Mid-Point Theorem

1. There will be total 10 MCQ in this test.
2. Please keep a pen and paper ready for rough work but keep your books away.
3. The test will consist of only objective type multiple choice questions requiring students to mouse-click their correct choice of the options against the related question number.
4. No negative marking for incorrect choice.
5. Please click on “Finish the Test” when you want to exit.
6. All the best – Happy Studying

1. D, E, F are midpoints of sides AB, BC and CA of ΔABC, if ar(ΔABC) = 64 cm² then, area of ΔBDE is:

32 cm²

12 cm²

8 cm²

16 cm²

Your comments:

Question 1 of 10

2. In the adjoining figure, ABCD and PQRC are rectangles, where Q is the midpoint of AC. Then DP is equal to

DP =1/3 DC

DP = PC

DP < PC

DP > PC

Your comments:

Question 2 of 10

3. In triangle ABC, E and F are the mid points of the sides AC and AB respectively. The altitude AP to BC intersects EF at Q. Then

A

B

C

D

Your comments:

Question 3 of 10

4. Figure shows that AD and BF are medians of ABC and BF DE then CE is equal to

1/2 AC

1/5 AC

1/4 AC

1/3 AC

Your comments:

Question 4 of 10

5. In , D, E and F are respectively mid points of BC, CA and AB. If the lengths of side AB, BC and CA are 7 cm, 8 cm and 9 cm respectively, the perimeter of DEF will be

19 cm.

15 cm

20 cm

12 cm

Your comments:

Question 5 of 10

6. If D, E and F are the mid points of the sides BC, CA and AB of an equilateral triangle ABC, then triangle DEF is

Equilateral triangle

Right tangled triangle

Isosceles triangle

Scalene triangle

Your comments:

Question 6 of 10

7. The quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rectangle is a

Square

Rhombus

Trapezium

Rectangle

Your comments:

Question 7 of 10

8. ABCD is a parallelogram in which P, Q, R and S are the mid points of sides AB, BC CD and DA respectively. Then,

PQ=1/2 CA

PQ > CA

PQ < CA

PQ = CA

Your comments:

Question 8 of 10

9. ABCD is a trapezium in which AB || DC and let E be the midpoint of AD. Let F be a point on BC such that EF || AB. Then MF is

1/2 AB

1/4 DC

1/3 AB

1/2 DC

Your comments:

Question 9 of 10

10. In the adjoining figure, the side AC of ABC is produced to E such that CE = 1/2 AC. If D is the midpoint of BC and ED produced meets AB at F, and CP, DQ are drawn parallel to BA, then FD is

FD = 1/3 FE

FD = 1/4 FE

FD = 1/2 FE

FD = 2/3 FE

Your comments:

Question 10 of 10

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