Question

Compare the area of right angled triangle $\Delta ABC$ and $\Delta PQR$ in which $\angle A = 30^\circ $ , $\angle B = 90^\circ $ , $AC = 4cm$ ,
$\angle P = 60^\circ $, $\angle Q = 90^\circ $ and $PQ = 4\,cm$:

Answer 1

Hint:
We will use trigonometric ratios to get the values of the lengths of all the other required sides of these.
For two given triangles, We will first use \[tan\theta = \dfrac{{perpendicular}}{{Base}}\] for $\Delta PQR$. Then we will use $\cos \theta = \dfrac{{base}}{{hypotenuse}}$
 and $\sin \,\theta = \dfrac{{perpendicular}}{{hypotenuse}}$ for $\Delta ABC$ . After getting the values of all the required sides, we will directly
calculate the area of triangles. Then to compare we will divide the areas of and to get the answer.

Complete step by step solution:
According to the question we are given two triangles, $\Delta ABC$ and $\Delta PQR$ in which
 $\angle A = 30^\circ $
 $\angle B = 90^\circ $
 $AC = 4\,cm$
 $\angle P = 60^\circ $
 $\angle Q = 90^\circ $
 $PQ = 4\,cm$

We know that,
 \[ \Rightarrow \tan \theta = \dfrac{{perpendicular}}{{Base}}\]

Now, in $\Delta PQR$ , We get
 \[ \Rightarrow \;tan{60^ \circ } = \dfrac{{QR}}{{PQ}}\]

We know that \[tan\,{60^ \circ }\] is \[\sqrt 3 \] and \[PQ = 4\] , Hence, we get
 $ \Rightarrow \sqrt 3 = \dfrac{{QR}}{4}$

By cross-multiplying, we get
 $ \Rightarrow QR = 4\sqrt 3 $

We know that area of a triangle is given by
 $ \Rightarrow Area = \dfrac{1}{2} \times base \times height$

Since, \[base = QR\] and \[height = PQ\] , we get
 $ \Rightarrow Area = \dfrac{1}{2}\, \times 4\sqrt 3 \times 4$

On simplification we get,
 $ \Rightarrow Area = 2\, \times 4\sqrt 3 $

On multiplication we get,
 $ \Rightarrow Area = 8\sqrt 3 $ … (1)
Hence, the area of $\Delta PQR$ is $8\sqrt 3 $

For $\Delta ABC$ , We know that,
 $\cos \theta = \dfrac{{base}}{{hypotenuse}}$

Now, in $\Delta ABC$ , we get
 $ \Rightarrow \cos {30^ \circ } = \dfrac{{AB}}{{AC}}$
We know that \[\cos {30^ \circ }\] is \[\dfrac{{\sqrt 3 }}{2}\] and \[AC = 4\] , Hence, we get
 $ \Rightarrow \dfrac{{\sqrt 3 }}{2}\, = \,\dfrac{{AB}}{4}$

By cross-multiplying, we get
  $ \Rightarrow AB = 2\sqrt 3 cm$

Also, we know that
 $\sin \theta = \dfrac{{perpendicular}}{{hypotenuse}}$

Hence, for $\Delta ABC$ , we have
 $ \Rightarrow \sin {30^ \circ } = \dfrac{{BC}}{{AC}}$

We know that \[\sin {30^ \circ }\] is \[\dfrac{1}{2}\] and \[AC = 4\] , Hence, we get
 $ \Rightarrow \dfrac{1}{2} = \,\dfrac{{BC}}{4}$

By cross-multiplying, we get
 $ \Rightarrow BC = 2cm$

We know that area of a triangle is given by
 $Area = \dfrac{1}{2} \times base \times height$

Since, \[base = {\text{ BC}}\] and \[height = AB\] , we get
 $ \Rightarrow Area = \dfrac{1}{2} \times 2 \times 2\sqrt 3 $

On simplification we get,
 $ \Rightarrow Area = 2\sqrt 3 $ …(2)
Hence, the area of $\Delta ABC$ is $2\sqrt 3 $

Now, dividing (1) by (2), we get
 $ \Rightarrow \dfrac{{ar(\Delta PQR)}}{{ar(\Delta ABC)}} = \dfrac{{8\sqrt 3 }}{{2\sqrt 3 }}$

On simplification we get,
 $ \Rightarrow \dfrac{{ar(\Delta PQR)}}{{ar(\Delta ABC)}} = 4$

Therefore,
 $ \Rightarrow ar(\Delta PQR) = 4 \times ar(\Delta ABC)$

Hence, $\Delta PQR > \Delta ABC$ is the required answer.

Note:
There is an alternate solution by using similarity of triangles,

In $\Delta ABC$ and $\Delta RQP$
 $\angle A = \angle R = 30^\circ $
 $\angle B = \angle Q = 90^\circ $
 $\angle C = \angle P = 60^\circ $

Hence, by AAA similarity of triangles, we get
 \[\Delta ABC\tilde = \Delta RQP\]

We know that ratio of the areas of two similar triangles is equal to the square of the ratio of its corresponding sides
Since \[\Delta ABC\tilde = \Delta RQP\] we get
 \[ \Rightarrow \dfrac{{ar(\Delta ABC)}}{{ar(\Delta RQP)}} = {\left( {\dfrac{{AC}}{{PR}}} \right)^2}\] … (1)

We know that,
 $\cos \theta = \dfrac{{base}}{{hypotenuse}}$

Hence for $\Delta RQP$ , we get
 $ \Rightarrow \cos {60^ \circ } = \dfrac{{PQ}}{{PR}}$

We know that $\cos {60^ \circ } = \dfrac{1}{2}$ and $PQ = 4$ , we get
 $ \Rightarrow \dfrac{1}{2} = \dfrac{4}{{PR}}$

By cross-multiplying, we get
 $ \Rightarrow PR = 8$ … (2)

Also, we are given that
 $ \Rightarrow AC = 4$ … (3)

Using (2) and (3) in (1), we get
 \[ \Rightarrow \dfrac{{ar(\Delta ABC)}}{{ar(\Delta RQP)}} = {\left( {\dfrac{4}{8}} \right)^2}\]

On simplification the brackets we get,
 \[ \Rightarrow \dfrac{{ar(\Delta ABC)}}{{ar(\Delta RQP)}} = {\left( {\dfrac{1}{2}} \right)^2}\]

On squaring the terms we get,
 \[ \Rightarrow \dfrac{{ar(\Delta ABC)}}{{ar(\Delta RQP)}} = \dfrac{1}{4}\]

By cross-multiplying, we get
 \[ \Rightarrow 4 \times ar(\Delta ABC) = 1 \times ar(\Delta RQP)\]

Hence,
 $ \Rightarrow ar(\Delta PQR) = 4 \times ar(\Delta ABC)$

Hence, $\Delta PQR > \Delta ABC$ is the required answer