Class 10 RD Sharma Solutions Chapter 4 Triangles Exercise 4.5 | Set 2 (original) (raw)
Last Updated : 23 Jul, 2025
Chapter 4 of the Class 10 RD Sharma Mathematics textbook, "Triangles," covers the properties and theorems related to triangles, including similarity, congruence, and the Pythagorean theorem. Exercise 4.5 focuses on applying these properties to solve problems related to triangles.
RD Sharma Solutions for Class 10 - Mathematics - Chapter 4 Triangles - Exercise 4.5 | Set 2
This section provides detailed solutions for Exercise 4.5 from Chapter 4 of the Class 10 RD Sharma Mathematics textbook. The exercise involves problems that require students to apply various properties and theorems of triangles, such as criteria for similarity and congruence, to find unknown lengths and angles in given triangles. Solutions are presented step-by-step to aid understanding and mastery of the concepts.
**Question 13. The perimeters of two similar triangles are 25 cm and 15 cm, respect. If one side of the first triangle is 9 cm, what is the corresponding side of the other triangle?
**Solution:
Given: perimeter of two similar triangles are 25 cm, 15 cm and one side 9 cm
To find : the other side.
Let the two triangles be ABC & PQR.
Let one of its side is (AB) = 9 cm and the other side of other triangle be PQ.
∆ABC ∼ ∆PQR. [Since the ratio of corresponding sides of similar triangles is same as the ratio of their perimeters.]
AB/PQ = BC/QR = AC/PR = 25/15
9/PQ = 25/15
25 PQ = 15 × 9
PQ = (15×9)/25
PQ =(3× 9 )/5
PQ = 27/5
PQ = 5.4 cm
Hence, the corresponding side of other ∆ is 5.4 cm.
**Question 14. In ΔABC and ΔDEF, it is being given that AB = 5 cm, BC = 4 cm, CA = 4.2 cm, DE = 10 cm, EF = 8 cm, and FD = 8.4 cm. If AL ⊥ BC, DM ⊥ EF, find AL: Dm.
**Solution:
Given,
AB = 5 cm, BC = 4 cm, CA = 4.2 cm, DE = 10 cm, EF = 8 cm, and FD = 8.4 cm
AL ⊥ BC, DM ⊥ EF
To find: AL: DM
Now In ∆ABC and ∆DEF,
AB/DE=BC/EF=AC/DF=1/2 [both triangles are similar]
[In similar triangle the ratio of corresponding altitude is same as the ratio of the corresponding sides.]
\therefore AL: DM = 1: 2
**Question 15. D and E are the points on the sides AB and AC respectively, of a ΔABC such that AD = 8 cm, DB = 12 cm, AE = 6 cm, and CE = 9 cm. Prove that BC = 5/2 DE.
**Solution:
AE/EC=6/9=2/3.......(1)
AD/DB=8/12=2/3......(2)
from (1) and (2) we know that
\therefore AD/DB=AE/EC
\therefore ED∥CB
Now In△AED and △ACB,
ED∥CB
∠AED=∠ACB
∠ADE=∠ABC
△AED∼△ACB [by AAA criteria]
\therefore AE/AC=ED//CB
AE/EC+AE= ED/CB [\therefore AC=AE+EC]
6/6+9= ED/CB
6/15=2/5=ED/CB
\therefore CB=5/2ED
Hence proved
**Question 16. D is the midpoint of side BC of a ΔABC. AD is bisected at the point E and BE produced cuts AC at the point X. Prove that BE: EX = 3 : 1
**Solution:
Given: In ∆ABC, D is the midpoint of BC, E is the midpoint of AD. BE produced meets AC at X.
To prove : BE: EX = 3 : 1
Now In Δ BCX and ΔDCY,
∠CBX = ΔCBY (corresponding angles)
∠CXB = ΔCYD (corresponding angles)
\therefore ΔBCX∼ΔDCY (AA similarity)
\therefore BC/DC = BX/ DY = CX/CY [ corresponding sides of two similar triangles are proportional]
BX/DY = BC/DC
BX/DY = 2DC/DC [As D is the midpoint of BC]
BX/DY = 2/1………...(i)
Now In ΔAEX and ΔADY,
∠AEX = ΔADY (corresponding angles)
∠AXE = ΔAYD (corresponding angles)
\therefore ΔAEX ∼ ΔADY (AA similarity)
\therefore AE/AD = EX/DY = AX/AY [ corresponding sides of two similar triangles are proportional]
EX/DY = AE/AD
EX/DY = AE/2AE [ D is the midpoint of BC]
EX/DY = ½…………..…(ii)
On Dividing eqn. (i) by eqn. (ii) we get,
BX/EX = 4/1
BX = 4EX
BE + EX = 4EX
BE = 4EX - EX
BE = 3EX
BE /EX = 3/1
BE : EX = 3:1
Hence proved
**Question 17. 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 rectangle contained by AB and BC.
**Solution:
Given : ABCD is a parallelogram and APQ is a straight line meeting BC at P and DC produced at Q.
