5 Marks Question

Question 15 Marks

Find the length of each side of a rhombus whose diagonals are 24cm and 10cm long.

Answer

In an rhombus, the diagonals are perpendicular bisectors of each other, and side are equal to eachother.
So, $\text{AO}=\frac{1}{2}\text{AC}=12\text{cm}$
$\text{OD}=\frac{1}{2}\text{BD}=5\text{cm}$
In right-angled $\triangle\text{AOD},$
$\text{AD}^2=\text{AO}^2+\text{OD}^2$
$\Rightarrow\text{AD}^2=\text{12}^2+\text{5}^2$
$\Rightarrow\text{AD}^2=\text{144}+\text{25}$
$\Rightarrow\text{AD}^2=\text{169}$
$\Rightarrow\text{AD}=\text{13}\text{cm}$
So, the length of the each side of the rhombus is 13cm.

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Question 25 Marks

State the SSS-criterion for similarity of trianglrs.

Answer

If the corresponding sides of two triangles are proportional then their corresponding angles are equal, and hence the two triangles are similar.

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Question 35 Marks

If $\triangle\text{ABC}\sim\triangle\text{DEF}$ such that 2AB = DE and BC = 6cm, find EF.

Answer

Given $\triangle\text{ABC}\sim\triangle\text{DEF}$ and 2AB = DE and BC = 6cm
$\Rightarrow\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}$
$\Rightarrow\frac{1}{2}=\frac{6}{\text{EF}}$
$\Rightarrow\text{EF}=3\text{cm}$

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Question 45 Marks

For the following statments state whether true (T) or false(F):
In a $\triangle\text{ABC},\text{AB}=6\text{cm},\angle\text{A}=45^\circ$ and $\text{AC}=8\text{cm}$ and in a $\triangle\text{DEF},\text{DF}=9\text{cm},\angle\text{D}=45^\circ$and $\text{DE}=12\text{cm},$ then $\triangle\text{ABC}\sim\triangle\text{DEF}.$

Answer

False.
Solution:
Given that,
$\angle\text{A}=45^\circ,\text{AB}=6\text{cm}$ and $\text{AC}=8\text{cm}$
$\angle\text{D}=45^\circ,\text{DF}=9\text{cm}$ and $\text{DE}=12\text{cm}$
Consider, $\triangle\text{ABC}$ and $\triangle\text{DFE},$
$\angle\text{A}=\angle\text{D}=45^\circ$
$\frac{\text{AB}}{\text{DE}}=\frac{6}{12}=\frac{1}{2}$
$\frac{\text{AC}}{\text{DF}}=\frac{8}{9}$
$\Rightarrow\frac{\text{AB}}{\text{DF}}\not=\frac{\text{AC}}{\text{DF}}$
Thus, the triangles are not similar.

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Question 55 Marks

In the given figure, D is the midpoint of side BC and $\text{AE}\perp\text{BC}.$ If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that.

$\text{b}^2=\text{p}^2+\text{ax}+\frac{\text{a}^2}{4}$

Answer

Given: D is the midpoint of side $\text{BC},\text{AE}\perp\text{BC},\text{BC}=\text{a},\text{AC}=\text{b},\text{AB}=\text{c},\text{ED}=\text{x},\text{AD}=\text{p}$ and $\text{AE}=\text{h}$
In $\triangle\text{AEC},\angle\text{AEC}=90^\circ$
$\text{AD}^2=\text{AE}^2+\text{ED}^2$ (by pythagoras theorem)
$\Rightarrow\text{p}^2=\text{h}^2+\text{x}^2$
In $\triangle\text{AEC},\angle\text{AEC}=90^\circ$
$\text{AC}^2=\text{AE}^2+\text{EC}^2$ (by pythagoras theorem)
$\text{b}^2=\text{h}^2+\Big(\text{x}+\frac{\text{a}}{\text{2}}\Big)^2=\big(\text{h}^2+\text{x}^2\big)+\text{ax}+\frac{\text{a}^2}{4}$
$\therefore\text{b}^2=\text{p}^2+\text{ax}+\frac{\text{a}^2}{4}\dots(1)$

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Question 65 Marks

In a trapezium ABCD, it is given that AB || CD and AB = 2CD. Its diagonals AC and BD intersect at the point O such that $\text{ar}(\triangle\text{AOB})=84\text{cm}^2.$ Find $\text{ar}(\triangle\text{COD}).$

Answer

The diagonals of a trapezium divide each other proportionally.
$\angle\text{CDO}=\angle\text{OBA}$ ...(alternate angles)
$\angle\text{COD}=\angle\text{AOB}$ ...(vertically opposite angles)
$\Rightarrow\triangle\text{COD}=\triangle\text{AOB}$ ...(AA criterion for similarity)
$\Rightarrow\frac{\text{ar}(\triangle\text{COD})}{\text{ar}(\triangle\text{AOB})}=\frac{\text{CD}^2}{\text{AB}^2}$
$\Rightarrow\frac{\text{ar}(\triangle\text{COD})}{84}=\frac{1^2}{2^2}$
$\Rightarrow\text{ar}(\triangle\text{COD})=21\text{cm}^2$

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