5 Marks Questions

Question 15 Marks

The perimeter of a certain sector of a circle of radius $5.6 m$ is $20.0 m$ . Find the area of the sector.

Answer

$2 r+\frac{2 \pi r \theta}{360}=20$
$\Rightarrow 11.2+2 \times \frac{22}{7} \times 5.6 \times \frac{\theta}{360}=20$
Solving, we get $\theta=90^{\circ}$
$\therefore \text { Area of sector }=\frac{22}{7} \times 5.6 \times 5.6 \times \frac{90}{360}$
$\quad=24.64 m^2$

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Question 25 Marks

From the top of a $45 m$ high light house, the angles of depression of two ships, on the opposite side of it, are observed to be $30^{\circ}$ and $60^{\circ}$. If the line joining the ships passes through the foot of the light house, find the distance between the ships. $($Use $\sqrt{3}=1.73 )$

Answer

$\tan 60^{\circ}=\sqrt{3}=\frac{45}{y}$
$\Rightarrow y=15 \sqrt{3}$
$\tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{45}{x}$
$\Rightarrow x=45 \sqrt{3}$
Distance between two ships $= x+y=60 \sqrt{3}$
$ =60 \times 1.73=103.8 m$

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Question 35 Marks

Sides $AB$ and $AC$ and median $AD$ to $\triangle ABC$ 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 $N$ such that $PM = MN$ and join $NR$
$\triangle ADB \cong \triangle EDC$
$\therefore AB=EC$
Similarly, $P Q=N R$
Since, $\frac{A B}{P Q}=\frac{A C}{P R}=\frac{A D}{P M}$
$\Rightarrow \frac{EC}{NR}=\frac{AC}{PR}=\frac{\frac{AE}{\frac{2}{PN}}}{\frac{2}{2}}$
$\therefore \triangle AEC \sim \triangle PNR$
$\Rightarrow \angle 1=\angle 2$
Similarly, $\angle 3=\angle 4$
Hence $\angle 1+\angle 3=\angle 2+\angle 4$ or $\angle A=\angle P$
Also, $\frac{ AB }{ PQ }=\frac{ AC }{ PR }$
$\therefore \triangle ABC \sim \triangle PQR$

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Question 45 Marks

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.

Answer

Correct Given, to prove, figure, construction Correct proof

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Question 55 Marks

Tara scored $40$ marks in a test, getting $3$ marks for each right answer and losing $1$ mark for each wrong answer. Had $4$ marks been awarded for each correct answer and $2$ marks been deducted for each wrong answer, then Tara would have scored $50$ marks. Assuming that Tara attempted all questions, find the total number of questions in the test.

Answer

Let number of correct answers be x and number of incorrect answers be $y$
$3 x-y=40$
$4 x-2 y=50$
Solving, we get $x=15, y=5$
$\therefore$ Total number of questions $=20$

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Question 65 Marks

Using graphical method, solve the following system of equations :
$
3 x+y+4=0 \text { and } 3 x-y+2=0
$

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Maths 5 Marks Questions

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