# RD Sharma Solutions Chapter 4 Triangles Exercise 4.5 Class 10 Maths

 Chapter Name RD Sharma Chapter 4 Triangles Book Name RD Sharma Mathematics for Class 10 Other Exercises Exercise 4.1Exercise 4.2Exercise 4.3Exercise 4.4Exercise 4.6Exercise 4.7 Related Study NCERT Solutions for Class 10 Maths

### Exercise 4.5 Solutions

1. In fig. 4.136, Î”ACB ～Î”APQ. If BC = 8cm, PQ = 4cm, BA = 6.5cm and AP = 2.8 cm, find CA and AQ.

Solution

Given Î”ACB ～Î”APQ

2. A vertical stick 10 cm long casts a shadow 8 cm long. At the same time a shadow 30 m long. Determine the height of the tower.

Solution

3. In Fig 4.137, AB||QR. Find the length of PB.

Solution

We have, Î”PAB and Î”PQR
∠P = ∠P  [common]
∠PAB = ∠PQR [corresponding angles]
Then, Î”PAB～Î”PQR  [ By AA similarity]

4. In fig. 4.138, XY ||BC. Find the length of XY

Solution

We have, XY ||BC
In Î”AXY and Î”ABC
∠A = ∠A  [common]
∠AXY = ∠ABC [corresponding angles]
Then, Î”AXY ～ Î”ABC [By AA similarity]

5. In a right angled triangle with sides a and b and hypotenuse c, the altitude drawn on the hypotenuse is x. Prove that ab = cx.
Solution

We have ∠C = 90° and CD ⊥ AB

6. In Fig. 4.139, ∠ABC = 90° and BD ⊥ AC. If BD = 8 cm and AD = 4cm, find CD.
Solution

We have, ∠ABC = 90° and BD ⊥ AC
Now, ∠ABD + ∠DBC - 90°  ...(i) [∵ ∠ABC - 90°]
And ∠C + ∠DBC - 90° ...(ii) [By angle sum prop. in Î”BCD]
Compare equations (i) & (ii)
∠ABD = ∠C ...(iii)
In Î”ABD and Î”BCD
∠ABD = ∠C  [ From (iii)]
∠ADB = ∠BDC  [Each 90° ]
Then, Î”ABD ～Î”BCD [By AA similarity]

7. In Fig. 4.14, ∠ABC = 90° and BD ⊥ AC. If AB = 5.7cm, BD = 3.8cm and CD = 5.4cm, find BC.
Solution

We have, ∠ABC = 90° and BD⊥ AC
In Î”ABC and Î”BDC
∠ABC = ∠BDC  [Each 90°]
∠C = ∠C [Common]
Then, Î”ABC ～ Î”BDC  [By AA similarity]

8. In Fig. 4.141, DE||BC such that AE = (1/4)AC. If AB = 6 cm, find AD.
Solution

We have, DE||BC, AB = 6cm and AE = 1/4 AC
∠A = ∠A  [Common]
Then, Î”ADE ～Î”ABC  [By AA similarity]

9. In fig., 4.142, PA, QB and RC are each perpendicular to AC. Prove that 1/x + 1/z + 1/y
Solution
We have, PA ⊥ AC, QB ⊥ AC and RC⊥AC
Let, AB = a and BC = b
In Î”CQB and Î”CPA
∠QCB = ∠PCA  [Common]
∠QBC = ∠PAC [Each 90°]
Then, Î”CQB ～Î”CPA  [By AA similarity]

10. In below fig, ∠A = ∠CED, Prove that Î”CAB ～Î”CED. Also, find the value of x.
Solution

11. The perimeters of two similar triangles are 25cm and 15cm respectively. If one side of first triangle is 9 cm, what is the corresponding side of the other triangle ?

Solution

Assume ABC and PQR to be 2 triangles
We have,
Î”ABC ～ Î”PQR
Perimeter of Î”ABC = 25cm
Perimeter of Î”PQR = 15 cm
AB = 9 cm
PQ = ?
Since, Î”ABC ～Î”PQR
Then, ratio of perimeter of triangles = ratio of corresponding sides

12. In Î”ABC and Î”DEf, it is being given that : AB = 5cm, BC = 4cm and CA = 4.2cm;
DE = 10cm, EF = 8cm and FD = 8.4 cm, If AL  ⊥ BC and DM ⊥ EF, find AL: DM.
Solution

13. D and E are the points on the sides AB and AC respectively of a Î”ABC such that ; Ad = 8cm, DB = 12cm, AE = 6 cm and CE = 9 cm. Prove that BC = 5/2 DE.
Solution

14. D is the mid - point of side BC of a Î”ABC. AD is bisected at the point E and BE produced cuts AC at the point X. Prove that BE : EX = 3 : 1
Solution

15.  ABCD is a parallelogram and APQ is a straight line meeting BC at P and DC produced at Q. Prove that the rectangle obtained by BP and DQ is equal to the AB and BC.
Solution

16. In Î”ABC, AL and CM are the perpendiculars from the vertices A and C to BC and AB respectively. If AL and CM intersect at O, prove that :
(i) Î”OMA and Î”OLC
(ii) OA/OC = OM/OL
Solution

17. In Fig below we have AB || CD|| EF. If AB = 6 cm, Cd = x cm, EF = 10 cm, BD = 4 cm and DE = y cm, calculate the values of x and y .
Solution

We have AB ||CD||EF. If AB = 6 cm, CD = x cm, Ef = 10 cm, BD = 4cm and DE = y cm
In Î”ECD and Î”EAB

18. ABCD is a quadrilateral in which AD = BC. If P, Q, R, S be the mid- points of AB, AC, CD and BD respectively, show that PQRS is a rhombus.
Solution

AD = BC and P, Q, R and S are the mid - points of sides AB, AC, CD and BD respectively,
show that PQRS is a rhombus.
In Î”BAD, by mid - point theorem
In Î”CAD, by mid - point theorem
Compare (i) and (ii)
PS ||QR and PS = QR
Since one pair of opposite sides is equal as well as parallel then PQRS is a parallelogram     ....(iii)
Now, In Î”ABC, by mid - point theorem
PQ || BC and PQ= 1/2 BC ...(iv)

19. In Fig. below, if AB ⊥ BC, DC⊥BC and DE⊥AC, Prove that  Î”CED ～Î”ABC.
Solution

20. In an isosceles Î”ABC, the base AB is produced both the ways to P and Q such that AP × BQ = AC2. Prove that Î”APC ～Î”BCQ.
Solution

21. A girl of height 90 cm is walking away from the base of a lamp - post at a speed of 1.2 m/sec. If the lamp is 3.6m above the ground, find the length of her shadow after 4 seconds.
Solution

∴ Length of shadow  = 1.6m

22. Diagonals AC and BD of a trapezium ABCD with AB|| DC intersect each other at the point O. Using similarity criterion for two triangles, show that  OA/OC = OB/OD .
Solution

23. If Î”ABC and Î”AMP are two right triangles, right angled at B and M respectively such that ∠BAC. Prove that
(i) Î”ABC ～ Î”AMP
(ii) CA/PA = BC/MP
Solution

24. A vertical stick of length 6m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28m long. Find the height of the tower.
Solution

25. In below Fig., Î”ABC is right angled at C and DE ⊥ AB. Prove that Î”ABC ～Î”ADE and Hence find the lengths of AE and DE.
Solution