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PQRS is a trapezium , in which PQ is parallel to RS , and PQ = 3(RS).

The diagonal of the trapezium intersect each other at X,

Then the ratio of the area of the Triangle PXQ and RXS. I tried but could not get it.

Does the area will be in the ratio of the base i.e. 1:3

Thanks in advance.

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3 Answers

up vote 1 down vote accepted

Since triangles are similar , as Andre correctly observed , and similar ratio is : $1 : 3$ we can write :

$|PQ| = 3k ~\text { and }~ h_1=3k$

$|RS| = k ~\text { and }~ h_2=k$

where $h_1 ~\text { and } ~h_2$ are corresponding heights of triangles .

Now , since :

$A_1=\frac{|PQ| \cdot h_1}{2} ~\text { and } A_2=\frac{|RS| \cdot h_2}{2}~$ it follows that :

$A_1 : A_2 = 9 : 1$

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The triangles $RXS$ and $PXQ$ are similar, with similarity ratio $1:3$. (The fact that the triangles are similar follows from a little angle chasing. Use properties of transversals of parallel lines.) So the big triangle has area $3^2$ (that is, $9$) times the area of the small triangle.

In general, if we have two similar figures, and the linear dimensions of the second are $\lambda$ times the linear dimensions of the first, then the area of the second is $\lambda^2$ times the area of the first. If you take a rectangle and scale it up by a factor of $3$, then the area gets multiplied by $9$. The same is true for the area of any region. This is why if, for example, we know that a circle of radius $1$ has area $\pi$, then it is automatic that a circle of radius $r$ must have area $\pi r^2$.

Remark: Note that if we are in three dimensions, and two figures are similar, with linear scaling factor $\lambda$, then the scaling factor for volume is $\lambda^3$. If your body gets scaled up so that all your linear dimensions are multplied by a factor of $3$, then your weight will get multiplied by $27$.

Something that you might find interesting is that (whether or not the trapezoid is symmetrical) the two "side" triangles that you have not mentioned always have equal area, and each has area $3$ times the area of $\triangle RXS$.

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First draw perpendicular line to PQ and RS through X. It intersects PQ and RS on A and B respectively. That is, hight of trapezium is AB.

We have to find the ratio of the area of the Triangle PXQ and RXS.


Here Triangles PXQ and RXS are similar
hence PQ:RS = QX:XR -----(1)

and also AXQ and BXR are similar
hence QX:XR = AX:XB ------(2)

From (1) and (2),
PQ:RS =AX:XB

the ratio of the area of the Triangle PXQ and RXS = PQ.AX : RS.BX
                                                  = (PQ:RS).(AX:BX)
                                                  = (3:1).(3:1)
                                                  = 9:1
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