# Free Triangles 01 Practice Test - 10th Grade

### Question 1

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:

11 m

31 m

15 m

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=√BD2−DE2 =√(17)2−82

=√289−64=√225=15 m∴AC=BE=15 m

### Question 2

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 –

ABC

AEF

BFD

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.

⇒△DEF≅△AEF

⇒△DEF≅△DEC

⇒△DEF≅△BFD by using SSS postulate.

### Question 3

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

1:3

3:1

1:9

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.

### Question 4

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

49:64

7:8

16:49

#### 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

### Question 5

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]

(x−3)2x=(x−2)(2x+3)(x−3)×(2x+3) = (x−2)×2x

2x2 - 3x - 9 = 2x2 - 4x

x=9

So AC=2x=18

### Question 6

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

### Question 7

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

BC.AB

BC2

BD2

AB.AC

#### SOLUTION

Solution :B

Consider △ADB and △ABC

∠BAD = ∠BAC [common angle]

∠BDA = ∠ABC [ 90∘]

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

So, ABAC=ADAB ⇒ AB2 = AC.AD ---(I)

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 △ADB, AB2 = AD2 + DB2 and in △BDC, CB2 = CD2 + DB2 [Pythagoras theorem]

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

AB2 - BC2 = AD2 - CD2 ⇒ AB2 + CD2 = AD2 + BC2 --------------(IV)

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

⟹ AB2BC2 - 1 = AD2BC2 - CD2BC2

⟹ ADDC - 1 = AD2−CD2BC2 [Using (III)]

⟹AD−DCDC = (AD−CD)(AD+CD)BC2

⟹1DC = ACBC2

⟹ 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.

ACBC = BCDC

BC2 = AC×DC

### Question 8

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

AB×BC=AD×DC

AD2 + CD2 = BD2 + AB2

AB2 + CD2 = BC2 + AD2

#### SOLUTION

Solution :C

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

AB2=AD2+BD2 ...(1)

BC2=CD2+BD2 ...(2)

Subracting (2) from (1), we get,

AB2=AD2+BD2BC2=CD2+BD2(−) (−) (−)–––––––––––––––––––––AB2−BC2=AD2−CD2

Rearranging, we get,

⇒AB2+CD2=AD2+BC2

### Question 9

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

#### 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.

### Question 10

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 = _____.

√2d

d√5

√5d

2√2d

#### 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=AB√5=d√5