Geometry Quiz

The angle bisectors of a parallelogram form a :

• Trapezium

• Rectangle

• Rhombus

• Kite

Tags

CCSS.HSG.CO.C.11

The angles of a quadrilaterals are in the ratio 3:4:5:6. The respective angles of the quadrilaterals are

• 60⁰, 80⁰, 100, 120⁰

• 120⁰, 100°, 80°, 60°

• 120°, 60°, 80, 100⁰

• 80⁰, 100⁰, 120°, 60°.

Answer explanation

QUICK APPROACH: Check options in the same ratio.

If diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a:

• Parallelogram

• Square

• Rhombus

• Trapezium

Tags

CCSS.HSG.CO.C.11

The line segment joining the mid points of the two sides of triangle is parallel to the third side and ......... of it

• equal

Tags

CCSS.HSG.SRT.B.4

Three angles of a quadrilateral are 75°, 90° and 75°. The fourth angle is

• 90⁰

• 95⁰

• 105⁰

• 120⁰

Tags

CCSS.HSG.C.A.3

A diagonal of a rectangle is inclined to one side of the rectangle at 25°. The acute angle between the diagonals is

• 55⁰

• 50⁰

• 40⁰

• 25⁰

Answer explanation

In ∆AOB,
∠OAB + ∠ABO + ∠AOB = 180°
⇒ 25° + 25° + ∠AOB = 180°
⇒ ∠AOB + 50° = 180°
⇒ ∠AOB = 180° − 50°
⇒ ∠AOB = 130° which is an obtuse angle

Now, ∠AOB + ∠AOD = 180° (angles on a straight line)
⇒ 130° + ∠AOD = 180°
⇒ ∠AOD = 180° − 130°
⇒ ∠AOD = 50° which is an acute angle

Hence, the acute angle between the diagonals is 50°.

Tags

CCSS.HSG.CO.C.11

ABCD is a rhombus such that ∠ACB = 40°, then ∠ADB =

• 45⁰

• 50⁰

• 40⁰

• 60⁰

Answer explanation

In rhombus ABCD, ∠ACB=40∘

∴∠BCD=2×∠ACB=2×40∘=80∘

But

∠BCD+∠ADC=180∘ (Sum of consecutive angles of ||gm)

⇒80∘+∠ADC=180∘

⇒∠ADC=180∘−80∘=100∘

∴∠ADB=(1/2)∠ADC

=(1/2)×100∘

=50∘

Tags

CCSS.HSG.CO.C.11

The quadrilateral formed by joining the midpoints of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if

• PQRS is a rectangle

• diagonals of PQRS are perpendicular

• PQRS is a rhombus

• diagonals of PQRS are equal.

Answer explanation

Given:
PQRS is a quadrilateral
The quadrilateral formed by joining the mid points of the sides of a quadrilateral PQRS, is a rectangle.
Let the rectangle be ABCD.
A is the mid-point of PQ,

B is the mid-point of QR,

C is the mid-point of RS

and D is the mid-point of PS.

Using Mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.

In ∆PQR,
AB || PR

and

In ∆QRS,
BC || QS

Since, AB⊥BC
Therefore, PR⊥QS


Hence, the quadrilateral formed by joining the mid points of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if diagonals of PQRS are perpendicular.

The quadrilateral formed by joining the midpoints of the sides of a quadrilateral PQRS, taken in order, is a rhombus, if

• PQRS is a rhombus

• diagonals of PQRS are perpendicular

• PQRS is an parallelogram

• diagonals of PQRS are equal.

Answer explanation

Given:
PQRS is a quadrilateral
The quadrilateral formed by joining the mid points of the sides of a quadrilateral PQRS, is a rhombus.
Let the rhombus be ABCD.
A is the mid-point of PQ, B is the mid-point of QR, C is the mid-point of RS and D is the mid-point of PS.

Using mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.

In ∆PQR,
AB = 12PR

and In ∆QRS,
BC = 12QS

Since, AB = BC (sides of a rhombus are equal)
Therefore, 12PR = 12QS
⇒ PR = QS
​

Hence, the quadrilateral formed by joining the mid points of the sides of a quadrilateral PQRS, taken in order, is rhombus, if diagonals of PQRS are equal.

If angles A, B, C and D of the quadrilateral ABCD, taken in order are in the ratio 3:7:6:4, then ABCD is a

• parallelogram

• kite

• rhombus

• trapezium

Answer explanation

Given, ratio of angles of quadrilateral ABCD is 3: 7: 6: 4

Let angles of quadrilateral ABCD be 3x, 7x, 6x and 4x, respectively.

We know that, sum of all angles of a quadrilateral is 360∘.

∴3x+7x+6x+4x=360∘

⇒20x=360∘

⇒x=360∘/20∘

x=18∘
∴ Angles of the quadrilateral are
∠A=3×18=54∘

∠B=7×18=126∘

∠C=6×18=108∘
And ∠D=4×18=72∘
From figure,

∠BCE=180∘−∠BCD [linear pair axiom]
⇒∠BCE=180∘−108∘=72∘
Since, the corresponding angles are equal .
∴BC||AD
Now, sum of co-interior angles,
∠A+∠B=126∘+54∘=180∘
and ∠C+∠D=108∘+72∘=180∘
Hence, ABCD is a trapezium.