3 Mark Question

Question 13 Marks

In the given figure, AOB is a straight line and OC is ray such that $\angle\text{AOC}=(3\text{x}+20)^\circ$ and $\angle\text{BOC}=(2\text{x}-10)^\circ.$ Find the value of x and hence find (i) $\angle\text{AOC}$ and $\angle\text{BOC.}$

Answer

Here, $3\text{x}+20+2\text{x}-10=180$
$\Rightarrow5\text{x}+10=180$
$\Rightarrow5\text{x}=170$
$\Rightarrow\text{x}=34$
$\angle\text{AOC}=(3\times34+20)^\circ$
$=(102+20)^\circ$
$=122^\circ$
$\angle\text{BOC}=(2\times34-10)^\circ$
$=(68-10)^\circ$
$=58^\circ$

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Question 23 Marks

Two legs of a right triangle are 8cm and 15cm long. Find the length of the hypotenuse of the triangle.

Answer

Since it is a right triangle, by using the Pythagoras theoram:
Length of the hypotenuse $=\sqrt{8^2+15^2}$
$=\sqrt{64+225}$
$=\sqrt{289}$
$=\pm17\text{cm}$
The length of the side can not be negative.
$\therefore$ Length of the hypotenuse = 17cm

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Question 33 Marks

In the given figure, x : y = 2 : 3 and $\angle\text{ACD}=120^\circ.$ Find the values of x,y and z.

Answer

Let x = 2k and y = 3k
$\therefore$ 2k + 3k = 120 (exterior angle property)
⇒ 5k = 120°
⇒ k = 24°
$\therefore$ x = 2 × 24° = 48° and y = 3 × 24° = 72°
$\text{In }\triangle\text{ABC}:$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow48^\circ+72^\circ+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{C}=180^\circ-120^\circ$
$\Rightarrow\angle\text{C}=60^\circ$
$\therefore\text{z}=60^\circ$

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Question 43 Marks

In the given figure, $\text{AB}||\text{CD},\angle\text{ABO}=60^\circ$ and $\angle\text{CDO}=40^\circ.$ Then, find $\angle\text{BOD}.$

Answer

$\angle\text{ABO}=60^\circ$
$\angle\text{CDO}=40^\circ$
$\angle\text{ABO}=\angle\text{BOC}=60^\circ$ [alternate angles]
$\angle\text{CDO}=\angle\text{DOC}=40^\circ$ [alternate angles]
$\angle\text{BOD}=\angle\text{BOC}+\angle\text{DOC}$
$=60^\circ+40^\circ=100^\circ$

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Maths 3 Mark Question

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