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Geometry Exercise 1

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Geometry 1

  • This online quiz will test your knowledge of Geometry in Quantitative Aptitude.
  • This Online Test is useful for academic and competitive exams.
  • Multiple answer choices are given for each question in this test. You have to choose the best option.
  • After completing the test, you can see your result.
  • There is no negative marking for wrong answers.
  • There is no specified time to complete this test.

The sum of the interior angles of a polygon is 1620°. The number of sides of the polygon are

The sum of the interior angles of a polygon of n sides is given by the expression (2n – 4) π/2

⇒ geometry-37703.png

geometry-37697.png

⇒ 2n = 22

⇒ n = 11

Thus the no. of sides of the polygon are 11.

A cyclic parallelogram having unequal adjacent sides is necessarily a :

It is a rectangle.

(In a cyclic parallelogram each angle is equal to 90°. So, it is definitely either a square or a rectangle. Since the given cyclic parallelogram has unequal adjacent sides, it is a square.)

If the angles of a triangle are in the ratio 5 : 3 : 2, then the triangle could be :

Let the angles of the triangle be 5x, 3x and 2x.

Now, 5x + 3x + 2x = 180°

⇒ 10x = 180

⇒ x = 18

⇒ Angles are 36, 54 and 90°

Given ∆ is right angled.

If one of the diagonals of a rhombus is equal to its side, then the diagonals of the rhombus are in the ratio:

Let the diagonals of the rhombus be x and y and the its sides be x
geometry-38023.png
Now, geometry-38017.png
⇒ geometry-38011.png
⇒ 3x² = y²

⇒ geometry-38005.png
⇒ y : x geometry-37999.png

In the given figure, ∠ ABC and ∠ DEF are two angles such that BA ⊥ ED and EF ⊥ BC, then find value of ∠ ABC + ∠ DEF.
37204.png

Since the sum of all the angle of a quadrilateral is 360°

We have ∠ ABC + ∠ BQE + ∠ DEF + ∠ EPB = 360°

∴ ∠ ABC + ∠ DEF = 180°

[since BPE = EQB = 90° ]

In the given figure given below, E is the mid-point of AB and F is the midpoint of AD. if the area of FAEC is 13, what is the area of ABCD?
38361.png

As F is the mid-point of AD, CF is the median of the triangle ACD to the side AD.

Hence area of the triangle FCD = area of the triangle ACF.

Similarly area of triangle BCE = area of triangle ACE.

∴ Area of ABCD = Area of (CDF + CFA + ACE + BCE)

= 2 Area (CFA + ACE) = 2 × 13 = 26 sq. units.

Give that segment AB and CD are parallel, if lines ℓ, m and n intersect at point O. Find the ratio of θ to ∠ODS
38355.png

Let the line m cut AB and CD at point P and Q respectively

∠ DOQ = x (exterior angle)

Hence, Y + 2x (corresponding angle)

∴ y = x ...(1)

Also . ∠ DOQ = x (vertically opposite angles)

In ∆ OCD, sum of the angles = 180⁰

∴ y + 2y + 2x + x =180°

⇒ 3x + 3y = 180°

⇒ x + y = 60 ...(2)

From (1) and (2)

x = y = 30 = 2y = 60

∴ ∠ ODS = 180 – 60 = 120°

∴ θ = 180 – 3x = 180 – 3(30) = 180 – 90 = 90°.

∴ The required ratio = 90 : 120 = 3 : 4.

