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Chapter 4: Triangles Exercise – 4.5

Question: 1

In fig. given below ΔACB ∼ ΔAPQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm, and AP = 2.8 cm find CA and AQ.

Congruent triangle

Solution:

Given,

ΔACB ∼ ΔAPQ

BC = 8 cm, PQ = 4 cm, BA = 6.5 cm, and AP = 2.8 cm

We need to find CA and AQ

Since, ΔACB ∼ ΔAPQ

AQ = 3.25 cm

CA = 2.8 × 2

CA = 5.6 cm

Therefore, CA = 5.6 cm and AQ = 3.25 cm.

Question: 2

In fig. given, AB ∥ QR, find the length of PB.

Parallelogram triangle

Solution:

Given,

AB ∥ PB

AB = 3 cm, QR = 9 cm and PR = 6 cm

We need to find out PB,

PB = 2 cm

Question: 3

In fig. given, XY∥ BC. Find the length of XY.

Parallelogram triangle

Solution:

Given,

XY ∥ BC

AX = 1 cm, XB = 3 cm, and BC = 6 cm

We need to find XY,

Since, ΔAXY ∼ ΔABC

XY = 1.5 cm

Question: 4

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: 

Right angled triangleLet the ΔABC be a right angle triangle having sides a and b and hypotenuse c. BD is the altitude drawn on the hypotenuse AC

We need to prove ab = cx

Since, the altitude is perpendicular on the hypotenuse, both the triangles are similar

a/x = c/b

xc = ab

∴ ab = cx

Question: 5

In fig. given, ∠ABC = 90 and BD⊥AC. If BD = 8 cm, and AD = 4 cm, find CD.

ABC right Angle

Solution:

Given,

∠ABC = 90 and BD⊥AC
When, BD = 8 cm, AD = 4 cm, we need to find CD.

Since, ABC is a right angled triangle and BD⊥AC.

So, ΔDBA ∼ ΔDCB (A-A similarity)

BD2 = AD × DC

(8)2 = 4 × DC

DC = 64/4 = 16 cm

∴ CD = 16 cm

Question: 6

In fig. given, ∠ABC = 90 and BD ⊥ AC. If AC = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, Find BC.

Triangle ABC

Solution:

Given:  BD ⊥ AC. AC = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, and ∠ABC = 90.

We need to find BC,

Since, ΔABC ∼ ΔBDC

BC = 8.1 cm

Question: 7

In the fig. given, DE ∥ BC such that AE = (1/4) AC. If AB = 6 cm, find AD.

Parallelogram triangle ABC

Solution:

Given, DE ∥ BC and AE = (1/4) AC and AB = 6 cm.

We need to find AD.
Since, ΔADE ∼ ΔABC ΔADE ∼ ΔABC

4 × AD = 6

AD = 6/4

AD = 1.5 cm

Question: 8

In the fig. given, if AB ⊥ BC, DC ⊥ BC, and DE ⊥ AC, prove that ΔCED ∼ ΔABC

Triangle

Solution: 

Given, AB ⊥ BC, DC ⊥ BC, and DE ⊥ AC

We need to prove that ΔCED∼ ΔABC

Now,

In ΔABC and ΔCED

∠B = ∠E = 90   (given)

∠A = ∠ECD  (alternate angles)

So, ΔCED∼ΔABC (A-A similarity)

Question: 9

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 

Solution:

Given trapezium ABCD with AB∥DC. OC is the point of intersection of AC and BD.

We need to prove

Now, in ΔAOB and ΔCOD

∠AOB  = ∠COD    (VOA)

∠OAB   = ∠OCD   (alternate angles)

Therefore, ΔAOB ∼ ΔCOD

Therefore,(corresponding sides are proportional)

Question: 10

If Δ ABC and Δ AMP are two right angled triangles, at angle B and M, respectively. Such that ∠MAP = ∠BAC. Prove that:

(i) ΔABC∼ΔAMP

Solution: 

Right angled triangle(i)  Given Δ ABC and Δ AMP are the two right angled triangle.

∠MAP = ∠BAC   (given)

∠AMP∠AMP = ∠B = 90

ΔABC ∼ ΔAMP    (A-A similarity)

(ii) ΔABC – ΔAMP

So, (corresponding sides are proportional)

Question: 11

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

Solution: 

TowerWe need to find the height of PQ.

Now, ΔABC ∼ ΔPQR  (A – A similarity)

PQ = 37.5 m

Question: 12

In fig. given, ∠A  = ∠CED, prove that ΔCAB ∼ ΔCED. Also find the value of x.

Similar triangle

Solution:

Comparing Δ CAB and ΔCED ΔCAB and ΔCED

 (Similar triangles have corresponding sides in the same proportions)

x = 6 cm

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

We need to find the other side.

Let the corresponding side of other triangle be x cm

Since ratio of perimeter = ratio of corresponding side

25 × x = 9 × 15

x = 135/25

x = 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

We need to find AL: DM

Since, both triangles are similar,

Here, we use the result that 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:

TriangleGiven AD = 8 cm, AE = 6 cm, and CE = 9 cm

We need to prove that,

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:

ABC is a triangle in which D is the midpoint of BC, E is the midpoint of AD. BE produced meets AC at X.

