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Properties: 1 Mark
Sum: 2 Marks
The given conditions make a right-angled triangle because one of its angles is${90}^{∘}$

Using the angle sum property of a triangle,
$\mathrm{âˆ}O+\mathrm{âˆ}A+\mathrm{âˆ}B={180}^{∘}$
$\mathrm{âˆ}O+{90}^{∘}+\mathrm{âˆ}B={180}^{∘}$
$\mathrm{âˆ}O+\mathrm{âˆ}B={90}^{∘}$
Sum of the other two angles will be ${90}^{∘}$

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Each part: 1 Mark
(a) We only need to know R (Right angle), H (Hypotenuse) and S (Side)to construct a right-angled triangle. The right angle is not included between the hypotenuse and the other side.
(b)If the hypotenuse and any one leg of one right-angled triangle are equal to the corresponding hypotenuse and leg of another right-angled triangle, the two triangles are congruent.
It means that if we have two right-angled triangles with thesame length of the hypotenuseand thesame length for one of the other two legs, then both right triangles are identical.

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Angle: 0.5 Mark
Type: 0.5 Mark
The angle between the two handsis 90 degrees.
It is an isosceles right-angled triangle.

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The minimum number of triangles that can be drawn if two sides and one angle has been given is $0$ since the angle might not be an included one.

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As stated in question line segment $BD$ is angle bisector so, $\angle ABD=\angle DBC$
And, $\angle ABD+\angle DBC=\angle ABC$
So,$\angle \left(ABC\right)=2\angle \left(ABD\right)$

Answer: Option A. -> 5cm
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A
The triangle will be a right angled triangle as $CB$ is perpendicular to $AB$ .So, the third side can be easily calculated by using Pythagoras theorem.

Let the length of the third side be $zcm$.
$z^2=3^2+4^2$
$z=\sqrt{25}$
$z=5\text{cm}$

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As stated in question line segment $BD$ is angle bisector so, $ABD=DBC$.
So, $\mathrm{âˆ}ABC=4\mathrm{âˆ}ABE$

Answer: Option A. -> ∠1 = ∠8
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A
When two lines intersect each other, four angles are formed. The pair of angles which lie on the opposite sides of the point of intersection are called vertically opposite angle.If two parallel lines are cutby a transversal:

(i) Alternate interior angles are equal i.e., $\mathrm{âˆ}$2 =$\mathrm{âˆ}$8 and $\mathrm{âˆ}$3 = $\mathrm{âˆ}$5.
(ii) Alternate exterior angles are equal i.e., $\mathrm{âˆ}$1 = $\mathrm{âˆ}$7 and $\mathrm{âˆ}$4 = $\mathrm{âˆ}$6.
(iii) Corresponding angles are equal i.e., $\mathrm{âˆ}$1 = $\mathrm{âˆ}$5, $\mathrm{âˆ}$4 = $\mathrm{âˆ}$8, $\mathrm{âˆ}$2 = $\mathrm{âˆ}$6 and $\mathrm{âˆ}$3 = $\mathrm{âˆ}$7.
(iv) Co-interior angles are supplementary i.e., $\mathrm{âˆ}$2 + $\mathrm{âˆ}$5 = 180${}^{∘}$ and $\mathrm{âˆ}$3 + $\mathrm{âˆ}$8 =180${}^{∘}$
$\mathrm{âˆ}$1 = $\mathrm{âˆ}$8 is incorrect.

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Construction: 3Marks
Draw a line l.
Choose any point on l and draw a perpendicular at that point.
On the perpendicular cut a point X which is 5.5cm away from l.
Then through X construct a line parallel to l.

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Construction: 2 Marks
Steps: 1 Mark
1) Draw $\stackrel{Â}{PQ}=6.5cm$
2) With P as centre, draw an arc with radius equal to 7 cm.
3) With Q as centre, draw another arc with radius 8 cm to cut the previous arc at R.
4) Join R to P and Q. PQR is the required triangle.