A. 10/3, 5/3

B. 3/4, 8/5

C. 20/3, 10/3

D. 9/10, 8/3

= 2 * (AD)

∴ 81 + 169 = 2 * (AD)

∴ 250 = 2 * (AD)

∴ 125 = AD

∴ 100 = AD

∴ 10 = AD

∴ Median = 10.

Since G divides AD in the ratio 2: 1

So, GA = 2/3 * AD = 2/3 * 10 = 20/3,

GD = 1/3 * 10 = 10/3

A. 10.5

B. 8.5

C. 7.6

D. 9.25

By the theorem of Pythagoras, QR

∴ QR = √289 = 17

Area of the triangle = 1/2 * Product of perpendicular sides = ½ * 8 *15 = 60.

Also area = 1/2 * QR * PS = 60 = 1/2 * 17 * PS = 60

∴ PS = 120/17

Again, PT is the median to the hypotenuse.

∴ PT = 1/2 * hypotenuse = 1/2 * 17 = 8.5

A. 31.5 sq. units

B. 35sq.units

C. 32 sq. units

D. 33.75 sq. units

As ∠ A = ∠ C is cyclic quadrilateral AEDC

ED must be || to AC, Let a = x, b = 2x, c = 3x

∆ EBD is similar to ∆ABC

Ar(∆EBD) / Ar(∆ABC) = a

If area ∆ EBD = A

⇒ Area ∆ ABC = 9A

Thus area of quadrilateral AEDC = 9A – A = 8A

8A = 28 Sq. units

⇒ ∆ ABC = 9A

= 31.5 sq units

A. 15:14

B. 14:15

C. 12:11

D. 7:5

AC= &redic;AB

∴(20 - r) + (21 - r) = 29 ⇒ r = 6

∆ OEC & ∆ OFC are congruent

Area of (Quad FOEC) ≡ 2 (Area ∆ OEC)

= 2 * 1/2 * 15 * 6

= 90 cm

Similarly ∆ AOD & ∆ AOF are congruent ≡ ∆ AOF (RHS)

Area of (Quad ADOF) = 2 (Area ∆AOD)

= 2 * 1/2 * 14 * 6 = 84 cm

Area(QuardFOEC) / Area(QuardADOF) = 90/84 = 15/14

A. 100°

B. 90°

C. 60°

D. 120°

NS is drawn parallel to ML from N

⇒ NS = ML = 6, SO = LO – LS

= LP – MN = 17 – 7 = 10

In ∆ NSO, NS = 6, SO = 10, NO = 8

⇒ NSO is right angled at ∠ N

Thus ∠ SNO = 90 0

⇒ ∠ LPO = 90 0

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A. 120 cm^{2}

B. 100 cm^{2}

C. 104 cm^{2}

D. 106 cm^{2}

Centroid G divide median in the ratio of 2 : 1

AM = 13 ⇒ AG = 2/3 (13) = 26/3 cm

BN = 12 cm,

Area (∆ABN) = 1/2 * BN * AG = 1/2 * 12 * 26/3

= 52 cm

Median BN divides ∆ABC into two triangle S of equal area

∴ Area (∆ABC) = 2(52) = 104 cm

A. 1

B. 2

C. More than one possible value of c exists

D. 3

ab = 4

ab could be 4 * 1 or 2 * 2

221

223 these are the possible triangles.

222

441

44c is a triangle if c is 1.

ab could be 4 * 1 or 2 * 2

221

223 these are the possible triangles.

222

441

44c is a triangle if c is 1.

22c will be a triangle if c is 1, 2 or 3 (trial and error).

- 221 is an acute. 1
^{2}+ 2^{2}> 2^{2} - 222 is an equilateral. (acute)
- 223 is an obtuse. 2
^{2}+ 2^{2}< 3^{2} - 144 is an acute. 1
^{2}+ 4^{2}> 4^{2}

Only the triangle 223 is obtuse. So, the third side has to be 3.

A. 6

B. 7

C. 5

D. 8

This is a counting question. Let us assume x ≤ y ≤ z.

The sum of two sides should be greater than the third.

x = 1, Possible triangle: 1, 7, 7

x = 2, possible triangle: 2, 6, 7

x = 3, possible triangles: 3, 5, 7 and 3, 6, 6

x= 4, possible triangles: 4, 5, 6 and 4, 4, 7

x = 5, possible triangle: 5, 5, 5

Therefore, there are 7 triangles possible.

x = 1, Possible triangle: 1, 7, 7

x = 2, possible triangle: 2, 6, 7

x = 3, possible triangles: 3, 5, 7 and 3, 6, 6

x= 4, possible triangles: 4, 5, 6 and 4, 4, 7

x = 5, possible triangle: 5, 5, 5

Therefore, there are 7 triangles possible.

Suggested Action:

A. 6

B. 11

C. 23

D. 17

Two possibilities: 2 equal sides could add up to 12 or sum of 2 unequal sides = 12.

i.e. Sum of two equal sides = 12

Sum of two unequal sides = 12

⇒ If sum of two equal sides were 12, sides of the triangle should be 6, 6, x.

What are the values that 'x' can take?

Value of x could range from 1 to 11.

So, 11 integer values exist.

⇒ 2 unequal sides adding to 12, could be: 1 + 11, 2 + 10, 3 + 9, 4 + 8 or 5 + 7.

So, how many isosceles triangles are possible with the above combinations?

There are six possibilities of isosceles triangles with the above combination are:

1, 11, 11

2, 10, 10

3, 9, 9

4, 8, 8

5, 7, 7

5, 5, 7

⇒ Triplets such as (1, 1, 11), (2, 2, 10), etc are eliminated as the sum of two smaller values is less than the largest value. These cannot form a triangle.

So, 11 + 6 = 17 possibilities.

A. 31.5

B. 32

C. 33

D. 32.5

r = 2

R = 7 =(1/2 of hypotenuse)

Hypotenuse= 14

r =(a+b−h)/2

2 =(a+b−14)/2

a + b – 14 = 4

a + b = 18

a

a

a

2a

a

Now the 2 roots of this equation will effectively be a, 18 – a. Product of the roots = 64.

Therefore, Area =1/2* product of roots = 32 sq. cms