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 

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
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, ABQR. 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]
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]
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]
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, DEBC such that AE = (1/4)AC. If AB = 6 cm, find AD.
We have, DEBC, AB = 6cm and AE = 1/4 AC
In Î”ADE and Î”ABC
∠A = ∠A [Common]
∠ADE = ∠ABC [Corresponding angles]
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
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.
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
DE = 10cm, EF = 8cm and FD = 8.4 cm, If AL ⊥ BC and DM ⊥ EF, find AL: DM.
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.
(i) Î”OMA and Î”OLC
(ii) OA/OC = OM/OL
We have AB CDEF. 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.
show that PQRS is a rhombus.
In Î”BAD, by mid  point theorem
PS  AD and PS = 1/2AD ...(i)
In Î”CAD, by mid  point theorem
QR  AD and QR = 1/2 AD ...(ii)
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)
(i) Î”ABC ～ Î”AMP
(ii) CA/PA = BC/MP