Class 10 RD Sharma Solutions – Chapter 4 Triangles – Exercise 4.5 | Set 2
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?
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.
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.]
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.
from (1) and (2) we know that
Now In△AED and △ACB,
△AED∼△ACB [by AAA criteria]
AE/EC+AE= ED/CB [AC=AE+EC]
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
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)
ΔBCX∼ΔDCY (AA similarity)
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)
ΔAEX ∼ ΔADY (AA similarity)
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
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.
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]
∆ABP ~ ∆QDA [By AA similarity]
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
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
AL ⊥ BC and CM ⊥ AB
(i) ΔOMA ∼ ΔOLC
Now In Δ OMA and ΔOLC,
∠MOA = ∠LOC [Vertically opposite angles]
∠AMO = ∠CLO [Each 90°]
ΔOMA ~ ΔOLC [By AA similarity]
OA/OC=OM/OL [Corresponding parts of similar Δ are proportional]
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.
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
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]
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
PQRS is a rhombus.
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
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]
ΔAPC∼ΔBCQ [By SAS similarity criterion]
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.
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,
∠DBE=∠ABC [Common angle ]
∆BDE∼∆BAC [ by AA similarity]
BE/BC=DE/AC=BD/AB [Corresponding sides are proportional]
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.
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 ]
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.
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
AB = 13 cm
Now In ΔABC & ΔADE,
∠BAC = ∠DAE [common]
∠ACB = ∠AED [each 90°]
ΔABC∼ΔADE [by AA similarity criterion]
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
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]
ΔPAC ~ ΔQBC [by AA criteria]
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]
ΔRCA ~ΔQBA [by AA criteria]
RC/QB = AC/AB [ Corresponding parts of similar triangles are proportional]
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]
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.
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]
∆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.