To prove : BP x DQ = AB x BC
Now In ∆ABP and ∆QDA,
∠B = ∠D [Opposite angles of parallelogram]
∠BAP = ∠AQD [Alternate interior angles]
\therefore ∆ABP ~ ∆QDA [By AA similarity]
\therefore AB/QD = BP/DA [the corresponding sides of similar triangles are proportional]
DA = BC [opposite sides of a parallelogram are equal]
AB/QD = BP/BC
AB × BC = QD × BP
Hence proved
Question 18. In ΔABC, AL and CM are the perpendiculars from the vertices A and C to BC and AB respect. If Al and CM intersect at O, prove that:
****(i) ΔOMA ∼ ΔOLC**
****(ii)OA/OC=OM/OL**
**Solution:
Given,
AL ⊥ BC and CM ⊥ AB
To prove:
(i) ΔOMA ∼ ΔOLC
(ii)OA/OC=OM/OL
Now In Δ OMA and ΔOLC,
∠MOA = ∠LOC [Vertically opposite angles]
∠AMO = ∠CLO [Each 90°]
\therefore ΔOMA ~ ΔOLC [By AA similarity]
\therefore OA/OC=OM/OL [Corresponding parts of similar Δ are proportional]
Hence proved
**Question 19. ABCD is a quadrilateral in which AD = BC. If P, Q, R, S be the midpoints of AB, AC, CD, and BD respect. Show that PQRS is a rhombus.
**Solution:
Given,
ABCD is a quadrilateral in which AD = BC and P, Q, R, S are the mid points of AB, AC, CD, BD, respectively.
To prove: PQRS is a rhombus
Now,
In ΔABC, P and Q are the mid points of the sides B and AC respectively
PQ ∥ BC and PQ = 1/2 BC. [By the midpoint theorem]
In ΔADC, Q and R are the mid points of the sides AC and DC respectively
By the mid-point theorem, we get,
QR ∥ AD QR = 1/2 AD [By the mid-point theorem]
\therefore QR = 1/2 BC [AD = BC]
Now In ΔBCD,
RS∥BC and RS = 1/2 AD [By the mid-point theorem]
RS= 1/2 BC [AD = BC]
From above eqns.
PQ = QR = RS
\therefore PQRS is a rhombus.
Hence proved
**Question 20. In an isosceles ΔABC, the base AB is produced both ways to P and Q such that AP x BQ = AC2. Prove that Δ APC ∼ Δ BCQ
**Solution:
Given : ΔABC is isosceles ∆ and AP x BQ = AC²
To prove : ΔAPC∼ΔBCQ.
we know that,
ΔABC is an isosceles triangle AC = BC.
AP x BQ = AC² (given)
AP x BQ = AC x AC
AP x BQ = AC x BC
AP/BC = ABQ……….(1).
so, ∠CAB = ∠CBA [as AC = BC]
180° – ∠CAP = 180° – ∠CBQ [angles opposite to equal sides are EQUAL]
∠CAP = ∠CBQ ………..(2)
In ∆APC and ΔBCQ
AP/BC = AC/BQ [From equation 1]
∠CAP = ∠CBQ [From equation 2]
\therefore ΔAPC∼ΔBCQ [By SAS similarity criterion]
Hence proved
**Question 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:
AD=1.2 m/sec ×4 sec =4.8 m =480 cm
Suppose the length of the shadow of the girl be x cm when she at position D. Therefore,
BD = x cm
In ∆BDE and ∆BAC,
∠BDE=∠BAC [each90°]
∠DBE=∠ABC [Common angle ]
\therefore ∆BDE∼∆BAC [ by AA similarity]
BE/BC=DE/AC=BD/AB [Corresponding sides are proportional]
90/360=x/480+x
1/4= x/480+x
4x=480+x
4x−x=480
3x=480
x=480/3=160
Hence, the length of her shadow after 4 seconds is 160 cm.
**Question 22. A vertical stick of length 6 m 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:
Let AB be stick and DE be tower.
A 6 m long stick casts a shadow of 4 m
So, AB = 6 m and BC = 4 m
Let DE casts shadow at the same time, So EF = 28 m
Let height of tower DE = x
Now in ∆ABC and ∆DEF,
∠B = ∠E (each 90°)
∠C = ∠F (shadows at the same time)
∆ABC ~ ∆DEF [AA criterion]
AB/DE=BC/EF [ Corresponding sides are proportional ]
\therefore 6/x=4/28
x=42 m
Hence height of tower is 42 m
Question 23. In the fig. given, ΔABC is a right-angled triangle at C and DE ⊥ AB. Prove that ΔABC ∼ ΔADE. **hence find the lengths of AE **and DE.
**Solution:
Given : ΔACB is right-angled triangle at C = 90°.
According to the figure : BC = 12 cm , AD=3 cm, DC = 2 cm.
AC = AD + DC = 3 +2= 5 cm
Now In ∆ACB,
AB² = AC² + BC² [by Pythagoras theorem]
AB² = 5² + 12²
AB² = 25 + 144 = 169
AB= √169 = 13
\therefore AB = 13 cm
Now In ΔABC & ΔADE,
∠BAC = ∠DAE [common]
∠ACB = ∠AED [each 90°]
\therefore ΔABC∼ΔADE [by AA similarity criterion]
\therefore AB/AD = BC/DE = AC/AE [ corresponding sides of two similar triangles are proportional]
13/3 = 12/ DE = 5/AE
13/3 = 12/DE
13 DE = 12×3
DE = 36/13
13/3 = 5/AE
13 AE = 5×3
AE = 15/13
Hence, the length of DE= 36/13 & AE = 15/13
**Question 24. In the figure, PA, QB and RC are each perpendicular to AC. Prove that 1/X+1/Z=1/Y
Given:
PA, QB and RC are each perpendicular to AC
To prove : 1/X+1/Z=1/Y
Now In ΔPAC and ΔQBC,
∠PCA = ∠QCB [common angle]
∠PAC = ∠QBC [Corresponding angles]
\therefore ΔPAC ~ ΔQBC [by AA criteria]
\therefore PA/QB = AC/BC [ Corresponding parts of similar triangles are proportional]
x/y = AB/BC
y/x = BC/AC
Now In ΔRCA and ΔQBA,
∠RAC = ∠QAB [common angle]
∠RCA = ∠QBA [Corresponding angle]
\therefore ΔRCA ~ΔQBA [by AA criteria]
\therefore RC/QB = AC/AB [ Corresponding parts of similar triangles are proportional]
z/y= AC/AB
y/z= AB/AC
y/z + y/z = BC + AC/ AC = 1 [by adding both eq we get]
y/z + y/z = 1
1/x + 1/z = 1/y [multiplying both sides by y]
Hence proved
**Question 25. In fig. given, 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:
Given : AB || CD || EF , AB = 6 cm, CD = x cm, BD = 4 cm, and DE = y cm and EF = 10 cm.
Now In ∆ECD and ∆EAB,
∠CED = ∠AEB [common]
∠ECD = ∠EAB [corresponding angles]
\therefore ∆ECD ~ ∆EAB ……....(1) [By AA similarity]
∴ EC/EA = CD/AB [Corresponding parts of similar triangles are proportional]
EC / EA = x/6 ……………(2)
In ∆ACD and ∆AEF
∠CAD = ∠EAF [common]
∠ACD = ∠AEF [corresponding angles]
∆ACD ~ ∆AEF [By AA similarity]
∴ AC/AE = CD/EF [Corresponding parts of similar triangles are proportional]
AC /AE = x/10 ……………..(3)
EC/EA+ AC /AE = x/6 + x/10 [by Adding eq 2 & 3]
(EC + AC) /AE =( 5x + 3x)/30
AE / AE = 8x /30
1 = 8x/30
x = 30/8
x = 3.75 cm
∆ECD ~ ∆EAB [From eq (i)]
DC/AB = ED /EB [Corresponding parts of similar triangle are proportional]
3.75/6 = y/y+4
6y = 3.75(y+4)
6y = 3.75y + 15
6y - 3.75y = 15
2.25y = 15
y = 15/2.25
y = 6.67 cm
Hence, x = 3.75 cm and y = 6.67 cm.