AB ⊥ BC and BD ⊥ AC. CE bisects the angle C. ∠A = 30°. Then, what is ∠CED?
37192.png

In ∆ ABC, = ∠C = 180 - 90 - 30 = 60°

geometry-37256.png

Again in ∆ DEC, = ∠ CED = 180 - 90 - 30 = 60°

In triangle ABC, angle B is a right angle. If (AC) is 6 cm, and D is the mid - point of side AC. The length of BD is 37186.png

In a right angled ∆, the length of the median is ½ the length of the hypotenuse .
Hence BD = ½ AC = 3 cm

ABCD is a square of area 4, which is divided into four non overlapping triangles as shown in the fig. Then the sum of the perimeters of the triangles is 37346.png

geometry-37421.png
ABCD is square a² = 4 ⇒ a = 2

geometry-37408.png
perimeters of four triangles

= AB + BC + CD + DA + 2(AC + BD)

geometry-37396.png

The sides of a quadrilateral are extended to make the angles as shown below :
38040.png
What is the value of x?

Sum of all the interior angles of a polygon taken in order is 360°.

geometry-38144.png

i.e., x + 90 + 115 + 75 = 360

⇒ x = 360° – 280° = 80°

⇒ x = 80°

Instead of walking along two adjacent sides of a rectangular field, a boy took a short cut along the diagonal and saved a distance equal to half the longer side. Then the ratio of the shorter side to the longer side is

geometry-38138.png

According to question,

geometry-38132.png

geometry-38125.png

geometry-38119.png

geometry-38113.png

geometry-38106.png

geometry-38100.png

If two parallel lines are cut by two distinct transversals, then the quadrilateral formed by these four lines will always be a :

The quadrilateral obtained will always be a trapeziam as it has two lines which are always parallel to each other.

geometry-37390.png

In the following figure, find ∠ADC.
37466.png

∠ACB = ∠BAC (Angles opposite equal sides are equal)

geometry-37642.png

Similarly, ∠ADC = ∠CAD

∴ ∠ACB = ∠BAC

geometry-37636.png

⇒ ∠ADC = ∠CAD

geometry-37630.png

In a triangle ABC, the internal bisector of the angle A meets BC at D. If AB = 4, AC = 3 and ∠A = 60°, then the length of AD is

geometry-38333.png
Using the theorem of angle of bisector,
geometry-38326.png
geometry-38320.png
In ∆ABD, by sine rule,

geometry-38314.png ...(1)
In ∆ABC, by sine rule;

geometry-38308.png
⇒ geometry-38302.png [putting the value of sin B from (1)]
geometry-38296.png

In the figure AG = 9, AB = 12, AH = 6, Find HC.
37958.png

m ∠ AHG = 180 – 108 = 72⁰

∴ ∠ AHG = ∠ ABC .....(same angle with different names)

∴ ∆ AHG – ∆ABC .....(AA test for similarity)

AH/AB = AG/AG;

⇒ 6/12 = 9/AC

∴ AC = geometry-37890.png = 18

∴ HC = AC – AH = 18 – 6 = 12

In ∆ABC, DE || BC and AD/DB = 3/5. If AC = 5.6 cm, find AE.
38180.png

In ∆ABC, DE || BC

By applying basic Proportionality theorem,

geometry-38239.pngButgeometry-38233.png

∴ geometry-38226.png

⇒ geometry-38219.png

⇒ geometry-38212.png

⇒ geometry-38206.png

⇒ 8AE = 3 × 5.6

⇒ AE = 3 × 5.6/8

∴ AE = 2.1 cm.

In the given fig. AB || QR, find the length of PB.
38174.png

∆PAB ~ ∆PQR

geometry-38200.png

∴ PB = 2cm

In a triangle ABC, the lengths of the sides AB, AC and BC are 3, 5 and 6 cm, respectively. If a point D on BC is drawn such that the line AD bisects the angle A internally, then what is the length of BD?

geometry-37853.png

As AD bisects ∠BAC, we have

geometry-37847.png

⇒ geometry-37841.png

⇒ geometry-37835.png

⇒ geometry-37829.png

⇒ geometry-37823.png

⇒ geometry-37817.png

= 2.25 cm

The number of tangents that can be drawn to two non-intersecting circles is :

Four tangents can be drawn to two non-intersecting circles in the following manner :

geometry-38709.png

Now check your Result..

Your score is

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