area of triangle $APD = \dfrac{1}{2} \times AD \times GP$

Similarly, In triangle $BPC$, $BC$ is the base of the triangle and $HP$ is the height of the triangle.

So, area of triangle $BPC = \dfrac{1}{2} \times BC \times HP$

Now, adding the area of both triangles we get,

$\Rightarrow ar\left( {APD} \right) + ar\left( {BPC} \right) = \left( {\dfrac{1}{2} \times AD \times GP} \right) + \left( {\dfrac{1}{2} \times BC \times HP} \right)$

Taking $\dfrac{1}{2}$ common from both in right hand side we get,

$\Rightarrow ar\left( {APD} \right) + ar\left( {BPC} \right) = \dfrac{1}{2} \times \left( {\left( {AD \times GP} \right) + \left( {BC \times HP} \right)} \right)$

Since,$ABCD$ is a rectangle so opposite sides are equal.

$\therefore AD = {\rm B}C$

$\Rightarrow ar\left( {APD} \right) + ar\left( {BPC} \right) = \dfrac{1}{2} \times \left( {\left( {AD \times GP} \right) + \left( {AD \times HP} \right)} \right)$

Taking $AD$ common from both we get,

$\Rightarrow ar\left( {APD} \right) + ar\left( {BPC} \right) = \dfrac{1}{2} \times AD \times \left( {GP + {\rm H}P} \right)$

From figure it is obvious that $GH = GP + HP$

$\therefore ar\left( {APD} \right) + ar\left( {BPC} \right) = \dfrac{1}{2} \times AD \times GH$

Now in rectangle $ABCD$, $AB = CD = GH$ because opposite sides are equal and parallel.

Thus, we can write

$\Rightarrow ar\left( {APD} \right) + ar\left( {BPC} \right) = \dfrac{1}{2} \times AD \times AB$

Since $ar\left( {ABCD} \right) = AB \times AD$ proved above.

We can finally write $ar\left( {APD} \right) + ar\left( {BPC} \right) = ar\left( {ABCD} \right)$

Since the right side of the result of (1) and of this are equal so, we can equate the left-hand side then,

$\Rightarrow ar\left( {APB} \right) + ar\left( {CPD} \right) = ar\left( {APD} \right) + ar\left( {BPC} \right) = \dfrac{1}{2}ar\left( {ABCD} \right)$

**Hence, (2) is proved.**

**Note:**

- In this type of problem make suitable construction which converts the given figure into a known figure whose area and perimeter can be written easily

- If the right-hand side of two equations is equal then we can say that the left-hand side must be equal.

- This relation is also valid in the case of a square.

Recently Updated Pages

Differentiate between Shortterm and Longterm adapt class 1 biology CBSE

How do you find slope point slope slope intercept standard class 12 maths CBSE

How do you find B1 We know that B2B+2I3 class 12 maths CBSE

How do you integrate int dfracxsqrt x2 + 9 dx class 12 maths CBSE

How do you integrate int left dfracx2 1x + 1 right class 12 maths CBSE

How do you find the critical points of yx2sin x on class 12 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Define limiting molar conductivity Why does the conductivity class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Name 10 Living and Non living things class 9 biology CBSE

The Buddhist universities of Nalanda and Vikramshila class 7 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE