Geometry Questions and Answers PDF for SSC CGL Tier 1, SSC CHSL, 2018, 2019 | Triangle Quiz 7

Directions: Read the following questions carefully and choose the right answer:
1
In the above figure, if area of triangle ABC is 64 sq. units, then find the area of triangle PQR, where D, E and F are mid points of sides of ΔABC and P, Q and R are midpoints of sides of ΔDEF.

» Explain it
A
Given that,

D, E and F are midpoints of BC, CA and AB and P, Q and R are midpoints of EF, FD and DE

we know that,

Area of ΔABC = 4 ΔDEF

But area of ABC = 64 sq. cm.

4 ΔDEF = 64 ⇒ ΔDEF =  64  = 16 sq. units
4

And area ΔDEF = 4 ΔPQR
⇒ 4 ΔPQR = 16 =  16  = 4 sq. units.
4

Hence, option A is correct.

2
Two medians PS and RT of ΔPQR intersect at G at right angles. If PS = 9 cm and RT = 6 cm, then the length of RS in cm is
» Explain it
C
PS = 9 cm



⇒ GS =  1  × 9 = 3 cm
3

RT = 6 cm
⇒  RG =  2  × 6 = 4 cm
3

∴   RS =  32 + 42  = 9 + 16  = 5 cm

Hence, option C is correct.

3
In the adjoining figure, if BC = a, AC = b, AB = c and ∠CAB = 120° , then the correct relation is :

» Explain it
C
Since ∠A is an obtuse angle in ∆ ABC, so

BC2 = AB2 + AC2 + 2 AB . AD

= AB2+ AC2 + 2 AB 1  AC
2

[∵   AD = AC cos 60° =  1  AC ]
2

= AB2 + AC2 + AB . AC

   a2 = b2 + c2 + bc.

Hence, option C is correct.
 
4
The length of side AB and side BC of a scalene triangle ABC are 12 cm and 8 cm respectively. The value of angle C is 59°. Find the length of side AC.
» Explain it
C


Given, AB = 12 cm, BC = 8 cm,

∠C = 59°

Let ∠A = Θ

∴   ∠B = 180° – (59° + Θ) = 121° – Θ

Now, let us see the choices. If AC = 12 cm, triangle would not be scalene. Hence, option A is ruled out. If AC = 10 cm, AB will become the largest side and ∠C the largest angle. But ∠C = 59°. Hence option B is ruled out. So, AC is either 14 cm or 16 cm. In any case, ∠B will be the largest angle and ∠A (say Θ) the smallest:

Also, ∠B = 180° – (59° + Θ) = 121° – Θ

By sine formula,

8 cm  =  12 cm  =  AC
sin Θ sin 59° sin (121° – Θ)

Thus,  8 cm  ≈  12 cm
sin Θ sin 60°

or, sin Θ ≈  8 cm × sin 60°  =  2  × 
3
 =  1  = 0.577
12 cm 3 2 3

∴   cos Θ = 
1 –  1
3
 = 
2
3

∴  sin (121° – Θ) ≈ sin (120° – Θ) = sin 120° cosΘ – cos120° sin Θ
=  
3
 × 
2
3
 –  –1  ×  1  =  1  +  1  = 0.996
2 2 3 2
2 3

Now,   AC  ≈  8 cm
sin (121° – Θ) sin Θ

or, AC =  8 cm sin(120° – Θ)  =  8 × 0.996  = 13.809 ≈ 14 cm
sin Θ 0.577

Hence, option C is correct.

5
The cordinates of the incentre of the triangle whose sides are 3x – 4y = 0, 5x + 12y = 0 and y – 15 = 0, are
» Explain it
B
We have




3x – 4y ≡ 0     ...(i)

5x + 12y ≡ 0     ....(ii)

y – 15 ≡ 0    ...(iii)

From (i) and (ii), A = (0, 0)

From (i) and (iii), B = (20, 15)

From (ii) and (iii), C = (–36, 15)

BC =   (20 + 36)2 + (15 – 15)2  = 56

AB =  (20)2 + (15)2  = 25

AC =  362 + 152  = 39

Let (α, β) be the incentre co-ordinates of ΔABC
α =  56 × 0 + 39 × 20 + 25(–36)
56 + 39 + 25

=   –900 + 780  =  –120  = –1
120 120

β =   56 × 0 + 39 × 15 + 25(15)  = 8
56 + 39 + 25

∴   (α, β) (–1, 8)

Hence, option B is correct.
 

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Geometry Questions for SSC 2018 2019 Exams i.e. SSC CGL Tier 1, SSC CHSL, CGL Tier 2

 
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At Smartkeeda you familiarized yourself with the key concepts and improved your problems-solving abilities. Practice and tests are important to optimize your preparation. Take the tests on Testzone to improve your problem-solving skills. Questions that have appeared in the previous SSC CGL and SSC CHSL exams are also in the given quizzes. Through them you can understand where the focus lies in an examination environment. The detailed solutions and short tricks of the questions may provide some alternate strategies that can help you improve your speed accuracy. You can practice individual exercise practice and you can analyze yourself topic wise as Trigonometry, Height and Distance, Quadrilateral and Polygon, Triangle, Circle, and Lines and Angles. Here are some Important Geometry Chapters Links:
 
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I hope this above article and exercise will help you to crack your SSC CGL 2018 or 2019 and SSC CHSL 2018 as well.
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