Math Olympiad Test: Triangles - 2 - Class 10 MCQ

Here altitude from A to BC is AD, we know in isosceles altitude on non-equal sides also median, 
so, BD = CD = BC​/2 = 14​/2 = 7cm
Now, we applying Pythagoras theorem in △ABD,
AB² = BD² + AD²
(25)² = (7)² + AD²
AD² = 625 − 49
AD² = 576
AD = 24cm.

Ratio of perimeters = ratio of sides 

∵ DE ║ BE
∴ DADE ~ DABC

⇒
⇒ adjusting this expression, we have
⇒ 
⇒ 
⇒ 21x² - 12x = 15x² − 6x − 8 + 20 x
⇒ 6x² − 26x + 8 = 0
⇒ 6x² −24x − 2x + 8 = 0
⇒ 6x(x − 4) − 2(x − 4) = 0

When, 
∴ 

In ΔPQR and ΔPST

⇒ According to Thales’ theorem

Subtracting unity from both sides, we have,

⇒ 21x² - 12x = 15x² - 6x + 20x - 8
⇒ 6x² - 26x + 8 = 0
⇒ 3x² - 13x + 4 = 0
⇒ 3x² - 12x - x + 4 = 0
⇒ 3x(x - 4) - 1(x - 4) = 0
⇒ (x - 4) (3x - 1) = 0

∴ x = 4

In ΔABE,
AE² + AB² = BE²
⇒ 9² + AB² = 15²
⇒ AB² = 225 - 81
= 144
⇒ AB = 12 m
Similarly,
In DBCD
⇒ BC² = BD² - DC² = 15² - 12² = 92
BC = 9m
∴ width of street = (AB + BC) = 12 + 9 = 21 m

Ratio of areas = (ratio of altitudes)²

∠ CDO + ∠DCO = 115°
⇒ ∠DCO = 115° - 70° = 45°
∠OAB = 45° (∴ DODC ~ DOBA)

A is the point of cleavage C is the point where top of three touches the ground
In ΔABC
AB² + BC² = AC² (Pythagoras’ Theorem)
6² + 8² = AC²
= AC = 10 m
∴ Total (original) height = 10 m + 6 m = 16 m

Let the angles be 2x and 3x respectively
∴ 2x + 3x = 110°
⇒ 5x = 110°
⇒ x = 22°
∴ Angles are 44°, 66°, and 70°

Let the side of rhombus be x cm
Now, we know that the diagonals of rhombus bisect each other at right angles.
∴ By using Pythagoras’ Theorem

= 64 + 225 = 289
⇒ x = 17 cm
∴ Perimeter =  4 × x = 4 × 17 = 68 cm.

∵  BC ║ DE
∴ 


[Ar(DADE) = Ar(DABC) + Ar(¨BCED)

⇒ BC = 10 cm

In ΔABC
AC² = AB² + BC² (Pythagoras’ theorem)

⇒ AC = 

Both Ds are right angled, and also the elevation of coming sunlight will be same.
∴ All the angles of the ΔABC and ΔPQR will be correspondingly identical.
∴ ΔABC ~ ΔPQR
⇒ AB/PQ = BC/QR 
⇒ 12/PQ = 8/40 
⇒ PQ = 12 × 5 = 60m

In ΔABD

AB² = BD² + AD²
⇒ BD = 

= 12 cm
∵ D bisects BC
∴ BC = 2 × BD
= 2 × 12 = 24 cm

We know that if, ΔABC ~ ΔPQR, then

or, Applying componendo and dividendo, it can be clearly seen that,