# Free Triangles 01 Practice Test - 10th Grade

If the distance between the top of two trees of height 20 m and 28 m is 17 m, then the horizontal distance between the trees is:

A.

11 m

B.

31 m

C.

15 m

D.

9 m

#### SOLUTION

Solution : C

Let AB and CD be two trees such that AB=20 m, CD=28 m & BD=17 m

Draw BE  parallel to AC. Then,
ED=8 m.
By applying Pythagoras' theorem,
BE2+DE2=BD2.

BE=BD2DE2          =(17)282
=28964=225=15 m

AC=BE=15 m

D, E, F are the mid points of the sides BC, CA and AB respectively of an equilateral  ABC. Then  DEF is congruent to triangle –

A.

ABC

B.

AEF

C.

BFD

D.

CDE

#### SOLUTION

Solution : B, C, and D

Consider AEF
AE = AF (As ABC is equilateral and E and F are midpoints of sides AC and AB respectively)
AFE=AEF=A=60o
AEF is equilateral.

Similarly we can prove that DEC  and BFD are equilateral.
Since DEF share a common side with each of the other three triangles, DEF is also equilateral.

DEFAEF
DEFDEC
DEFBFD by using SSS postulate.

The ratio of the corresponding sides of two similar triangles is 1 : 3. The ratio of their corresponding heights is _________.

A.

1:3

B.

3:1

C.

1:9

D.

9:1

#### SOLUTION

Solution : A

If two triangles are similar, then the ratio of the corresponding sides are equal.
Hence,
the ratio of heights = the ratio of sides = 1 : 3.

The areas of two similar triangles are 49 cm2 and 64 cm2 respectively. The ratio of their corresponding sides is ____.

A.

49:64

B.

7:8

C.

16:49

D. None of the above

#### SOLUTION

Solution : B

The ratio of areas of two similar triangles is equal to ratio of squares of corresponding sides.
4964=(Side of  first 1st triangle)2(Corresponding side of 2nd triangle)2

78=Side of  first 1st triangleCorresponding side of 2nd triangle

Ratio of corresponding sides =7:8

If LM  AB, AL=x-3, AC=2x, BM=x-2, BC=2x+3. What is value of AC?

__

#### SOLUTION

Solution :

In ABC,

It's given LM  AB,
So, ALCA=BMCB      [By BPT]
(x3)2x=(x2)(2x+3)

(x3)×(2x+3) = (x2)×2x

2x2 - 3x - 9 = 2x24x

x=9
So AC=2x=18

A ladder is placed against a wall such that its foot is at a distance of 2.5 m from the wall and its top reaches 6 m above the ground. Find the length of the ladder.

__

#### SOLUTION

Solution :

By applying Pythagoras theorem:

AB2+BC2=AC2

62+2.52=AC2

We get, AC = 6.5m

ABC is a right angled triangle, right angled at B. BD is perpendicular to AC. What is AC . DC?

A.

BC.AB

B.

BC2

C.

BD2

D.

AB.AC

#### SOLUTION

Solution : B

∠BDA = ∠ABC                [ 90]

Therefore by AA similarity criterion, ADB and  ABC are similar.

Similarly, BDC and  ABC are similar.

So, BCAC =  DCBC  ⇒  BC2 = AC.DC   ---(II)

Dividing (I) and (II) and cancelling out AC, we get,   (ABBC)2 =   ADDC  ----(III)

Also, In ADBAB2AD2 + DB2 and in  BDCCB2 = CD2DB2 [Pythagoras theorem]

Subtracting these two equations above and cancelling off DB2 on both sides, we get

Dividing this equation with BC2 on both sides,  AB2BC2 +  CD2BC2 =  AD2BC2 +  BC2BC2   ⇒  AB2BC2CD2BC2AD2BC2 + 1

AB2BC2 - 1 = AD2BC2 - CD2BC2

1DCACBC2

BC2 = AC.DC

Alternatively,

Consider ABC and BDC,

ABC=BDC=90

C=C    [Common angle]

Therefore by AA similarity criterion, ABC and  BDC are similar.

ACBCBCDC

BC2 = AC×DC

ABC is a triangle, right angled at B. BD is a perpendicular to AC as shown. Which of the following is true?

A.

B.

AD2 + CD2 = BD2 +  AB2

C.

AB2 + CD2 = BC2 +  AD2

D. None of the above

#### SOLUTION

Solution : C

Since ΔADB and ΔDBC are right triangles, using Pythagoras theorem, we can write:

BC2=CD2+BD2  ...(2)

Subracting (2) from (1), we get,

Rearranging, we get,

Refer to the following figure. Three squares are constructed on each side of the triangle as shown, with the length of each square equal to the side on which it is constructed. If the largest side is 13, sum of the areas of the three squares is __. The angle opposite to the blue colour square is the right angle.

#### SOLUTION

Solution :

Let the other two sides of the triangle be a and b. Since the largest side is the hypotenuse, we have, by Pythagoras theorem,

Hypotenuse2 = a2 +  b2

132 = a2 + b2 = 169           ------------------- (I)

Now,

Sum of the areas of the three squares = (side of yellow square)2 + (side of brownish square)2  +

(side of blue square)2

Sum of the areas of the three squares =a2  + b2 + 132 = 132 + 132 (from (I))

Sum of the areas of the three squares = 169 + 169 = 338.

AB is the diameter of the circle with centre O. P is a point on the circle such that PA = 2PB. If AB = d, then BP = _____.

A.

2d

B.

d5

C.

5d

D.

22d

#### SOLUTION

Solution : B

Let O be the centre of the circle. The angle subtended by an arc at the center is twice the angle subtended by it at any point on the remaining part of the circle. (here that point is P).

AOB=180
APB=AOB2=90

APB is a triangle, right angled at P.

By pythagoras theorem,
AB2=AP2+PB2=(2PB)2+PB2        =5 PB2

AB=5 PB

PB=AB5=d5