# Similar Triangles Worksheet

Similar Triangles Worksheet
• Page 1
1.
ABCD is a square with each side 2 cm long. $\stackrel{‾}{\mathrm{BD}}$ and $\stackrel{‾}{\mathrm{AC}}$ are extended to E and F such that $\stackrel{‾}{\mathrm{EF}}$ || $\stackrel{‾}{\mathrm{DC}}$. EF = 8 cm. Find the area of CDEF.

 a. 12 square cm b. 15 square cm c. 14 square cm d. 16 square cm

#### Solution:

ΔOEF is right angled at O.
[Diagonals of square are perpendicular bisectors.]

Area of CDEF = Area of ΔOEF - area of ΔODC

Draw ONM perpendicular to CD and EF.

ON / OM = FN / CM
[ΔOMC ~ ΔONF.]

OM = BC2 = 22 = 1

FN = EF2 = 82 = 4

ON = OM × FN / CM = 1 × 41 = 4
[From step 6, the value of FN = 4. OM = CM = 1.]

Area of CDEF = 1 / 2 × EF × ON - 1 / 2 × CD × OM
[Step 2.]

= 1 / 2 × 8 × 4 - 1 / 2 × 2 × 1
[Substitute EF = 8, ON = 4, CD = 2, OM = 1.]

= 15 square cm

2.
On a given tennis court, the distance between the base line and the net is 10 m and the distance between the service line and the net is 5 m. The net is 0.9 m high. A player serves the ball directly above the base line. (Assume the ball travels in a straight line.) What is the minimum height that the ball can be hit at so as to ensure the ball lands in the service area?

 a. 1.65 m b. 2.7 m c. 5.9 m d. 1.8 m

#### Solution:

Mark the figure with A, B, C, D, E as shown below.

Consider triangles ADE and ABC. They are similar.

DE / BC=AE / AC

0.9y=515

y = 0.9 × 155 = 2.7 m

So, the minimum height that the ball can be hit is 2.7 m.

3.
Which of the following are always similar?
I. All equilateral triangles. II. All circles III. All regular hexagons. IV. All isosceles triangles.V. All parallelograms.
 a. I, and IV only b. I, II, and III only c. I, II, III, IV, and V d. I, III, and V only

#### Solution:

The interior angles for any two equilateral triangles are always congruent. The sides are also proportional. So, any two equilateral triangle will be similar.
{Each angle measure is 60, all the sides are equal.]

All circles are similar figures.

The interior angles for any two regular hexagons are always congruent. The sides are also proportional. So, any two regular hexagons will be similar.
{Each angle measure is 120, all the sides are equal.]

The corresponding angles of two isosceles triangles need not be congruent. So, two isosceles triangles need not be similar.

The corresponding angles of two parallelograms need not be congruent. So, two parallelograms need not be similar.

So, only 1, 2 and 3 are right.

4.
ΔAPQ $\sim$ ΔACB. If AQ = 2 cm, PC = 5 cm, QB = 13 cm and BC = 15 cm, then find the length of $\stackrel{‾}{\mathrm{AP}}$ rounded to the nearest decimal.

 a. 2.27 cm b. 1.77 cm c. 1.27 cm d. 0.77 cm

#### Solution:

AP / AC = AQ / AB
[ΔAPQ ~ ΔACB.]

AP / AP+5 = 2 / 2+13
[Substitution.]

15(AP) = 2(AP) + 10
[Cross multiplication.]

13(AP) = 10
[Simplify.]

AP = 0.77
[Divide by 13 on both sides.]

The length of AP is 0.77 cm.

5.
ΔABC $\sim$ ΔPQR, find the measure of $\angle$ABC.

 a. 70 b. 30 c. 80 d. 40

#### Solution:

mPQR = 180 - (30 + 70)
[Triangle Angle - Sum Theorem.]

mPQR = 80
[Simplify.]

mABC = mPQR
[Corresponding angles of similar triangles are Congruent.]

mABC = 80
[Substitute.]

6.
In the figure, $\stackrel{‾}{\mathrm{AB}}$ || $\stackrel{‾}{\mathrm{CD}}$, $x$ = 75 and $y$ = 145. Find $m$$\angle$OCD.

 a. 80 b. 70 c. 45 d. 35

#### Solution:

mAOB = mCOD
[Vertically opposite angles.]

mOAB = mODC
[Alternate angles.]

ΔAOB ~ ΔODC
[AA similarity postulate.]

mAOB = 180 - mAOC

mAOB = 180 - 145 = 35
[Substitute.]

mABO = 180 - (35 + 75)
[Triangle Sum Theorem.]

mABO = 70
[Simplify.]

mOCD = mABO
[Corresponding angles of similar triangles are congruent.]

mOCD = 70
[Substitute.]

7.
ΔAED $\sim$ ΔACB and $x$ = 46, $y$ = 74. Find the measure of $\angle$ABC.

 a. 74 b. 46 c. 60 d. 56

#### Solution:

mADE = 180 - (mDAE + mDEA)
[Triangle Angle sum Theorem.]

mADE = 180 - (46 + 74) = 60
[Substitute.]

[Corresponding angles of congruent triangles are congruent.]

mABC = 60
[Substitute.]

8.
$\angle$CFD = 35°, find $\angle$FAE.

 a. 35o b. 72o c. 55o d. 60o

#### Solution:

CDF AEF = 90o

CFD AFE
[Vertically opposite angles.]

ΔAEF ~ ΔCDF
[AA similarity postulate.]

FCD = 180o - (CFD + CDF)
[Triangle Sum Theorem.]

FCD = 180o - (35o + 90o)
[Substitute.]

FCD = 55o
[Simplify.]

FAE FCD
[Corresponding angles of similar triangles are congruent.]

FAE = 55o

9.
CD and GH are angular bisectors of $\angle$ACB and $\angle$FGE and ΔABC $\sim$ ΔFEG. IF $\frac{AC}{FG}$ = $\frac{3}{4}$, then find $\frac{CD}{GH}$.

 a. $\frac{16}{9}$ b. $\frac{3}{4}$ c. $\frac{4}{3}$ d. $\frac{9}{16}$

#### Solution:

ΔDBC ~ ΔHEG
[AA similarity postulate.]

Area of ΔDBC Area of ΔHEG = CD2 GH2
[The areas of two similar triangles are proportional to the square of their corresponding sides.]

Area of ΔADC Area of ΔFHG = CD2 GH2
[The areas of two similar triangles are proportional to the square of their corresponding sides.]

Area of ΔDBC Area of ΔHEG = Area of ΔADC Area of ΔFHG
[Transitive property.]

= Area of ΔDBC + Area of ΔADC Area of ΔHEG + Area of ΔFHG
[Property of proportions.]

= Area of ΔABC Area of ΔFEG

= AC2 FG2
[The areas of two similar triangles are proportional to the square of their corresponding sides.]

CD2 GH2 = AC2 FG2
[Transitive property.]

CD / GH = AC / FG
[Simplify.]

CD / GH = 3 / 4
[Substitute.]

10.
ABCD is a parallelogram $\stackrel{‾}{\mathrm{FD}}$ || $\stackrel{‾}{\mathrm{AB}}$, $\angle$BAE = 50o, $\angle$CFB = 54o. Find $\angle$BEA.

 a. 76o b. 50o c. 40o d. 54o

#### Solution:

CFB = ABE = 54o
[Alternate angles.]

BAE = FCB = 50o
[Opposite angles of a parallelogram are congruent.]

ΔABE ~ ΔCFB
[AA similarity postulate.]

BEA = 180o - (ABE + BAE)
[Triangle Angle Sum Theorem.]

BEA = 180o - (54o + 50o)
[Substitution.]

BEA = 76o
[Simplify.]