PE 2 - Triangles

In ∆ADC, 
Since AD = CD ⇒ ∠CAD = ∠ACD = x
Also, ∠CDB = 2x (Exterior angle theorem)

Now, ∠ACD + ∠DCB + ∠BCE = 180°
⇒ x + y + 75 = 180
⇒ x + y = 105   ….(1)

In ∆DCB,
Since, CD = CB ⇒ ∠DBC = 2x

∴ 2x + 2x + y = 180
⇒ 4x + y = 180   …(2)

Solving (1) and (2) we get
x = 25° and y = 80°

⇒ ∠DBC = 2x = 50°

Hence, option (c).

Since DE = BE, ⇒ ∠DBE = ∠BDE = x (say) 

Since EF = EC, ⇒ ∠EFC = ∠ECF = y (say) 

Hence, ∠DEB = 180°– 2x and ∠FEC = 180°– 2y.

⇒ ∠DEF = 180° - ∠DEB - ∠FEC

⇒ ∠DEF = 180° - (180 – 2x) – (180 – 2y)

⇒ ∠DEF = 2(x + y) - 180

Also, a + b + 40° = 180° or a + b = 140°. 

∴ ∠DEF = 2 × 140 – 180 = 100°

Hence, 100.

In a right triangle, we know that, midpoint of hypotenuse is the circum-center of the triangle. 

⇒ DA = DB = DC

∴ In triangle ABC, AB = BD = DA
⇒ ∆ABD is an equilateral triangle.

⇒ ∠CBD = 90 – 60 = 30°.

Hence, option (d).

If a², b² and c² are three sides of a triangle then sum of any two sides of the triangle must be greater than third i.e. a² + b² > c². 

Now for a triangle with sides a, b and c, it is 

Acute triangle if, a² + b² > c²
Right triangle if, a² + b² = c²
Obtuse triangle if, a² + b² < c²

∴ Triangle with sides a, b and c will be an acute triangle.

Hence, option (a).

The height of an equilateral triangle of side 'a' = $\frac{\sqrt{3}}{2}a$

It is given that, $\frac{\sqrt{3}}{2}a$ = 18

⇒ a = 36/√3 = 12√3

Therefore, area of the equilateral triangle = $\frac{\sqrt{3}}{4}{a}^2$

= $\frac{\sqrt{3}}{4}\times {\left(12\sqrt{3}\right)}^2$ = 108√3 cm²

Hence, option (d).

For constant perimeter an equilateral triangle has the maximum area.

∴ The fencing of 400 m should be divided equally i.e., 400/2 = 200 m each.

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Now we get an equilateral triangle with sides 200 m each.

∴ Area of the triangle = √3/4 × (200)² = 10,000√3

Hence, option (c).

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In ∆BAC and ∆BPQ

$\frac{BP}{BA}=\frac{BQ}{BC}=\frac{3}{5}$ and ∠B is common

⇒ ∆BAC ~ ∆BPQ (S-A-S Rule)

⇒ $\frac{Area of ∆BPQ}{Area of ∆BAC}={\left(\frac{3}{5}\right)}^2=\frac{9}{25}$   …(1)

In ∆BRS and ∆BPQ

$\frac{BR}{BP}=\frac{BS}{BQ}=\frac{3}{5}$ and ∠B is common

⇒ ∆BAC ~ ∆BRQ (S-A-S Rule)

⇒ $\frac{Area of ∆BRS}{Area of ∆BPQ}={\left(\frac{3}{5}\right)}^2=\frac{9}{25}$   …(2)

From (1) and (2) we get,

Area of ∆BAC : Area of ∆BPQ : Area of ∆BRS = 625 : 225 : 81

Since area of ∆BAC and ∆BRS are integers, hence the minimum possible area of ∆BAC = 625 sq. units.

Hence, 625.

We know in any triangle sum of two sides is always greater that the third side.

Also, for an obtuse angled triangle square of the longest side is greater than sum of square of the two shorter sides.

Case I: Let 14 be the largest side. 
⇒ 7 + x > 14
⇒ x > 7

Also, 14² > 7² + x²
⇒ x² < 196 – 49
⇒ x² < 147
⇒ x < √147

⇒ x = 8 or 9 or 10 or 11 or 12

Case II: Let x be the largest side. 
⇒ 7 + 14 > x
⇒ x < 21

Also, x² > 7² +14²
⇒ x² > 196 + 49
⇒ x² > 245
⇒ x > √245

⇒ x = 16 or 17 or 18 or 19 or 20

∴ x can take 10 integral values.

Hence, 10.

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Area ∆ABC = $\frac{1}{2}\times x\times h$ = 60 sq.cm.
⇒ x × h = 120   …(1)

Also, in right triangle ADC,
13² = ${\left(\frac{x}{2}\right)}^2$ + h²

⇒ x² + 4h² = 676

⇒ x² + 4 × (120/x)² = 676

It would be better to check options now.

∴ x = 10 or 24, are the roots of the above equation.

Hence, option (d).

We know, in a triangle, line joining the mid-points of two sides is parallel to the third side and half of it.

∴ Sides of the triangle formed by joining midpoints of the sides of a triangle will be half of the original triangle.

Perimeter of the original triangle = 3 × 32 = 96
Perimeter of the triangle formed by joining the mid-points of the previous triangle = 3 × 16 = 48
Perimeter of the triangle formed by joining the mid-points of the previous triangle = 3 × 8 = 24
.
.
.

∴ Total perimeter = 96 + 48 + 24 + …

= $\frac{96}{1-{\displaystyle \frac{1}{2}}}$ = 96 × 2 = 192 cm.

Hence, option (a).