Ex-4.7 Triangles, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

1 If the sides of a triangle are 3 cm, 4 cm, and 6 cm long, determine whether the hiangle is a right-angled triangle.

Sol:

We have,

Sides of triangle AB= 3 cm

BC= 4 cm

AC= 6 cm

∴ AB² = 3² = 9

BC² = 4² = 16

AC² = 6² = 36

Since, AB² + BC²  ≠ AC²

Then, by converse of Pythagoras theorem, triangle is not a right Mangle.

2.      The sides of certain triangles are given below. Determine which of them right triangles are.
 (i) a = 7 cm, b = 24 cm and c = 25 cm
 (ii) a = 9 cm, b =16 cm and c = 18 cm
 (iii) a = 1.6 cm, b= 3.8 cm and c = 4 cm
 (iv) a = 8 cm, b = 10 cm and c = 6 cm 

Sol: We have, 

a= 7 cm, b = 24 cm and c = 25 cm

∴ a² = 49, b²= 576 and c² = 625
Since, a² + b² = 49 + 576

= 625

= c²

Then, by converse of Pythagoras theorem, given triangle is a right triangle.

We have,

a= 9 cm, b = 16 cm and c = 18 cm

∴ a² = 81,b²= 256 and c² = 324

Since, a² + b² = 81 + 256 = 337 ≠ c²

Then, by converse of Pythagoras theorem, given triangle is not a light triangle.

We have,

a=1.6 cm,b= 3.8 cmandC= 4 cm

∴ a² = 64, b² = 100 and c² = 36

Since, a² + c² = 64 + 36 = 100 = b²

Then, by converse of Pythagoras theorem, given Mangle is a right triangle.

3. A man goes 15 metres due west and then 8 metres due north. How far is he from the starting point?

Sol:

Let the starting point of the man be 0 and final point be A.

∴ In ΔABO, by Pythagoras theorem AO² = AB² + B0²

^⇒ AO² = 8² + 15²

^⇒ AO² = 64 + 225 = 289

^⇒ AO == 17m

∴ He is 17m far from the starting point.

4. A ladder 17 m long reaches a window of a building 15 m above the ground. Find the distance of the foot of the ladder from the building.

Sol:

In ΔABC, by Pythagoras theorem AB² + BC² = AC²

⇒ 15² + BC² = 17²

⇒ 225 + BC² = 17²

= BC³ = 289 — 225

⇒  BC² = 64

⇒  BC = 8

∴ Distance of the foot of the ladder from building = 8m

5. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.

Sol:  

Let CD and AB be the poles of height 11 and 6 m. Therefore CP = 11 — 6 = 5 m

From the figure we may observe that AP = 12m In triangle APC, by applying Pythagoras theorem AP² + PC²  AC²

12²+ 5² = AC²

AC² = 144 + 25 = 169

AC = 13

Therefore distance between their tops = 13m.

6. In an isosceles triangle ABC, AB = AC = 25 cm, BC = 14 cm. Calculate the altitude from A on BC.

Sol:

We have

AB = AC = 25 cm and BC = 14 cm

In ΔABD and ΔACD

∠ADB = ∠ADC               [Each 90°]

AB = AC                                  [Each 25 cm]

AD = AD                                   [Common]

Then, ΔABD ≌ ΔACD              [By RHS condition]

, BD = CD = 7 cm                     [By c.p.c.t]

In AADB, by Pythagoras theorem

AD² + BD² = AB²

⇒  AD² + 7² = 25²

⇒   AD² = 625 — 49 = 576

⇒  AD = = 24 cm

7. The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its tip reach?

Sol:

Let, length of ladder be AD = BE = I m In ΔACD, by Pythagoras theorem

AD² = AC² + CD²
l² = 8² + 6^{2           .............(ii)}

In ΔBCE, by pythagoras theorem BE² = BC² + CE²

⇒ l² = BC² + 8^{2      .............(ii)}

Compare (i)and (ii) BC² + 8² = 8² + 6²

⇒ BC² = 6²

⇒ BC = 6m

8. Two poles of height 9 m and 14 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.

Sol:

We have,

AC= 14m, DC= 12m and ED = BC = 9m

Construction: Draw EB ⊥ AC

∴ AB = AC — BC = 14 — 9 = 5m

And, EB = DC = 12m

In ΔABE, by Pythagoras theorem,

AE² = AB² + BE²

⇒ AE² = 5² +12²

⇒ AE² = 25 + 144 =

⇒  AE= VW) = 13m

∴  Distance between their tops = 13 m

9. Using Pythagoras theorem determine the length of AD in terms of b and c shown in Fig. 4.219

Sol:

We have,

In ABAC, by Pythagoras theorem BC² = An² + AC²

⇒ BC² = c² + b²

⇒ BC = 

In ΔABD and MBA

∠B = ∠B                          [Common]

∠ADB = ∠BAC               [Each 90°]

Then, ΔABD ∼ ΔCBA [By AA similarity]

       [Corresponding parts of similar A are proportional]

10. A triangle has sides 5 cm, 12 cm and 13 cm. Find the length to one decimal place, of the perpendicular from the opposite vertex to the side whose length is 13 cm.

Sol:

Let, AB = 5cm, BC = 12 cm and AC = 13 cm. Then, AO² = AB² + BC² . This proves that ΔABC is a right triangle, right angles at B. Let BD be the length of perpendicular from B on AC.

Now, Area ΔABC = 1/2(BC X BA)

= 1/2 (12 X 5)

= 30 cm² 

Also, Area of ΔABC = 1/2 AC X BD = 1/2 (13 X BD)

⇒ (13 XBD) = 30 X 2

⇒ BD = 60/13 cm