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 a circle with radius

4.5 cm

and draw two tangents to the circle from the external point at a distance

7.5 cm

from the centre. Find the length of the tangents and find the relation between them.

Accepted Answer

Steps of Construction :
(a) First draw a circle with centre $O$ and radius $7.5 cm$ .

(b) Take a point $P$ at a distance $7.5 cm$ from the centre $O$ . Join $O$ and $P$ .

(c) Bisect $O P$ at $T$ .

(d) Draw a circle with $T$ as centre and radius $O T$ , which passes through P and intersects the circle with centre $O$ at $R$ and $Q$ .

(e) Join $P$ and $Q, P$ and $R$ .

Then, $P Q$ and $P R$ are the required tangents to the circle with centre $O$ from an external point $P$ .

In $△ O P Q, ∠ O Q P = 90 °$ [The radius in a circle is perpendicular to the tangent].

So, $△ O P Q$ is a right-angled triangle, whose $O Q = 4.5 cm$ and $O P = 7.5 cm$ .

Applying Pythagoras theorem in $△ O P Q$ , we get, $P Q^{2} + O Q^{2} = O P^{2}$

$\Rightarrow P Q^{2} = O P^{2} - O Q^{2}$

$\Rightarrow P Q^{2} = {(7.5)}^{2} - {(4.5)}^{2}$

$\Rightarrow P Q^{2} = 56.25 - 20.25$

$\Rightarrow P Q^{2} = 36$

$\Rightarrow P Q = \sqrt{36} = 6 cm$ .

Similarly, we can prove that $P R = \sqrt{36} = 6 cm$

Therefore, the required length of the tangents are $6 cm$ each. The relation between the tangents is that their lengths are the same.

Construct a circle with radius $4.5 cm$ and draw two tangents to the circle from the external point at a distance $7.5 cm$ from the centre. Find the length of the tangents and find the relation between them.

Important Questions on Construction

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