Name: Construction of Triangle When Base Angles and Perimeter Is Given
Uploaded: 2019-07-19
Duration: 5 min 46 s
Description: A perimeter and the three angles are a must for a triangle to exist. But you don't necessarily need all of these to construct a triangle. Let’s learn how in...

Question

Construct an angle of

45 °

at the initial point of a given ray and justify the construction.

Accepted Answer

We need to construct an angle of $45 °$ at the initial point of a given ray and justify its construction.

Steps of construction:

$1 .$ Draw a ray $O A$ .

$2 .$ Taking $O$ as the centre and any radius, draw an arc cutting $O A$ at $B$ .

$3 .$ Now, taking $B$ as the centre and with the same radius as before,
draw an arc intersecting the previously drawn arc at point $C$ .

$4 .$ With $C$ as the centre and the same radius, draw an arc cutting the arc at $D$ .

$5 .$ With $C$ and $D$ as the centres and radius more than $\frac{1}{2}$ $C D$ , draw two arcs intersecting at $P$ .

$6 .$ Join $O P$ .

Thus, $∠ A O P = 90^{o}$

Now we should draw the bisector of $∠ A O P$ .

$7 .$ Let $O P$ intersect the original arc at point $Q$ .

$8 .$ Now, taking $B$ and $Q$ as the centres, draw two arcs intersecting at $R$ and join $O R$ .

Thus, $∠ A O R = 45^{o}$

Justification:

We need to prove $∠ A O R = 45 °$ .

Join $O C$ & $O B$ .

Thus, $O B = B C = O C$

$∴ ∆ O C B$ is an equilateral triangle.

$∴ ∠ B O C = 60 °$

Join $O D, O C$ and $C D$ .

Thus, $O D = O C = D C$

$∴ ∆ D O C$ is an equilateral triangle

$∴ ∠ D O C = 60 °$

Join $P D$ and $P C$

Now, in $∆ O D P$ and $∆ O C P$ ,

$O D = O C$ (Radius of same arcs)

$D P = C P$ (Arc of same radii)

$O P = O P$ (Common)

$∴ ∆ O D P ≅ ∆ O C P$ ( $SSS$ Congruency)

$∴ ∠ D O P = ∠ C O P$ (CPCT)

So, we can say that,

$∠ D O P = ∠ C O P = \frac{1}{2} ∠ D O C$

$∠ D O P = ∠ C O P = \frac{1}{2} \times 60 ° = 30 °$ ( $∵ ∠ D O C = 60 °$ )

Now,

$∠ A O P = ∠ B O C + ∠ C O P$

$∠ A O P = 60 ° + 30 °$ $= 90 °$

Now, join $QR$ and $BR$ .

In $∆ O Q R$ and $∆ O B R$ ,

$O Q = OB$ (Radius of same arcs)

$QR = B R$ (Arc of same radii)

$OR = O R$ (Common)

$∴ ∆ O Q R ≅ ∆ O B R$ ( $SSS$ Congruency)

$∴ ∠ O Q R = ∠ B O R$ (CPCT)

$∠ QOR = ∠ B O R = \frac{1}{2} ∠ A O P$

$∠ D O P = ∠ C O P = \frac{1}{2} \times 90 °$ $= 45 °$

Thus, $∠ A O R = 45 °$

Hence, justified.

EXERCISE

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