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

Important for :
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 / 4

= 16 sq. units

And area ΔDEF = 4 ΔPQR

⇒ 4 ΔPQR = 16 = 16 / 4

= 4 sq. units.

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 / 3) × 9 = 3 cm

RT = 6 cm

⇒ RG = (2 / 3) × 6 = 4 cm

∴, 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 . 1/2 AC

[∵ AD = AC cos 60° = 1/2 AC ]

= 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 / sin Θ) = (12 cm / sin 59º) = AC / {sin(121º – Θ)}

Thus, (8 cm / sin &Theta;) ≈ (12 cm / sin 60º)

or, sin Θ ≈ (8 cm × sin 60º) / 12 cm

= (2/3) × (√3/2) = (1/√3) = 0.577

∴ cos θ = √{1 – (1/3)} = √(2/3)

∴  sin (121° – Θ) ≈ sin (120° – Θ)

= sin 120° cosΘ – cos120° sin Θ

= (√3 / 2) × √(2 / 3) – (–1 / 2) × (1 /√ 3)

= (1 / √2) + (1 / 2√3) = 0.996

= Now, {AC / sin (121º – Θ)} / sin Θ

or, AC = {8 cm sin (120º – Θ)} / sin Θ

= {(8 × 0.996) / 0.0577} = 13.809 ≈ 14 cm

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 = –120 / 120 = –1

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

∴ (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

Is it possible to find the height of a tower by merely observing the length of its shadow and the position of the sun? Is it within our powers to deduce the dimensions of a cylindrical container by measuring the height and circumference of the heap of grain that has been poured out of it? One doesn’t need to be a Sherlock Holmes to crack these mysteries. Anyone with an elementary knowledge of geometry will be able to answer these questions. Geometry plays a major role in SSC Exams 2 to 3 questions generally being asked in SSC CGL Tier 1 and SSC CHSL and questions number increased as per the total number of question in particular exams like in SSC CGL Tier 2, that value increased by 5 to 10 questions.
With geometry, one enters, quite literally, into the real world – the world of dimensions. We are not dealing with abstract equations or mere numbers but grappling with concepts like size, shape, location, direction and orientation of objects that figure prominently in real life.

Coordinate geometry and trigonometry find wide applications in varied fields like aviation, defense, navigation and prospecting. Many abstract designs that modern artists and architects create, and we so admire, are nothing more than creatively put together collections of geometric shapes. The concepts of area and perimeter are vital not just to people dealing in real estate but to anyone planning to make optimum use of any available space. As you can see, there is hardly any aspect of the real world that is not influenced by geometry.

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: