# Free Triangles 02 Practice Test - 10th Grade

ABC is such thatAB=3 cm,BC=2 cm and CA=2.5 cm. DEF is similar to ABC. If EF=4 cm, then the perimeter of DEF is ____.

A.

7.5 cm

B.

15 cm

C.

22.5 cm

D.

30 cm

#### SOLUTION

Solution : B

ABDE =ACDFBCEF2412

DE=2×AB=6 cm,DF=2×AC=5 cm.

Perimeter of ΔDEF=(DE+EF+DF)
=6+4+5=15 cm.

ABC is a triangle and DE is drawn parallel to BC cutting the other sides at D and E. If AB=3.6 cm,AC=2.4 cm and AD=2.1 cm, then AE =  ______.

A. 1.4 cm
B. 1.8 cm
C. 1.2 cm
D. 1.05 cm

#### SOLUTION

Solution : A By Basic Proportionality Theorem,
Subtituting values:
3.62.1=2.4AE
AE=2.1×2.43.6=1.4 cm

In the adjoining figure, if BC=a,AC=b,AB=c and CAB=120,
then which of the following is the correct relation? A.

a2 = b2c2 + 2bc

B.

a2 = b2c2 - 2bc

C.

a2 = b2c2 + bc

D.

a2 = b2c2 - bc

#### SOLUTION

Solution : C

In CDB,

BC2=CD2+BD2 [Pythagoras theorem]

BC2=CD2+(DA+AB)2

BC2=CD2+DA2+AB2+(2×DA×AB) ...(i)

CD2+DA2=AC2 ...(ii) [Pythagoras Theorem]

Here, CAB=120 (given)

Substituting  the values from (ii) & (iii) in (i) we get,

BC2=AC2+AB2+(AC×AB)

a2=b2+c2+bc

Alternatively,

Since  A is an obtuse angle in  ABC so,

a2=b2+c2+bc

In  ABC, A and C are given, the number of triangles that can be constructed with this data is -

A.

One

B.

Two

C.

Three

D.

Infinite

#### SOLUTION

Solution : D

S=180o(A+C)

The angles of the traingle are known. With the given measurement of the angles, we can draw infinite number of traingles (Similar triangles).

In triangle ABC, D is a point on AB and E is a point on AC such that DE || BC. If ADAB = AEx, Then x is

___.

#### SOLUTION

Solution :

In ABC,

DE BC      (Given)

Therefore, ADAB = AEAC    [By Basic Proportionality Theorem]

Comparing the above with ADAB = AEx

x = AC

In below shown figure, PSSQ= PTTR andPST= PRQ. Then ΔPQR is a/an __ triangle. #### SOLUTION

Solution :

It is given that

PSSQ = PTTR

So, ST||QR     [By converse of Basic Proportionality Theorem]

, PST=PQR (Corresponding Angles)

Also, it is given that

PST = PRQ

So,PRQ =  PQR

Therefore, PQ = PR (Sides opposite the equal angles)

i.e., ΔPQR is an isosceles triangle.

ABCD is a square. Equilateral triangles ACF and ABE are drawn on the the diagonal AC and side AB respectively. Find area of ACF : area of ABE.

A.

2:1

B.

2:1

C.

4:1

D.

8:1

#### SOLUTION

Solution : B Let AB be a units long.

We know that diagonal of a square = 2× side length
AC=2×AB

Now,  ABE and ACF are equilateral triangles.

Each angle of both  ABE and ACF is 60.

ABE ACF (by AAA similarity criterion)

So, ar(ACF)ar(ABE)=AC2AB2=2a2a2=21

Find the ratio of area of the equilateral triangles whose sides are 3 and 4 units.

A.

9 : 25

B.

9: 16

C.

16: 25

D. None of the above

#### SOLUTION

Solution : B

Let the triangles be ABC and DEF

Then  AB=3 units and DE=4 units.

Each angle of both ABC and DEF is 60.

ABE ACF
( AAA similarity)

So,
ar(ABC)ar(DEF)=BC2EF2=3242=916

In a right ABC, a perpendicular BD is drawn on  to the largest side from the opposite vertex. Which of the following does not give the ratio of the areas of ABD and ACB? A.

(ABAC)2

B.

C.

(BDCB)2

D.

#### SOLUTION

Solution : D

Consider  ΔABD and ΔACB: ∠BDA =  ∠ABC     [ 90]

By AA similarity criterion,
ABD ~ ACB

Hence,

A ladder is resting on a wall of height       107m such that the foot of the ladder when placed 107m away from the wall, half of the ladder is extending above the wall. When the tip of the ladder is placed on the tip of the wall, how far is the foot of the ladder from the wall?

A.

100 7 m

B.

707 m

C.

70 m

D.

7011 m

#### SOLUTION

Solution : C Consider the above figure:

AD is the part of the ladder in the first case and AC is the ladder in the second case

AB = BD = 10 7m          [given]

By Pythagoras Theorem in  ABD,

= 2 (107)2

AC = 2AD = 20 27

Now,
In  ΔABC,
AC2 =  AB2 +  BC2

⇒ BC2=AC2AB2
=(2072)2 -  (107)2
=(107)2[(22)21]
= (107)2×7

BC=107×7=70 m