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 |

[All 5 triangles are congruent.]

Area of Δ ABC : Area of Δ BDE = 5Ã‚Â² : 1

[Triangles are similar. Area is proportional to square of the sides.]

= 25 : 1

Correct answer : (2)

2.

$\stackrel{\u203e}{\mathrm{DE}}$ || $\stackrel{\u203e}{\mathrm{FG}}$ || $\stackrel{\u203e}{\mathrm{BC}}$. Area of ΔADE = Area of DEGF = Area of BCGF . Find the ratio of $\stackrel{\u203e}{\mathrm{AD}}$ to $\stackrel{\u203e}{\mathrm{BF}}$.

a. | $\sqrt{3}-\sqrt{2}$ | ||

b. | $3\sqrt{3}-2\sqrt{2}$ | ||

c. | $2\sqrt{3}+3\sqrt{2}$ | ||

d. | $\sqrt{3}+\sqrt{2}$ |

[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 :

[Step 2, Area of similar triangles proportional to square of the sides.]

AF =

[Step 4.]

AD : AB = 1 :

[Step 3, Area of similar triangles proportional to square of the sides.]

AB =

[Step 5.]

BF = AB - AF =

[Step 5 and 7.]

[Step 8.]

=

[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 |

Ratio of the areas of the two triangles = 2

[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 $\stackrel{\u203e}{\mathrm{EF}}$.

a. | 3 cm | ||

b. | 8 cm | ||

c. | 5 cm | ||

d. | 13 cm |

[Given.]

[The perimeters of two similar triangles are proportional to their corresponding sides.]

[Substitute.]

EF =

[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 |

Ratio of areas = 5

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 |

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

Correct answer : (4)

7.

In the figure, $\stackrel{\u203e}{\mathrm{ST}}$ || $\stackrel{\u203e}{\mathrm{QR}}$ and $\frac{\mathrm{PS}}{\mathrm{SQ}}$ = $\frac{9}{10}$ . What is the ratio of the areas of ΔPST to the area of ΔPQR?

a. | $\frac{81}{361}$ | ||

b. | $\frac{361}{81}$ | ||

c. | $\frac{9}{10}$ | ||

d. | $\frac{10}{9}$ |

[

[

Correct answer : (-1)

8.

In the trapezoid PQRS, $\stackrel{\u203e}{\mathrm{PQ}}$ || $\stackrel{\u203e}{\mathrm{RS}}$ 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 |

[Given.]

[Alternate angles.]

[Alternate angles.]

ΔPOQ ~ ΔROS

[AAA postulate for similarity.]

[Substitute.]

Area of ΔROS =

[Simplify.]

Correct answer : (1)

9.

$m$$\angle $P = $m$$\angle $M. If the ratio of areas of ΔPQR and ΔMLK is 9 : 25 , then find $\frac{\mathrm{PT}}{\mathrm{MY}}$.

a. | 3 : 5 | ||

b. | 9 : 25 | ||

c. | 5 : 3 | ||

d. | 25 : 9 |

[Given.]

[Transitive properties of equality.]

[Alternendo: Property of Proportion.]

[Given.]

ΔPQR ~ ΔMLK

[The ratio of areas of similar triangles is equal to the ratio of square of the corresponding heights.]

[Substitute.]

[Simplify.]

Correct answer : (1)

10.

$\stackrel{\u203e}{\mathrm{SR}}$ || $\stackrel{\u203e}{\mathrm{PQ}}$, 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 |

[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.]

[Formula.]

[Substitute.]

[Vertically opposite angles.]

[Alternate angles.]

ΔMPQ ~ ΔMRS

[AA ~ postulate.]

[The areas of two similar triangles are proportional to the square on the corresponding sides.]

Area of ΔPMQ = 3 × (

[Substitute.]

Area of ΔPMQ = 12 sq.cm

[Simplify.]

Correct answer : (3)