Similarity in Right Angled Triangles

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)`

Choose the correct alternative: 

ΔABC and ΔDEF are equilateral triangles. If ar(ΔABC): ar(ΔDEF) = 1 : 2 and AB = 4, then what is the length of DE?

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 = `square` × breadth

∴ 192 = `square` × breadth

∴ Breadth = 12 cm

Also,

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

In ∆TLM,

By Pythagoras theorem

∴ TM² = TL² + `square`

∴ TM² = 12² + `square`

∴ TM² = 144 + `square`

∴ TM² = 400

∴ TM = 20

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`

In the following figure, AE = EF = AF = BE = CF = a, AT ⊥ BC. Show that AB = AC =  `sqrt3xxa`

Find the length of the hypotenuse in a right angled triangle where the sum
of the squares of the sides making right angle is 169.
(A)15 (B) 13 (C) 5 (D) 12

∆ABC is an equilateral triangle. Point P is on base BC such that PC = `1/3`BC, if AB = 6 cm find AP.

Complete the following activity to find the length of hypotenuse of right angled triangle, if sides of right angle are 9 cm and 12 cm.

Activity: In ∆PQR, m∠PQR = 90°

By Pythagoras Theorem,

PQ² + `square` = PR²    ......(I)

∴ PR² = 9² + 12² 

∴ PR² = `square` + 144

∴ PR² = `square`

∴ PR = 15

∴ Length of hypotenuse of triangle PQR is `square` cm.