Tag: areas of similar triangles

A vertical pole of $5.6m$ height casts a shadow $3.2m$ long. At the same time find the height of a pole which casts a shadow $5m$ long.

If ratio of heights of two similar triangles is $4:9$, then ratio between their areas is?

The area of two similar triangles ABC and PQR are $25\ cm^{2}\ & \  49\ cm^{2}$, respectively. If QR $=9.8$ cm, then BC is:

$\Delta ABC\sim\Delta PQR.$ If area$\left (ABC \right)= 2.25 m^{2}$, area$ \left (PQR \right)= 6.25 m^{2}$, $ PQ = 0.5 m $, then length of AB is:

In $ \triangle ABC\sim \triangle DEF$,  BC $ = $ 4 cm, EF $ =$ 5 cm and area($\triangle $ABC)$ = $ 80 $cm^2$, the area($\triangle$ DEF) is:

In $XYZ$ and $\triangle PQR,XYZ\leftrightarrow PQR$ is similarity, $XY=8,ZX=16,PR=8$. So $PQ+QR$=______.

Given $\Delta ABC-\Delta PQR$. If $\dfrac{AB}{PQ}=\dfrac{1}{3}$, then find $\dfrac{ar\Delta ABC}{ar\Delta PQR'}$.

A point taken on each median of a triangle divides the median in the ratio 1:3 reckoning from the vertex . then the ratio of the area of the triangle with vertices at these points  to that of the original triangle is :  

$\Delta DEF -\Delta ABC$; If DE $:$ AB $=2:3$ and ar($\Delta$DEF) is equal to $44$ square units, then find ar($\Delta$ABC) in square units.

Given, $\Delta$ABC$-\Delta$PQR. If $\dfrac{ar(\Delta ABC)}{ar(\Delta PQR)}=\dfrac{9}{4}$ and $AB=18$cm, then find the length of PQ.