Quadrilaterals Test

Result

Quadrilaterals Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

The quadrilateral formed by joining the mid-points of the sides $$AB,BC,CD,DA$$ of a quadrilateral $$ABCD$$ is

A
a trapezium but not a parallelogram

B
a quadrilateral but not a trapezium

C
a parallelogram only

D
a rhombus

Solution
Question 2
1 / -0

If $$ABC$$ is a triangle, right-angled at $$B$$ and $$M,N$$ are midpoints of $$AB$$ and $$BC$$ respectively, then what is $$4({AN}^{2}+{CM}^{2})$$ equal to?

A
$$3{AC}^{2}$$

B
$$4{AC}^{2}$$

C
$$5{AC}^{2}$$

D
$$6{AC}^{2}$$
Question 3
1 / -0

$$ABC$$ is a triangle in which $$AB=AC$$. Let $$BC$$ be produced to $$D$$. From a point $$E$$ on the line $$AC$$ let $$EF$$ be a straight line such that $$EF$$ is parallel to $$AB$$. Consider the quadrilateral $$ECDF$$ thus formed. If $$\angle ABC={65}^{o}$$ and $$\angle EFD={80}^{o}$$, then what is $$\angle FDC$$ equal to?

A
$${42}^{o}$$

B
$${41}^{o}$$

C
$${37}^{o}$$

D
$${35}^{o}$$
Question 4
1 / -0

Which of the following statements is INCORRECT ?

A
Each diagonal of a quadrilateral divides it into two triangles.

B
Each side of a quadrilateral is less than the sum of the renraining three sides

C
A quadriiateral can have atmost three angles

D
A quadrilateral has two diagonals.

Solution
Question 5
1 / -0

In the figure given, $$M$$ is the midpoint of the side $$CD$$ of the parallelogram $$ABCD$$. What is $$ON:OB$$?

A
$$3:2$$

B
$$2:1$$

C
$$3:1$$

D
$$5:2$$
Question 6
1 / -0

The figure formed by joining the mid-points of the sides of a quadrilateral $$ABCD$$, taken in order, is a square only if,

A
$$ABCD$$ is a rhombus

B
diagonals of $$ABCD $$ are equal

C
diagonals of $$ABCD$$ are equal and perpendicular

D
diagonals of $$ABCD$$ are perpendicular.

Solution

In given figure,
$$ABCD$$ is a quadrilateral and $$P, Q, R$$ & $$S$$ are mid-pints of sides $$AB, BC, CD$$ and $$DA$$ respectively. Then, $$PQRS$$ is a square.
$$\therefore PQ = QR = RS = PS$$ --------- (1)
and $$PR = SQ$$
But $$PR = BC$$ and $$SQ = AB$$
$$\therefore AB = BC$$
Thus, all sides of quadrilateral $$ABCD$$ are equal.
Hence, quadrilateral $$ABCD$$ is either a square or a rhombus.
Now, in $$\triangle ADB$$,
By using Mid-point theorem,
SP$$\left|\right| DB; SP = \dfrac{1}{2} DB$$ ------ (2)
Similarly in $$\triangle$$ABC,
$$PQ \left|\right| AC; PQ = \dfrac{1}{2}AC$$ ----- (3)
From equation (1),
$$PS = PQ$$
From (2) and (3),
$$\dfrac{1}{2} DB = \dfrac{1}{2} AC$$
$$\therefore DB = AC$$

Thus, diagonals of $$ABCD$$ are equal and therefore quadrilateral $$ABCD$$ is a square. So, diagonals of quadrilateral also perpendicular.
Question 7
1 / -0

The basic elements of a Quadrilateral are _____.

A
4 vertices

B
4 sides

C
4 angles

D
All of these

Solution
Question 8
1 / -0

Which of the following statements is INCORRECT?

A
Each diagonal of a quadrilateral divides it into two triangles

B
Each side of a quadrilateral is less than the sum of the remaining three sides

C
A quadrilateral can have atmost three angles

D
A quadrilateral has two diagonals
Question 9
1 / -0

If length of each side of a rhombus $$PQRS$$ is $$8cm$$ and $$\angle {PQR}={120}^{o}$$, then what is the length (in cm) of $$QS$$?

A
$$4\sqrt {5}$$

B
$$6$$

C
$$8$$

D
$$12$$

Solution

$$PQS$$ is an equilateral triangle
Hence, $$QS=8cm$$
Question 10
1 / -0

$$ABCD$$ is a quadrilateral inscribed in a circle. Diagonals $$AC$$ and $$BD$$ are joined.
If $$\angle CAD = 60^{\circ}$$ and $$\angle BDC = 25^{\circ}$$. Find $$\angle BCD$$

A
$$85^{\circ}$$

B
$$120^{\circ}$$

C
$$115^{\circ}$$

D
$$95^{\circ}$$

Solution

$$\angle CAD=60^{\circ}$$
& $$\angle BDC=25^{\circ}$$
$$\angle CAD=\angle DBC=60^{\circ}$$ [As angles are in same segment].
In $$\triangle BCD$$,
$$\angle DBC+\angle BDC+\angle BCD=180^{\circ}$$
$$\Rightarrow 60^\circ+25^\circ+\angle BCD=180^{\circ}$$
$$\Rightarrow \angle BCD=180^{\circ}-85^{\circ}=95^{\circ}$$