PE 4 - Triangles

Let ABC be the triangle with h as the altitude and BC as base. Let PQRS be the rectangle inscribed in triangle ABC.

​​​​​​​

As height of rectangle = PS = x, and 

base of the rectangle = SR = 2x = PQ

Now, height of ∆APQ = height of triangle – height of rectangle = h – x

Now, ∆APQ and ∆ABC are similar.

∴ $\frac{\mathrm{PQ}}{\mathrm{BC}}$ = $\frac{\mathrm{h}-\mathrm{x}}{\mathrm{h}}$

∴ $\frac{2\mathrm{x}}{\mathrm{b}}$ = $\frac{\mathrm{h}-\mathrm{x}}{\mathrm{h}}$
∴ $\frac{2\mathrm{x}}{\mathrm{b}}$ = 1 - $\frac{\mathrm{x}}{\mathrm{h}}$
∴ $\frac{2\mathrm{x}}{\mathrm{b}}$ + $\frac{\mathrm{x}}{\mathrm{h}}$ = 1 
∴ x = $\frac{\mathrm{bh}}{\mathrm{b}+2\mathrm{h}}$

Hence, option (c).

Let us draw DG || AF as shown in the figure.

Since D is the mid-point of AB, AD = DB
Also, since E is the mid-point of CD, CE = ED

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In ∆CDG, since DG || EF, hence
CF : FG = CE : ED = 1 : 1
⇒ CF = FG   ...(1)

In ∆BAF, since DG || AF, hence
BG : GF = BD : DA = 1 : 1
⇒ BG = GF   ...(2)

From (1) and (2), we get
​​​​​​​BG = GF = FC

∴ BG = GF = FC = ⅓ BC = 2 cm.

Hence, option (c).

We know area of smaller triangle formed by joining mid-points of a triangle is one-fourth of bigger triangle.

∴ Area of S₂ = ¼ (Area of S₁) = 36/4
Area of S₃ = ¼ (Area of S₂) = 36/42
Area of S₄ = ¼ (Area of S₃) = 36/43
... and so on

∴ Sum of all the areas = 36 + $\frac{36}{4}$ + $\frac{36}{4^2}$ + $\frac{36}{4^3}$ + ... 

This is sum of an infinite GP whose first term is 36 and common ratio = 1/4

∴ Sum of all the areas = $\frac{36}{1-{\displaystyle \frac{1}{4}}}$ = $\frac{36\times 4}{3}$ = 48

Hence, option (a).

Let us draw perpendiculars from D to EF and A to BC as shown in the figure.

​​​​​​​

Since EF || BC and AE = EB, it means
​​​​​​​⇒ ​E and F are the mid-points of AB and AC respectively.
⇒ Also, AS = SR   ...(1)

Also EF || GH and DG = GE, it means
⇒ G and H are the mid-points of DE and DF respectively.
⇒ GH = ½ EF   ...(2)
⇒ ​​​​​​​Also, DQ = PQ   ...(3)

From (1) and (3), we get
AS = SR = PQ = DQ
⇒ AS = DQ

In ∆AEF and ∆DGH,
perpendicular is same i.e., AS = DQ,
while base GH of ∆DGH is half the base EF of ∆AEF

∴ Area of ∆DGH will be half the area of ∆AEF = ½ × 10 = 5 cm².

Hence, option (b).

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As we can see from the figure above, the folded triangle obtained will be a right isosceles triangle.

⇒ Area of folded triangle = 8 = ½ × a × a
⇒ a = 4

∴ Height of the original triangle = 4, and
Base of the original triangle = 2a = 8

Now, in ∆ABD,
AB² = AD² + BD²
⇒ AB² = 16 + 16 = 32
⇒ AB = 4√2

Hence, option (d).

Let the base side for altitude 12, 15 and 20 be a, b and c respectively.

∴ Area of triangle = ½ × 12 × a = ½ × 15 × b = ½ × 20 × c

⇒ 12a = 15b = 20c

⇒ a : b : c = 5 : 4 : 3

∴ a = 5k, b = 4k and c = 3k

Now, b² + c² = a²,
∴ The given triangle is a right triangle.

