In - PQR - - Q -90- seg QM is the median- PQ 2- QR 2-169- Draw a circumcircle of - PQR-

Since nbsp;∆PQR nbsp;is nbsp;a nbsp;right nbsp;angled nbsp;triangle.So, nbsp;PR2=PQ2+QR2 nbsp; nbsp; nbsp; nbsp;(Phythagoras nbsp;theore ...

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Standard X

Mathematics

Centroid of a Triangle

In Δ PQR , ∠ ...

Question

In $∆$ PQR, $∠$ Q = 90 $°$ , seg QM is the median. PQ² + QR² = 169. Draw a circumcircle of $∆$ PQR.

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Solution

$Since ∆ PQR is a right angled triangle . So, {PR}^{2} = {PQ}^{2} + {QR}^{2} (Phythagoras theorem) {PR}^{2} = 169 (Given) {PR}^{2} = 13^{2} PR = 13 units$

As we know that if a right triangle is inscribed in a circle, then the hypotenuse will be a diameter of the circle.
So, PR is the diameter of the circumcircle.
Radius = $\frac{13}{2} = 6.5 units$

Steps of construction:

1. Draw a circle with centre M and a radius of 6.5 units.
2. Draw the diameter PR of the circle.
3. Take a point Q on the circle.
4. Join PQ and RQ.

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Similar questions

Q. In

Δ P Q R

, if

P R^{2} = P Q^{2} + Q R^{2}

, prove that

∠ Q

is right angle.

Q. In a

Δ P Q R,

P R^{2} - P Q^{2} = Q R^{2}

and M is a point on side PR such that

Q M ⊥ P R

. Prove that

Q M^{2} = P M \times M R

Angles Q and R of a ΔPQR are 25° and 65°.

Write which of the following is true:

(i) PQ² + QR²= RP²

(ii) PQ² + RP²= QR²

(iii) RP² + QR²= PQ²

Question 6
Angles Q and R of a $Δ$ PQR are $25^{\circ}$ and $65^{\circ}$ .
Write which of the following is true:
(i) $P Q^{2} + Q R^{2} = R P^{2}$
(ii) $P Q^{2} + R P^{2} = Q R^{2}$
(iii) $R P^{2} + Q R^{2} = P Q^{2}$

Q. Question 1
In a

Δ P Q R,

P R^{2} - P Q^{2} = Q R^{2}

and M is a point on side PR such that

Q M ⊥ P R

. Prove that

Q M^{2} = P M \times M R

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In ∆PQR, ∠Q = 90°, seg QM is the median. PQ2 + QR2 = 169. Draw a circumcircle of ∆PQR.

In $∆$ PQR, $∠$ Q = 90 $°$ , seg QM is the median. PQ² + QR² = 169. Draw a circumcircle of $∆$ PQR.