5 Marks Questions

Question 15 Marks

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 5.8 cm and its base is of radius 2.1 cm , find the total surface area of the article.

Answer

$
\begin{aligned}
\begin{aligned}
\text { CSA of cylinder } & =2 \times \frac{22}{7} \times 2.1 \times 5.8 \\
& =76.56 cm^2
\end{aligned} \\
\begin{aligned}
\text { CSA of two hemisphere } & =4 \times \frac{22}{7} \times 2.1 \times 2.1 \\
& =55.44 cm^2
\end{aligned}
\end{aligned}
$
Total Surface Area of article $=76.56+55.44=132 cm^2$

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Question 25 Marks

Sides $A B$ and $A C$ and median $A D$ of a $\triangle A B C$ are respectively proportional to sides PQ and PR and median PM of another $\triangle PQR$. Show that $\triangle ABC \sim \triangle PQR$.

Answer

Produce AD to E such that $AD = DE$ and join EC .
Produce PM to L such that $PM = ML$ and join LR .
$
\begin{array}{c}
\therefore \triangle ABD \cong \triangle ECD \\
\therefore AB=EC \\
\text { Similarly }, PQ=LR \\
\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM} \\
\frac{EC}{LR}=\frac{AC}{PR}=\frac{2 AD}{2 PM}=\frac{AE}{PL} \\
\therefore \triangle AEC \sim \triangle PLR
\end{array}
$
$
\Rightarrow \angle 2=\angle 4
$
Similarly, $\angle 1=\angle 3$
Adding both, $\angle BAC =\angle QPR$
$
\therefore \triangle ABC \sim \triangle PQR
$

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Question 35 Marks

In the given figure, $\triangle FEC \cong \triangle GDB$ and $\angle 1=\angle 2$. Prove that $\triangle ADE \sim \triangle ABC$.

Answer

$\Delta FEC \cong \Delta GDB$
Therefore, $\angle 3=\angle 4$
In $\triangle ABC$,
$\angle 3 =\angle 4$
$\therefore AB =AC.......(i)$
In $\triangle ADE , \angle 1=\angle 2$
$AD=AE......(ii)$
Dividing $(ii)$ by $(i)$
$\frac{AD}{AB}=\frac{AE}{AC}$
$\Rightarrow DE \| BC$
$\angle 1=\angle 3 \text { and } \angle 2=\angle 4$
$\therefore \triangle ADE \sim \triangle ABC$

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Question 45 Marks

From a point on a bridge across the river, the angles of depressions of the banks on opposite sides of the river are $30^{\circ}$ and $60^{\circ}$ respectively. If the bridge is at a height of $4 m$ from the banks, find the width of the river.

Answer

Let $A B$ be the width of river
In right $\triangle PAQ$,
$\tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{4}{x}$
$\Rightarrow 4 \sqrt{3}=x$
In right $\triangle PBQ$,
$\tan 60^{\circ}=\sqrt{3}=\frac{4}{y}$
$\Rightarrow y=\frac{4}{\sqrt{3}}$
Width of river $=x+y=4 \sqrt{3}+\frac{4}{\sqrt{3}}=\frac{16}{3} \sqrt{3} m$

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Question 55 Marks

The age of a man is twice the square of the age of his son. Eight years hence, the age of the man will be $4$ years more than three times the age of his son. Find their present ages.

Answer

Let present age of son $= x$ years
and present age of man $=2 x^2$ years
$\text{A.T.Q.}$
$3(x+8)+4=2 x^2+8$
$\Rightarrow 2 x^2-3 x-20=0$
$\Rightarrow(2 x+5)(x-4)=0$
$x \neq-\frac{5}{2} \text { So, } x=4$
Present age of son $=4$ years
Present age of man $=32$ years

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Question 65 Marks

Find the value of $'k\ ' $for which the quadratic equation $( k +1) x ^2-6( k +1) x +3( k +9)=0, k \neq-1$ has real and equal roots.

Answer

For real and equal roots, $ D=b^2-4 a c=0$
$36(k+1)^2-4(k+1) \times 3(k+9)=0$
$\Rightarrow k^2-2 k-3=0$
$\Rightarrow(k-3)(k+1)=0$
$ k \neq-1 \text { So, } k=3$

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Maths 5 Marks Questions

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