⇒ The perpendicular on two smaller sides are equal to sides.

The two smaller sides will be 15 and 20

∴ Area of triangle = ½ × 15 × 20 = 150.

Hence, 150.

Let the base sides for altitudes 3, 4 and 6 be a, b and c respectively.

∴ Area of triangle (∆) = ½ × 3 × a = ½ × 4 × b = ½ × 6 × c

⇒ ∆ = ½ × 3 × a
⇒ $\frac{\mathrm{a}}{∆}$ = $\frac{2}{3}$   ...(1)

⇒ ∆ = ½ × 4 × b
⇒ $\frac{\mathrm{b}}{∆}$ = $\frac{1}{2}$   ...(2)

⇒ ∆ = ½ × 6 × c
⇒ $\frac{\mathrm{c}}{∆}$ = $\frac{1}{3}$   ...(3)

Adding (1), (2) and (3), we get

$\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{∆}$ = $\frac{2}{3}$ + $\frac{1}{2}$ + $\frac{1}{3}$ = $\frac{3}{2}$

⇒ $\frac{∆}{\mathrm{a}+\mathrm{b}+\mathrm{c}}$ = $\frac{2}{3}$

⇒ $\frac{∆}{2\mathrm{s}}$ = $\frac{2}{3}$

⇒ $\frac{\mathrm{r}}{2}$ = $\frac{2}{3}$

⇒ r = $\frac{4}{3}$

Hence, option (c).

In ∆ABD and ∆ACD, since​​​​​​​ height is same, the ratio of areas will be same as the ratio of bases.

⇒ $\frac{\mathrm{Area}(∆\mathrm{ABD})}{\mathrm{Area}(∆\mathrm{ACD})}$ = $\frac{\mathrm{BD}}{\mathrm{CD}}$ = $\frac{2}{3}$

Now, in ∆BED and ∆CED, since​​​​​​​​​​​​​​ height is same, the ratio of areas will be same as the ratio of bases.

⇒ $\frac{\mathrm{Area}(∆\mathrm{BED})}{\mathrm{Area}(∆\mathrm{CED})}$ = $\frac{\mathrm{BD}}{\mathrm{CD}}$ = $\frac{2}{3}$

Now, $\frac{\mathrm{Area}(∆\mathrm{ABD})}{\mathrm{Area}(∆\mathrm{ACD})}$ = $\frac{\mathrm{Area}(∆\mathrm{BED})}{\mathrm{Area}(∆\mathrm{CED})}$ = $\frac{2}{3}$

∴ $\frac{\mathrm{Area}(∆\mathrm{ABD})}{\mathrm{Area}(∆\mathrm{ACD})}$ = $\frac{\mathrm{Area}(∆\mathrm{BED})}{\mathrm{Area}(∆\mathrm{CED})}$ = $\frac{\mathrm{Area}(∆\mathrm{ABD})-\mathrm{Area}(∆\mathrm{BED})}{\mathrm{Area}(∆\mathrm{ACD})-\mathrm{Area}(∆\mathrm{CED})}$ = $\frac{2}{3}$

⇒ $\frac{\mathrm{Area}(∆\mathrm{ABE})}{\mathrm{Area}(∆\mathrm{ACE})}$ = $\frac{2}{3}$

Let Area(∆ABE) = 2A and Area(∆ACE) = 3A   ...(1)

Now, since AF : FC = 3 : 4

⇒ $\frac{\mathrm{Area}(∆\mathrm{AEF})}{\mathrm{Area}(∆\mathrm{CEF})}$ = $\frac{3}{4}$

From (1),
Area(∆AEF) = 3A × 3/7 = 9A/7 and Area(∆ACE) = 3A × 4/7 = 12A/7   ...(2)

From (1) and (2),
$\frac{\mathrm{EB}}{\mathrm{EF}}$ = $\frac{\mathrm{Area}(∆\mathrm{ABE})}{\mathrm{Area}(∆\mathrm{AFE})}$ = $\frac{2\mathrm{A}}{9\mathrm{A}/7}$ = $\frac{14}{9}$

Hence, option (b).