PE 3 - Triangles

Let A(∆ADE) = 1, hence A(DEGF) = A(FGCB) = 1

∴ A(∆AFG) = 2 and A(∆ABC) = 3

In the figure, DE || FG || BC, hence ∆ADE is similar to ∆AFG is similar to ∆ABC

Since, the 3 triangles are similar, the ratio of their areas will be same as the ratio of squares of their sides.

⇒ A(∆ADE) : A(∆AFG) : A(∆ABC) = 1 : 2 : 3 = AD² : AF² : AB²

⇒ AD : AF : AB = 1 : √2 : √3

Now, DF = AF - AD = √2 - 1
and, FB = AB - AF = √3 - √2

∴ AD : DF : FB = 1 : (√2 - 1) : (√3 - √2)

Hence, option (d).

Given, a² + b² + c² = ab + bc + ca

⇒ 2a² + 2b² + 2c² = 2ab + 2bc + 2ca

⇒ 2a² + 2b² + 2c² - 2ab - 2bc - 2ca = 0

⇒ (a² - 2ab + b²) + (c² - 2ac + a²) (b² - 2bc + c²) = 0

⇒ (a - b)² + (a - c)² + (b - c)² = 0

This is only possible when a = b = c

∴ The given triangle is an equilateral triangle.

Hence, option (c).

In a right triangle circumradius is half of the hypotenuse.

∴ Hypotenuse of the triangle = 2 × 3 = 6 cm.

Area = ½ × hypotenuse × (altitude on hypotenuse) = ½ × 6 × (altitude on hypotenuse)

Now, the area will be highest when altitude on the hypotenuse is highest. 

⇒ The highest value of hypotenuse will be equal to circumradius i.e., 3 cm.

∴ Highest possible area = ½ × 6 × 3 = 9 cm².

Hence, option (a).

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Area of larger circle = 1386 = πr²
​​​​​​​⇒ r = 21

The larger circle is incircle of triangle ABC.
In circle of an equilateral triangle = a/2√3 = 21
​​​⇒ side of the equilateral triangle ABC = 42√3

Now, let us draw a line DE which is tangent to both the circles. DE will also be parallel to BC.

Now ∆ADE will also be an equilateral triangle.
Height of ∆ADE = Height of ∆ABC - diameter of larger circle
= 42√3 × √3/2 - 42 = 63 - 42 = 21 cm.

∆ADE ≈ ∆ABC
Height of ∆ADE is one-third the height of ∆ABC
∴ Inradius of ∆ADE is one-third the inradius of ∆ABC = 1/3 × 21 = 7 cm.

∴ Area of smaller circle = 22/7 × 7² = 154 cm².

Hence, option (b).

Since we need to find the shortest side, it means 10 is the longest side of the triangle.

​​​​​​​

If we cosider the perpendicular from A to BC.

⇒ Area of triangle = 20 = ½ × 10 × AD
⇒ AD = 4

In right triangle ABD,
AB = 5, AD = 4 hence BD = 3 [3, 4, 5 is a pythagorean triplet]

Now, CD = 10 - 3 = 7

In right triangle ACD,
AC² = CD² + AD²
⇒ AC² = 49 + 16 = 65
​​​​​​​⇒ AC = √65

Hence, option (a).

In right triangle ABC, since AB and BC have to be integers it has to be a pythagorean triplet.
Since, AC = 17, the applicable pythagorean triplet is 8, 15 and 17.
Now, (AB, BC, AC) = (8, 15 and 17) or (15, 8 and 17).

In ∆ABC and ∆BCD
∠ABC = ∠BCD, &
∠BAC = ∠CBD
∴ ∆ABC is similar to ∆BCD

Case 1: (AB, BC, AC) = (8, 15 and 17)
$\frac{\mathrm{Area} \mathrm{of} ∆\mathrm{ABC}}{\mathrm{Area} \mathrm{of} ∆\mathrm{BCD}}$ = ${\left(\frac{\mathrm{AB}}{\mathrm{BC}}\right)}^2$ = ${\left(\frac{8}{15}\right)}^2$ = $\frac{64}{225}$

Case 2: (AB, BC, AC) = (15, 8 and 17)
$\frac{\mathrm{Area} \mathrm{of} ∆\mathrm{ABC}}{\mathrm{Area} \mathrm{of} ∆\mathrm{BCD}}$ = ${\left(\frac{\mathrm{AB}}{\mathrm{BC}}\right)}^2$ = ${\left(\frac{15}{8}\right)}^2$ = $\frac{225}{64}$

Hence, option (d).

∆PAB is similar to ∆BCR

 $\frac{\mathrm{PA}}{\mathrm{BC}}$ = $\frac{\mathrm{AB}}{\mathrm{CR}}$

⇒ $\frac{9-\mathrm{a}}{\mathrm{a}}$ = $\frac{\mathrm{a}}{12-\mathrm{a}}$

⇒ a² = 108 - 21a + a²

⇒ 21a = 108

⇒ a = 36/7

Hence, area of square ABCQ = (36/7)² = 1296/49 = 26.45 cm².

Hence, option (c).

In a right triangle inradius (r) = (sum of two smaller sides - hypotenuse)/2 and

Circumradius (R) = Hypotenuse/2

∴ 2(R + r) = 2R + 2r = b + (a + c - b) = a + c

Hence, option (d).

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We know centroid (O) divides the bigger triangle in 6 smaller triangles of equal areas.

∴ Area of ΔABC = 6 × (Area of ΔCOD) = 6 × 16 = 96 cm².

Area of ΔBEC = 1/2 of ΔABC = 1/2 of 96 = 48 cm².

Now, ΔBFD is similar to ΔBEC

∴ Area of ΔBFD : Area of ΔBEC = BD² : BC² = 1 : 4

⇒ Area of ΔBFD = 48/4 = 12 cm².

Now, Area of DOEF = Area of ΔBEC - Area of ΔBFD - Area of ΔCOD
= 48 - 12 - 16 
= 20 cm². 

Hence, option (c).

Since ∆EFG is an isosceles triangle such that EF = EG, hence angle bisector from E will also be perpendicular to FG.

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Now ∆EQG is a right triangle, hence QR is the altitude from right vertex.
⇒ EQ² = ER × EG = 21 × 30 = 630

Also, ∆EQP is a right triangle, hence QP is the altitude from right vertex.
⇒ EQ² = EP × EF
∴ EP × EF = 630

Hence, option (d).