Quadrilaterals Test

Result

Quadrilaterals Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

If the area of quadrilateral ABCD is zero, then the four points A,B,C,D are

A
collinear

B
not collinear

C
nothing can be said

D
none of these

Solution
Question 2
1 / -0

A square is drawn by joining the mid points of the given square a third square in the same way and this process continues indefinitely. If a side of the first square is $$16$$ cm , then the sum of the area of all of the squares

A
$$128$$ sq. cm

B
$$256$$ sq. cm

C
$$512$$ sq. cm

D
$$1024$$ sq. cm

Solution
Question 3
1 / -0

In the given figure, if $${ b }^{ \circ }-{ d }^{ \circ }={ 80 }^{ \circ },{ m }^{ \circ }={ 140 }^{ \circ }and\quad { m }^{ \circ }+{ e }^{ \circ }={ 2c }^{ \circ },then\quad ({ a }^{ \circ }+{ d }^{ \circ })-({ e }^{ \circ }+{ C }^{ \circ })$$

A
$${0 }^{ \circ }$$

B
$${ 40 }^{ \circ }$$

C
$${ 80 }^{ \circ }$$

D
$${ 50 }^{ \circ }$$

Solution
Question 4
1 / -0

If D is mid-point of the side BC of a triangle ABC and AD is perpendicular to AC then $$\dfrac{a^2 c^2}{b^2}$$ =

A
1

B
2

C
3

D
4

Solution
Question 5
1 / -0

Let M, N, P and Q be mid points of the edges AB, CD, AC and BD respectively of the tetrahedron ABCD. Further, MN is perpendicular to both AN and CD and PQ is perpendicular to both AC and BD. Then which of the following is/are correct :

A
AB = CD

B
BC = DA

C
AC = BD

D
AN = BN

Solution
Question 6
1 / -0

In right triangle ABC, right angle is at C, M is the mid-point Of hypotenuse AB, C is joined to M and produced to a point D such that DM = CM. point D is joined to point B show that :

A
$$\Delta AMC\cong \Delta BMD$$

B
$$\angle DBC$$ is a right angle

C
$$\Delta DBC\cong \Delta ACB$$

D
$$CM=\dfrac { 1 }{ 2 } AB$$

Solution
Question 7
1 / -0

A,B,C,D are mid points of sides of parallelogram PQRS. If $$ar\left( {PQRS} \right) = 36\,\,c{m^2}$$ then $$ar(ABCD)$$

A
$$24\,\,c{m^2}$$

B
$$18\,\,c{m^2}$$

C
$$30\,\,c{m^2}$$

D
$$36\,\,c{m^2}$$

Solution
Question 8
1 / -0

In a quadrilateral ABCD, the point P divides DC in the ratio 1:2 and Q is the mid point of AC. If $$\overline{AB}+\overline{2AD}+\overline{BC}-\overline{2DC}=k\overline{PQ}.$$ then $$k=$$

A
-6

B
-4

C
6

D
4

Solution
Question 9
1 / -0

PQRS 5 $$\angle Q=7\angle s,\quad \angle Q=....$$

A
$${ 105 }^{ 0 }$$

B
$${ 75 }^{ 0 }$$

C
$${ 90 }^{ 0 }$$

D
$${ 60 }^{ 0 }$$
Question 10
1 / -0

The mid point of the side of a triangle along with any of the vertices as the fourth point make a parallelogram of area=

A
$$\frac{1}{2}ar(ABC)$$

B
$$\frac{1}{3}ar(ABC)$$

C
$$\frac{1}{4}ar(ABC)$$

D
$$ar(ABC)$$

Solution