In trapizium abcd ab and cd are parallel sides.The diagonals ac and bd intesect at x .The ratio of area of triangle axb to that of cxd is 25:49 .find the ratio of area of triangle axd and trapizium abcd?
Dx/xb=cx/ax=5/7…….please help me…
[Math] Find the ratio of area of triangle axd and trapezium abcd
euclidean-geometrygeometry
Related Solutions
Consider the diagram below:
We are given that $\overline{AB} \parallel \overline{CD}$. If two parallel lines are cut by a transversal, then alternate interior angles are congruent. Thus, $\angle ABP \cong \angle CDP$ and $\angle CAB \cong \angle DCP$. Therefore, $\triangle ABP \sim \triangle CDP$.
We are also given that the area of $\triangle ABP$ is $32~\text{cm}^2$, while the area of $\triangle CDP$ is $50~\text{cm}$. Hence, the ratio of the areas of the similar triangles is $$\frac{A(ABP)}{A(CDP)} = \frac{32~\text{cm}^2}{50~\text{cm}^2} = \frac{16}{25}$$ The ratio of the areas of similar triangles is the square of the ratio of corresponding sides. Thus, $$\frac{|AB|}{|CD|} = \frac{|AP|}{|CP|} = \frac{|BP|}{|DP|} = \sqrt{\frac{16}{25}} = \frac{4}{5}$$ Let $E$ be the foot of the altitude from point $P$ to $\overline{AB}$; let $F$ be the foot of the altitude from point $P$ to $\overline{CD}$. Since all right angles are congruent, $\angle BEP \cong \angle DFP$. Since we have also established that $\angle EBP \cong \angle FDP$, $\triangle BEP \sim \triangle DFP$. Thus, $$\frac{|EP|}{|FP|} = \frac{|BP|}{|DP|} = \frac{4}{5}$$ The area of trapezoid $ABCD$ is \begin{align*} A(ABCD) & = \frac{1}{2}(|AB| + |CD|)(|EP| + |FP|)\\ & = \frac{1}{2}\left(|AB| + \frac{5}{4}|AB|\right)\left(|EP| + \frac{5}{4}|EP|\right)\\ & = \frac{1}{2}\left(\frac{9}{4}|AB|\right)\left(\frac{9}{4}|EP|\right)\\ & = \frac{81}{16} \cdot \frac{1}{2} \cdot |AB| \cdot |EP|\\ & = \frac{81}{16} \cdot A(\triangle ABP)\\ & = \frac{81}{16}(32~\text{cm}^2)\\ & = 162~\text{cm}^2 \end{align*}
When you can't figure out how something like this will happen, it often pays to invert the logic. What if we start with the tangent circles, and see what quadrilaterals we can get that way?
In addition to the circles, I've drawn in the centers $P,Q,R,S$. The two pairs of tangent circles give us that $X$ lies on the line segments $PR$ and $QS$. $X$ is unlabeled in the figure due to too much clutter, but it's the point in the middle where all those circles and lines intersect.
Now, we start chasing angles. From isosceles triangle $XSA$ and the circle $S$ is the center of, $\angle AXS=90^\circ-\frac12\angle XSA=90^\circ-\angle XDA$. Similarly, $\angle QXB=90^\circ-\angle BCX$. From the straight line due $QXS$, $\angle BXA=180^\circ-\angle QXB-\angle AXS=\angle XDA+\angle BCX$. Similarly, using triangles $CRX$ and $XPB$, $\angle CXB=\angle CDX+\angle XAB$. Add the two equations: $$\angle CXA=\angle CXB+\angle BXA=\angle XDA+\angle BCX+\angle CDX+\angle XAB=\angle CDA+\angle BCX+\angle XAB$$ Consider the quadrilateral $CXAB$. The sum of its angles is $360^\circ$, so if we add $\angle CXA+\angle ABC$ to both sides of the equation, we get $$2\angle CXA+\angle ABC=\angle CDA+360^\circ$$ $$\angle CXA=180^\circ+\frac{\angle CDA-\angle ABC}{2}$$ That is the key to our construction - we have found an angle through $X$ in terms of angles of the original quadrilateral. The locus of all points that make that particular angle is a circular arc between $A$ and $C$. If we construct that circle, and the corresponding circular arc for $\angle DXB=180^\circ+\frac{\angle DAB-\angle BCD}{2}$, the intersection will be the $X$ we seek.
I won't write out the explicit construction for those arcs here, but it's routine in theory. Note that the "inside" of the arc is toward the larger of the two angles, and it becomes a line segment in the case of equal angles.
Best Answer
Let the areas of $\triangle ABX$ and $CDX$ be $25x$ and $49x$ respectively.
Since $\triangle ABX\sim\triangle CDX$,
$$\displaystyle \frac{AX}{CX}=\frac{BX}{DX}=\sqrt{\frac{25}{49}}=\frac{5}{7}$$
We have
$$\frac{\textrm{area of }\triangle ABX}{\textrm{area of }\triangle ADX}=\frac{5}{7}$$
$$\textrm{area of }\triangle ADX=35x$$
and
$$\frac{\textrm{area of }\triangle ABX}{\textrm{area of }\triangle BCX}=\frac{5}{7}$$
$$\textrm{area of }\triangle BCX=35x$$
The area of the trapezium $ABCD$ is $25x+35x+35x+49x=144x$.
$$\frac{\textrm{area of }\triangle ADX}{\textrm{area of trapezium }ABCD}=\frac{35}{144}$$
In general, we have the following result:
Let $ABCD$ be a convex quadrilateral and $X$ be the point of intersection of its diagonals. Then we have
$$\textrm{area of }\triangle ABX\times\textrm{area of }\triangle CDX=\textrm{area of }\triangle BCX\times\textrm{area of }\triangle DAX$$
Furthermore, if $AB=CD$, then
$$\textrm{area of }\triangle BCX=\textrm{area of }\triangle DAX$$
and therefore,
$$\textrm{area of }\triangle ABX\times\textrm{area of }\triangle CDX=(\textrm{area of }\triangle DAX)^2$$