In a ∆ PQR, PR² - PQ² = QR² and M is a point on side PR such that QM ⊥ PR. Prove that QM² = PM × MR
Solution:
Given, in triangle PQR, PR² - PQ² = QR²
M is a point on the side PR such that QM ⊥ PR
We have to prove that QM² = PM × MR.
Now, PR² - PQ² = QR² can be written as PR² = PQ² + QR²
Which satisfies Pythagoras theorem of a right angle.
So, the triangle PQR is a right triangle with Q at the right angle.
We know that a perpendicular drawn from the vertex of the right angle of a right triangle to its hypotenuse divides the triangle into two triangles which are similar to the whole triangle and to each other.
So, the perpendicular QM divides the triangle PQR into two similar triangles PMQ and RMQ with M at right angles.
From the triangles PMQ and RMQ,
∠PMQ = ∠RMQ = 90°
∠MQR = ∠QPM = 90° - ∠R
AAA criterion states that if two angles of a triangle are respectively equal to two angles of another triangle, then by the angle sum property of a triangle their third angle will also be equal.
Therefore, the triangles PMQ and RMQ are similar.
By the property of area of similar triangles,
Area of the triangle RMQ/Area of the triangle PMQ = QM²/PM²
We know that the area of the triangle = 1/2 × base × height
Area of the triangle PMQ = 1/2 × PM × QM
Area of the triangle RMQ = 1/2 × RM × QM
So, (1/2 × RM × QM)/(1/2 × PM × QM) = QM²/PM²
Cancelling out common terms,
RM/PM = QM²/PM²
On cross multiplication,
RM × PM² = QM² × PM
RM × PM = QM²
Therefore, it is proved that QM² = RM × PM
✦ Try This: PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM.MR
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 6
NCERT Exemplar Class 10 Maths Exercise 6.3 Problem 1
In a ∆ PQR, PR² - PQ² = QR² and M is a point on side PR such that QM ⊥ PR. Prove that QM² = PM × MR
Summary:
In a ∆ PQR, PR² - PQ² = QR² and M is a point on side PR such that QM ⊥ PR. It is proved that QM² = PM × MR
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