Theorem of Similarity in Right Angled Triangle

Find the length of the side and perimeter of an equilateral triangle whose height is `sqrt3` cm.

A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of wall. Complete the given activity.

Activity: As shown in figure suppose

PR is the length of ladder = 10 m

At P – window, At Q – base of wall, At R – foot of ladder

∴ PQ = 8 m

∴ QR = ?

In ∆PQR, m∠PQR = 90°

By Pythagoras Theorem,

∴ PQ² + `square` = PR²    ...(I)

Here, PR = 10, PQ = `square`

From equation (I)

8² + QR² = 10²

QR² = 10² – 8²

QR² = 100 – 64

QR² = `square`

QR = 6

∴ The distance of foot of the ladder from the base of wall is 6 m.

From given figure, In ∆ABC, AD ⊥ BC, then prove that AB² + CD² = BD² + AC² by completing activity.

Activity: From given figure, In ∆ACD, by pythagoras theorem

AC² = AD² + `square`

∴ AD² = AC² – CD²   ...(I)

Also, In ∆ABD, by pythagoras theorem,

AB² = `square` + BD²

∴ AD² = AB² – BD²   ...(II)

∴ `square` – BD² = AC² – `square`

∴ AB² + CD² = AC² + BD²

From given figure, In ∆PQR, If ∠QPR = 90°, PM ⊥ QR, PM = 10, QM = 8, then for finding the value of QR, complete the following activity.

Activity: In ∆PQR, If ∠QPR = 90°, PM ⊥ QR,   ...(Given)

In ∆PMQ, By Pythagoras Theorem,

∴ PM² + `square` = PQ²   ...(I)

∴ PQ² = 10² + 8² 

∴ PQ² = `square` + 64

∴ PQ² = `square`

∴ PQ = `sqrt(164)`

Here, ∆QPR ~ ∆QMP ~ ∆PMR

∴ ∆QMP ~ ∆PMR

∴ `(PM)/(RM) = (QM)/(PM)`

∴ PM² = RM × QM

∴ 10² = RM × 8

RM = `100/8 = square`

And,

QR = QM + MR

QR = `square` + `25/2 = 41/2`

Prove that, in a right angled triangle, the square of the hypotenuse is
equal to the sum of the squares of remaining two sides.

ΔPQR, is a right angled triangle with ∠Q = 90°, QR = b, and A(ΔPQR) = a. If QN ⊥ PR, then prove that QN = `(2ab)/sqrt(b^4 + 4a^2)`

In right angled triangle PQR,
if ∠ Q = 90°, PR = 5,
QR = 4 then find PQ and hence find tan R.

From the information given in the figure, prove that PM = PN =  \[\sqrt{3}\]  × a