Maths M.C.Q (1 Marks)

Het a man be at $O$ and goes to $24m$ due west and then $7m$ due north.
Distance of man from starting point be $OB$
So,

In right $\triangle\text{ABO},$
$ \mathrm{OB}^2=A B^2+A O^2 $
$ \Rightarrow O B^2=(7)^2+(24)^2 $
$ \Rightarrow O B^2=49+576 $
$ \Rightarrow O B^2=625 $
$ \Rightarrow O B=25 \mathrm{~m}$
Thus, the distance of man from starting point is $25m$.

Het $AB$ and $CD$ be two poles and distance blw their root is $12m$.

$ED = CD - CE = 11 - 6 = 5cm$
In right $\triangle\text{ADE},$
$ A D^2=A E^2+D E^2 $
$ \Rightarrow A D^2=(12)^2+(5)^2 $
$ \Rightarrow A D^2=144+25 $
$ \Rightarrow A D^2=169 $
$ \Rightarrow A D=13 \mathrm{~m}$
Thus distance blw their tops is $13m$.

$\triangle\text{ABC}$ is an equilateral triangle and $\text{AD}\perp\text{BC}.$

In $\triangle\text{ABD},$ applying Pythagoras theorem, we get,
$\text{AB}^2=\text{AD}^2+\text{BD}^2$
$\text{AB}^2=\text{AD}^2+\Big(\frac{1}{2}\text{BC}\Big)^2\Big(\because\text{BD}=\frac{1}{2}\text{BC}\Big)$
$\text{AB}^2=\text{AD}^2+\Big(\frac{1}{2}\text{AB}\Big)^2\Big(\because\text{AB}=\text{BC}\Big)$
$\text{AB}^2=\text{AD}^2+\frac{1}{4}\text{AB}^2$
$3\text{AB}^2=4\text{AD}^2$
We got the result as $B$.

Given: The area of two similar triangles is $121cm^2$ and $64cm^2$ respectively. The median of the first triangle is $12.1cm$.
To find: Corresponding medians of the other triangle.
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their medians.
$\frac{\text{ar}(\text{triangle 1})}{\text{ar}(\text{triangle 2})}=\Big(\frac{\text{median 1}}{\text{median 2}}\Big)^2$
$\frac{121}{64}=\Big(\frac{12.1}{\text{median 2}}\Big)^2$
Taking square root on both side, we get,
$\frac{11}{8}=\frac{12.1\text{cm}}{\text{median 2}}$
$\Rightarrow$ median$2$ $= 8.8cm$
Hence the correct answer is $B$.

In right $\triangle\text{OAB},$
$ A B^2=O A^2+O B^2(\text { Pythagoras Theorem }) $
$ \Rightarrow A B^2=(10)^2+(10)^2(O A=O B=10 \mathrm{~cm})$
$ \Rightarrow A B^2=100+100=200$
$\Rightarrow\text{AB}=\sqrt{200}=10\sqrt{2}\text{cm}$
Thus, the length of the chord is $10\sqrt{2}\text{cm}.$
Hence, the correct answer is option $B$.

$\triangle\text{ABC}\sim\triangle\text{DEF}$
$\text{BC}=3\text{cm},\ \text{EF}=4\text{cm}$
$\text{ar}(\triangle\text{ABC})=54\text{cm}^2$
$\because\triangle\text{ABC}\sim\triangle\text{DEF}$
$\therefore\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{BC}^2}{\text{EF}^2}$
$\Rightarrow\frac{54}{\text{ar}(\triangle\text{DEF})}=\frac{3^2}{4^2}=\frac{9}{16}$
$\therefore\text{ar}(\triangle\text{DEF})=\frac{16\times54}{9}=96\text{cm}^2$

In triangle $ABC, E$ is a point on $AC$ such that $\text{BE}\perp\text{AC}.$
We need to find $AB^2+ BC^2+ AC^2$
Since $\text{BE}\perp\text{AC},$ $\text{CE} = \text{AE} = \frac{\text{AC}}{2}$ (In a equilateral triangle, the perpendicular from the vertex bisects the base.)
In triangle ABE, we have,
$ A B^2=B E^2+A E^2 $
$ \text { Since } A B=B C=A C $
$ \text { Therefore, } A B^2=B C^2=A C^2=B E^2+A E^2 $
$ \Rightarrow A B^2+B C^2+A C^2=3 B E^2+3 A E^2$
Since in triangle BE is an altitude, so $\text{BE}=\frac{\sqrt{3}}{2}\text{AB}$
$\text{BE}=\frac{\sqrt{3}}{2}\text{AB}$
$=\frac{\sqrt{3}}{2}\times\text{AC}$
$=\frac{\sqrt{3}}{2}\times2\text{AE}=\sqrt{3}\text{AE}$
$\Rightarrow\text{AB}^2+\text{BC}^2+\text{AC}^2=3\text{BE}^2+3\Big(\frac{\text{BE}}{\sqrt{3}}\Big)^2$
$=3\text{BE}^2+\text{BE}^2=4\text{BE}^2$
Hence option $C$ is correct.

In isosceles $\triangle\text{ABC},\ \text{AC}=\text{BC}$

$ \text { and } \mathrm{AB}^2=\mathrm{AC}^2+\mathrm{AC}^2=2 \mathrm{AC}^2$
$ =\mathrm{AC}^2+\mathrm{BC}^2(\mathrm{AC}=\mathrm{BC})$
By converse of Pythagoras Theorem,
$\angle\text{C}=90^\circ$

$M$ is the mid-point of $AB$.
$\therefore\text{BM}=\frac{\text{AB}}{2}$
$N$ is the mid-point of $BC$.
$\therefore\text{BN}=\frac{\text{BC}}{2}$
Now,
$\text{AN}^2+\text{CM}^2=\bigg(\text{AB}^2+\Big(\frac{1}{2}\text{BC}\Big)^2\bigg)+\bigg(\Big(\frac{1}{2}\text{AB}\Big)^2+\text{BC}^2\bigg)$
$=\text{AB}^2+\frac{1}{4}\text{BC}^2+\frac{1}{4}\text{AB}^2+\text{BC}^2$
$=\frac{5}{4}\Big(\text{AB}^2+\text{BC}^2\Big)$
$\Rightarrow4\Big(\text{AN}^2+\text{CM}^2\Big)=5\text{AC}^2$
Hence option $B$ is correct.

$\triangle\text{ABC}$ is an isosceles with $\angle\text{C}=90^\circ$

$AC = BC$
$AC = 6cm$
$A B^2=A C^2+B C^2 \text { (Pythagoras Theorem) }$
$(6)^2+(6)^2=36+36=72(A C=B C)$
$\text{AB}=\sqrt{72}=\sqrt{(36\times2}=6\sqrt{2}\text{cm}.$