ABC is a triangleWe need to prove BE: EX = 3:1

In Δ BCX  and  ΔDCY

∠CBX  = ΔCBY   (corresponding angles)

∠CXB  = ΔCYD   (corresponding angles)

ΔBCX ∼ ΔDCY   (angle-angle similarity)

We know that corresponding sides of similar sides of similar triangles are proportional

In ΔAEX and ΔADY,

∠AEX  = ΔADY  (corresponding angles)

∠AXE = ΔAYD  (corresponding angles)

ΔAEX – ΔADY  (angle-angle similarity)

We know that corresponding sides of similar sides of similar triangles are proportional

Dividing eqn. (i) by eqn. (ii)

BX = 4EX

BE + EX = 4EX

BE = 3EX

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.

Solution:

ABCD is a parallelogram and APQ is a straight line meeting BC at P and DC produced at Q.

ABCD is a parallelogram and APQ is a straight lineWe need to prove, the rectangle obtained by BP and DQ is equal to the rectangle contained by AB and BC. We need to prove that BP x DQ = AB x BC

In ΔABP and ΔQCP,

∠ABP = ΔQCP   (alternate angles as AB ∥ DC)

∠BPA = ΔQPC   (VOA)

ΔABP ∼ ΔQCP (AA similarity)

We know that corresponding sides of similar triangles are proportional

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

Solution:

Triangle ABC(i) In ΔOMA and ΔOLC,

∠AOM = ∠COL (VOA)

∠OMA = ∠OLC (90 each)

ΔOMA ∼ ΔOLC   (A-A similarity)

(ii) Since ΔOMA ∼ ΔOLC by A-A similarity, then

(Corresponding sides of similar triangles 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.

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.

PQRS is a rhombusTo prove,

PQRS is a rhombus

Proof,

In ΔABC, P and Q are the mid points of the sides B and AC respectively

By the midpoint theorem, we get,

PQ ∥ BC, PQ = 1/2 BC.

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 and QR = 1/2 AD = 1/2 BC    (AD = BC)

In ΔBCD,

By the mid-point theorem, we get,

RS∥BC and RS = 1/2 AD = 1/2 BC    (AD = BC)

From above eqns.

PQ = QR = RS

Thus, PQRS is a rhombus.

Question: 20

In an isosceles Δ ABC, 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 × BQ = AC2

Isosceles triangleWe need to prove that ΔAPC ∼ ΔBCQ.

Given ΔABC is an isosceles triangle AC = BC.

Now, AP × BQ = AC2 (given)

AP × BQ = AC × AC

Also, ∠CAB = ∠CBA   (equal sides have angles opposite to them)

180 – CAP = 180 – CBQ

∠CAP = ∠CBQ

Hence, ΔAPC ∼ ΔBCQ (SAS similarity)

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:

Given, girl’s height = 90 cm, speed = 1.2m/sec and height of lamp = 3.6 m

Triangle with length of the shadowWe need to find the length of her shadow after 4 sec.

Let, AB be the lamp post and CD be the girl

Suppose DE is the length of her shadow

Let, DE = x

and BD = 1.2 × 4

BD = 4.8 m

Now, in ΔABE and ΔCDE we have,

∠B = ∠D

∠E = ∠E

So, by A-A similarity criterion,

ΔABE ∼ ΔCDE

3x = 4.8

x = 1.6

Hence, the length of her shadow after 4 sec. Is 1.6 m

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:

Given length of vertical stick = 6 m

Vertical stick of triangle We need to find the height of the tower

Suppose AB is the height of the tower and BC is its shadow.

Now, ΔABC ∼ ΔPCR (B = Q and A = P)

AB = (28 × 6)/4

AB = 42 m

Hence, the 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.

ABC is a right angled triangle

Solution:

Given Δ ACB is right angled triangle and C = 90

We need to prove that ΔABC ∼ ΔADE and find the length of AE and DE.

ΔABC ∼ ΔADE

∠A = ∠A  (common angle)

∠C = ∠E   (90)

So, by A-A similarity criterion, we have

In  ΔABC ∼ ΔADE

Since, AB2 = AC2 + BC2

= 52 + 122

= 132

∴ DE = 36/13 cm

and   AE = 15/13 cm

Question: 24

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.

ABCDEF is a parallelogram triangle

Solution:

Given AB CD EF.

AB = 6 cm, CD = x cm, and EF = 10 cm.

We need to calculate the values of x and y

In ΔADB  and  ΔDEF,

∠ADB = ∠EDF (VOA)

∠ABD = ∠DEF  (alt.Interior angles)

Y = 40/6

Y = 6.67 cm

Similarly, in ΔABE, we have

6.7 × X = 6 × 4

X = 24/6.7

X = 3.75 cm

Therefore, x = 3.75 cm and y = 6.67 cm

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