To get the best deal on Tutoring, call 1-855-666-7440 (Toll Free)

Area and Perimeter of a Triangle Worksheet

Area and Perimeter of a Triangle Worksheet
  • Page 1
 1.  
The triangles BDE, EFG, GHI, IJK and KLC are congruent to each other. These are similar to ΔABC. What is the ratio of the area of ΔABC to the area of ΔBDE?


a.
5 : 1
b.
25 : 1
c.
125 : 1
d.
10 : 1


Solution:

BC = 5 BE
[All 5 triangles are congruent.]

Area of Δ ABC : Area of Δ BDE = 5² : 1
[Triangles are similar. Area is proportional to square of the sides.]

= 25 : 1


Correct answer : (2)
 2.  
DE || FG || BC. Area of ΔADE = Area of DEGF = Area of BCGF. Find the ratio of AD to BF.


a.
3 -2
b.
33 - 22
c.
23 + 32
d.
3 +2


Solution:

Area of Δ ADE = Area of DEGF = Area of BCGF
[Given.]

Area of Δ AFG = 2 Area of Δ ADE
[Step 1.]

Area of Δ ABC = 3 Area of Δ ADE
[Step 1.]

AD : AF = 1 : 2
[Step 2, Area of similar triangles proportional to square of the sides.]

AF = 2AD
[Step 4.]

AD : AB = 1 : 3
[Step 3, Area of similar triangles proportional to square of the sides.]

AB = 3AD
[Step 5.]

BF = AB - AF = (3 -2)AD
[Step 5 and 7.]

AD / BF =13 -2
[Step 8.]

= 3 +2
[Rationalise the denominator.]


Correct answer : (4)
 3.  
The similarity ratio of two similar triangles is 2 : 3. What is the ratio of their areas?
a.
3 : 2
b.
9 : 4
c.
2 : 3
d.
4 : 9


Solution:

If the similarity ratio of two similar triangles is a : b, then the ratio of their areas is a2 : b2.

Ratio of the areas of the two triangles = 22 : 32 = 4 : 9
[Step 1.]


Correct answer : (4)
 4.  
The perimeters of two similar triangles ABC and DEF are 30 cm and 18 cm respectively. If BC = 5 cm, then find the measure of EF.
a.
3 cm
b.
8 cm
c.
5 cm
d.
13 cm


Solution:

ΔABC ~ ΔDEF
[Given.]

Perimeter of ΔABC / Perimeter of Δ DEF = BC / EF
[The perimeters of two similar triangles are proportional to their corresponding sides.]

30 / 18 = 5EF
[Substitute.]

EF = 18 × 530
[Cross Product property.]

EF = 3 cm
[Simplify.]


Correct answer : (1)
 5.  
The ratio of the corresponding sides of two similar triangles is 5 : 11. What is the ratio of their areas?
a.
1:1
b.
121 : 25
c.
25 : 121
d.
5 : 11


Solution:

The ratio of areas of similar triangles is equal to the ratio of squares of the corresponding sides .

Ratio of areas = 52 : 112 = 25 : 121


Correct answer : (3)
 6.  
The similarity ratio of two similar triangles is 7 : 13. What is the ratio of their perimeters?
a.
49 : 169
b.
13 : 7
c.
169 : 49
d.
7 : 13


Solution:

If the similarity ratio of two similar triangles is a : b, then the ratio of their Perimeters is a : b.

Ratio of the perimeters of the two triangles = 7 : 13.


Correct answer : (4)
 7.  
In the figure, ST || QR and PS SQ = 9 10 . What is the ratio of the areas of ΔPST to the area of ΔPQR?

a.
81 361
b.
361 81
c.
9 10
d.
10 9


Solution:

In ΔPST and ΔPQR, PST = PQR
[ST || QR, Corresponding angles are congruent.]

P = P Þ ΔPST ~ ΔPQR
[ PS / SQ = 9 / 10.]

PS / PQ = 9 / 19

Area of ΔPSTArea of ΔPQR = PS2PQ2 = 92 192 = 81 / 361


Correct answer : (-1)
 8.  
In the trapezoid PQRS, PQ || RS and PQ = 2RS. If the area of ΔPOQ is 168 sq.cm, then find the area of ΔROS.

a.
42 sq.cm
b.
28 sq.cm
c.
12.96 sq.cm
d.
84 sq.cm


Solution:

PQ || RS
[Given.]

OSR = OQP
[Alternate angles.]

ORS = OPQ
[Alternate angles.]

ΔPOQ ~ ΔROS
[AAA postulate for similarity.]

Area of ΔPOQArea of ΔROS = PQ2SR2

168 cm2Area of ΔROS = (2RS)2(RS)2= 4
[Substitute.]

Area of ΔROS = 168 / 4 = 42 sq.cm
[Simplify.]


Correct answer : (1)
 9.  
mP = mM. If the ratio of areas of ΔPQR and ΔMLK is 9 : 25, then find PT MY.

a.
3 : 5
b.
9 : 25
c.
5 : 3
d.
25 : 9


Solution:

PQ / PR = 1 and ML / MK = 1
[Given.]

PQ / PR = ML / MK
[Transitive properties of equality.]

PQ / ML = PR / MK
[Alternendo: Property of Proportion.]

P = M
[Given.]

ΔPQR ~ ΔMLK

Area of ΔPQRArea of ΔMLK = PT2MY2
[The ratio of areas of similar triangles is equal to the ratio of square of the corresponding heights.]

PT2MY2 = 9 : 25
[Substitute.]

PT / MY = 3 : 5
[Simplify.]


Correct answer : (1)
 10.  
SR || PQ, area of ΔSMP = 6 sq.cm, area of ΔSQR = 9 sq.cm. What is the area of ΔPMQ ?


a.
6 sq.cm
b.
15 sq.cm
c.
12 sq.cm
d.
14 sq.cm


Solution:

Area of ΔPSR = Area of ΔSQR
[ΔPSR and ΔQSR have same base SR and are between the same parallel lines.]

Area of ΔPSR = 9 sq.cm
[Substitute.]

Area of ΔSMP + Area of ΔSMR = 9 sq.cm
[Area of ΔPSR = Area of ΔSMP + Area of ΔSMR.]

Area of ΔSMR = (9 - 6) sq.cm
[Substitute.]

Area of ΔSMR = 3 sq.cm

Area of ΔMQR = Area of ΔQSR - Area of ΔSMR

Area of ΔMQR = (9 - 3) sq.cm = 6 sq.cm
[Substitute and simplify.]

Area of ΔMQRArea of ΔSMR = 12 × MQ × RK12 × SM × RK
[Formula.]

6 / 3 = MQ / MS
[Substitute.]

SMR = PMQ
[Vertically opposite angles.]

MRS = Ð MPQ and MSR = PQM
[Alternate angles.]

ΔMPQ ~ ΔMRS
[AA ~ postulate.]

Area of ΔMPQArea of ΔMRS = MQ2 MS2
[The areas of two similar triangles are proportional to the square on the corresponding sides.]

Area of ΔPMQ = 3 × (6 / 3)2
[Substitute.]

Area of ΔPMQ = 12 sq.cm
[Simplify.]


Correct answer : (3)

*AP and SAT are registered trademarks of the College Board.