In triangle $\Delta ABC$, $AB=2\text{ cm}$, $BC=3\text{ cm}$, $CA=4\text{ cm}$. $D$ is the midpoint of $AC$. If a square is constructed on the side $BD$,what is the area of the square.
Now i used similarity between $\Delta ABD$ and $\Delta BCD$ but to no result..
[Math] In triangle ABC,AB=2cm BC=3cm CA=4cm.D is the midpoint of AC.if a square is constructed on the side BD,what is the area of the square
euclidean-geometrygeometryplane-geometry
Related Solutions
I think the following image will say more than any text. You can divide the image into smaller squares that will allow you immediately to calculate the ratio.
The ratio of the red area within the whole square is the same as the red area in the big green square to the area of the whole green square. And this is (counting in units of the 9 small squares): $(1+1/4+1/4) : 9 = 1.5 : 9 = 1:6$.
Since $LO$ is both median and the bisector of $\angle NLM$, $\triangle LMN$ is isosceles and $ LO \perp MN$.
So $MO = NO= \sqrt{5^2-3^2\ }=4$ and area $\triangle LMN = 4\times 3 =12 \text{cm}^2$
To see that $\triangle LMN$ is isosceles, consider that $\angle LON = 180°-\angle LOM$. Flip $\triangle LON$ to superimpose $\angle NLO$ and $\angle MLO$. Then the only way to make $|MO|=|NO|$ is to have $\angle LON = \angle LOM =\perp$
Best Answer
All you need is to find the length of the median $BD$. You can derive the following median formula using Cosine Law.
$$ \begin{align} &M_b=\frac{1}{2}\sqrt{2c^2+2a^2-b^2} \\\Rightarrow\ &BD=\frac{\sqrt{10}}{2} \\\Rightarrow\ &{\square}_{Area}=\frac{5}{2} \end{align} $$