Determine how many times are of PQR in terms of XYZ

[mathlearn]

If that's not clear enough, DIAGRAM this will do it.

Going Ahead I see that,

Line PQ is bisected by the parallel line which originates from X

Parallel line which originates from X is parallel to the line PQ

QY=YZ=ZR

and using the converse of the midpoint theorem

The straight line through the midpoint of one side of a triangle and parallel to another side,bisects the third side.

\(\displaystyle \therefore\) PX = XR (converse of the midpoint theorem)

And We know that the area of \(\displaystyle \triangle \)'s between same pair of parallel lines and lie on the same base are equal in area

As YZ=ZR

\(\displaystyle \triangle\)YZX is equal to the area of \(\displaystyle \triangle\) XZR

I updated the diagram

Can You help me to determine the area of \(\displaystyle \triangle\)PQR in terms of \(\displaystyle \triangle\)XYZ

Many Thanks :)

 

[Evgeny.Makarov]

Use the standard formula for the area: $S_{\triangle PQR}=\frac12PH\cdot QR$ where $H$ lies on $QR$ and $PH$ is the altitude from $P$. Also note that the altitude of $\triangle PQR$ is twice that of $\triangle XYZ$.

 

[mathlearn]

:)

I marked the altitudes and updated the diagram

Use the standard formula for the area: $S_{\triangle PQR}=\frac12PH\cdot QR$ where $H$ lies on $QR$ and $PH$ is the altitude from $P$.

The area of \(\displaystyle \triangle\) PQR can be stated as :

$S_{\triangle PQR}=\frac12PH\cdot QR$

And The area of \(\displaystyle \triangle\) XYZ

$S_{\triangle XYZ}=\frac12XH\cdot YZ$

Also note that the altitude of $\triangle PQR$ is twice that of $\triangle XYZ$.

$S_{\triangle PQR}=\frac122XH\cdot QR$

\(\displaystyle \therefore\)$S_{\triangle PQR}=XH\cdot QR$

Many Thanks :)

 

[Evgeny.Makarov]

The area of \(\displaystyle \triangle\) PQR can be stated as :

$S_{\triangle PQR}=\frac12PH\cdot QR$

And The area of \(\displaystyle \triangle\) XYZ

$S_{\triangle XYZ}=\frac12XH\cdot YZ$

The altitude from $X$ has a different base point on $QR$, i.e., not $H$.

\(\displaystyle \therefore\)$S_{\triangle PQR}=XH\cdot QR$

This is correct, but you need to find $\dfrac{S_{\triangle PQR}}{S_{\triangle XYZ}}$.

 

[mathlearn]

:)

QY=YZ=ZR $\left(given\right)$

$\therefore$ the triangles marked in red have the same area
(areas of triangles between same pair of parallel lines and equal base)

By Taking the area of one such triangle as "a",

And we see that the triangle PQR is divided into two . Therefore the area marked in orange strips should be equal to '3a'.

PX=XR (Converse of midpoint theorem)

$\therefore$ Total area= 6a or 6*XYZ

Correct I Guess ?

Many Thanks :)

 

[Evgeny.Makarov]

And we see that the triangle PQR is divided into two .

Not just that, but into two triangles with equal area since $PX=XR$.

I agree with the rest; it's a nice solution.

I would have solved it as follows. The altitude of $\triangle PQR$ is twice the altitude of $\triangle XYZ$, and the base is three times. Therefore, the ratio of the areas is $2\cdot 3=6$.