Similarity in Right Angled Triangles

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 + ${\displaystyle \square }$ = PR²    .....(I)

Here, PR = 10, PQ = ${\displaystyle \square }$

From equation (I)

8² + QR² = 10²

QR² = 10² – 8²

QR² = 100 – 64

QR² = ${\displaystyle \square }$

QR = 6

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

Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq.cm. Complete the following activity.

Activity: As shown in figure LMNT is a reactangle.
∴ Area of rectangle = length × breadth

∴ Area of rectangle = ${\displaystyle \square }$ × breadth

∴ 192 = ${\displaystyle \square }$ × breadth

∴ Breadth = 12 cm

Also,

∠TLM = 90°    ......[Each angle of reactangle is right angle]

In ∆TLM,

By Pythagoras theorem

∴ TM² = TL² + ${\displaystyle \square }$

∴ TM² = 12² + ${\displaystyle \square }$

∴ TM² = 144 + ${\displaystyle \square }$

∴ TM² = 400

∴ TM = 20

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 + ${\displaystyle \square }$

∴ AD² = AC² – CD²    ......(I)

Also, In ∆ABD, by pythagoras theorem,

AB² = ${\displaystyle \square }$ + BD²

∴ AD² = AB² – BD²    ......(II)

∴ ${\displaystyle \square }$ − BD² = AC² − ${\displaystyle \square }$

∴ AB² + CD² = AC²+ BD²

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

In the following figure, AE = EF = AF = BE = CF = a, AT ⊥ BC. Show that AB = AC =  ${\displaystyle \sqrt{3}\times a}$

∆ABC is an equilateral triangle. Point P is on base BC such that PC = ${\displaystyle \frac{1}{3}}$BC, if AB = 6 cm find AP.

A congruent side of an isosceles right angled triangle is 7 cm, Find its perimeter

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