ज्यामिति प्रश्नावली

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

⇒

⇒ 2n = 22

⇒ n = 11

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

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.)

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.

Let the diagonals of the rhombus be x and y and the its sides be x

Now,

⇒

⇒ 3x² = y²

⇒

⇒ y : x

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° ]

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.

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.

In ∆ ABC, = ∠C = 180 - 90 - 30 = 60°

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

In a right angled ∆, the length of the median is ½ the length of the hypotenuse .

Hence BD = ½ AC = 3 cm

ABCD is square a² = 4 ⇒ a = 2

perimeters of four triangles

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

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

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

⇒ x = 360° – 280° = 80°

⇒ x = 80°

According to question,

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

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

Similarly, ∠ADC = ∠CAD

∴ ∠ACB = ∠BAC

⇒ ∠ADC = ∠CAD

Using the theorem of angle of bisector,

In ∆ABD, by sine rule,

...(1)

In ∆ABC, by sine rule;

⇒ [putting the value of sin B from (1)]

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 = = 18

∴ HC = AC – AH = 18 – 6 = 12

In ∆ABC, DE || BC

By applying basic Proportionality theorem,

But

∴

⇒

⇒

⇒

⇒ 8AE = 3 × 5.6

⇒ AE = 3 × 5.6/8

∴ AE = 2.1 cm.

∆PAB ~ ∆PQR

∴ PB = 2cm

As AD bisects ∠BAC, we have

⇒

⇒

⇒

⇒

⇒

= 2.25 cm

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