# R.D. Sharma Solutions Class 10th: Ch 4 Triangles Exercise 4.7

## Chapter 4 triangles R.D. Sharma Solutions for Class 10th Math Exercise 4.7

**Exercise 4.7**

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

**Solution**

Side of traingles are:

AB = 3cm

BC = 4 cm

AC = 6 cm

Now, we know that the triangle can be right angled if it follows the pythagoras theorem.

∴ AB

^{2}= 3

^{2}= 9

BC

^{2}= 4

^{2}= 16

AC

^{2}= 6

^{2}= 36

Since, AB

^{2}+ BC

^{2}≠ AC

^{2}

Then, by converse of Pythagoras theorem, Triangle is not a right angled.

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 = l6 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

**Solution**

**Solution**

**Solution**

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 .

**Solution**

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

**Solution**

**Solution**

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.

**Solution**

**Solution**

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.

**Solution**

11. ABCD is a square. F is the mid-point of AB. BE is one third of BC. If the area of ∆FBE =

108 cm

12. In an isosceles triangle ABC, if AB = AC = 13 cm and the altitude from A on BC is 5 cm, find BC.

108 cm

^{2}, find the length of AC.**Solution**12. In an isosceles triangle ABC, if AB = AC = 13 cm and the altitude from A on BC is 5 cm, find BC.

**Solution**

13 . In a ∆ABC, AB = BC = CA = 2a and AD ⊥ BC. Prove that

(i) AD = a√3

(ii) Area (∆ABC) =√3a

^{2 }^{}

**Solution**

14. The lengths of the diagonals of a rhombus are 24 cm and 10 cm. Find each side of the rhombus.

**Solution**

15. Each side of a rhombus is 10 cm. If one of its diagonals is 16 cm find the length of the other diagonal.

**Solution**

**Solution**

17. In Fig., 4.221, ∠B < 90° and segment AD ⊥ BC, show that

(i) b

^{2}= h^{2}+ a^{2}+ x^{2}- 2ax
(ii) b

^{2}= a^{2}+ c^{2}- 2ax**Solution**

^{2}= 3BD

^{2}.

**Solution**

19. △ABD is a right triangle right angled at A and AC ⟂ BD. Show that:

(i) AB

^{2}= CB × BD

(ii) AC

^{2}= DC × BC

(iii) AD

^{2}= BD × CD

(iv) AB

^{2}/AC

^{2}= BD/DC

**Solution**

(i)

(ii)

(iii)

(iv)

20. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut ?

**Solution**

**Solution**

**Solution**

23. In right-angled triangle ABC in which ∠C = 90°, if D is the mid-point of BC, prove that AB

^{2}= 4 AD

^{2}- 3 AC

^{2}.

**Solution**

24. In Fig. 4.223, D is the mid-point of side BC and AE ⊥ BC. If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that:

(i) b

^{2}= p^{2}+ax+a^{4}
(ii) c

^{2}= p^{2}- ax+a^{2}/4
(iii) b

^{2}+c^{2}= 2p^{2}+ a^{2}/2**Solution**

(i)

(ii)

(iii)

25. In ∆ABC, ∠A is obtuse, PB ⊥AC and QC ⊥ AB. Prove that:

(i) AB × AQ = AC × AP

(ii) BC

(i) AB × AQ = AC × AP

(ii) BC

^{2}= (AC × CP + AB × BQ)**Solution**26. In a right ∆ABC right-angled at C, if D is the mid-point of BC, prove that CD

^{2}= 4(AD

^{2}- AC

^{2}) .

**Solution**

27. In a quadrilateral ABCD, ∠B = 90°, AD

^{2}= AB^{2}+ BC^{2}+ CD^{2}, prove that ∠ACD = 90°.**Solution**

**Solution**