Mathematics Questions and Answers – The Mid-Point Theorem

«
»

This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “The Mid-Point Theorem”.

1. What is the length of DE if DE || BC and D and E are midpoints of AB and AC?

a) 18cm
b) 15cm
c) 9cm
d) 20cm
View Answer

Answer: c
Explanation: Since In ΔABC, D and E are midpoints of the sides AB and AC and DE || BC, by mid-point theorem, DE = ½ BC = ½ x 18 = 9cm.
advertisement

2. Which of the following relation is correct if S is midpoint of PR?

a) QS = PR
b) QS = ½ PR
c) QS = 2PR
d) QS = ¼ PR
View Answer

Answer: b
Explanation: Draw ST || RQ meeting PQ at point T

∠STQ + ∠TQR = 180°  (Interior angles on the same side of transversal)
⇒ ∠STQ = 90°
Also, Since ST || RQ and S is midpoint of PR, by midpoint theorem,
T is midpoint of PQ i.e. PT = TQ  ————— (i)
and ST= ½ RQ
In ΔPTS and ΔTQS,
PT = TQ  (from equation i)
∠PTS = ∠STQ = 90°
TS = TS  (Common side)
Hence, ΔPTS ≅ ΔTQS  (by SAS congruence criterion)
⇒ PS = QS  (Corresponding parts of congruent triangles)
But, PS = ½ PR  (S is midpoint of PR)
⇒ QS = ½ PR.

3. What is the ration of AC/CF if AD and BE are the median and DF || BE?

a) 4
b) 1
c) 1/4
d) 1/2
View Answer

Answer: a
Explanation: Since In ΔBEC, D is the midpoint of the BC and DF || BE, by mid-point theorem,
DF = ½ BE and
F is midpoint of EC i.e. EF = FC  ——————–(i)
Also, AE = EC = ½ AC ——————-(ii) (BE is median to side AC)
From i, CF = ½ EC
⇒ CF = ½ (½ AC)  (from equation ii)
⇒ CF = ¼ AC.
advertisement
advertisement

4. Identify the correct relation from the given options if PQRS is a parallelogram and V is midpoint of PQ.

a) TU = ¼ PU
b) TU = PU
c) TU = ½ PU
d) TU = 2PU
View Answer

Answer: c
Explanation: Since In ΔPQU, V is the midpoint of the PQ and VR || PU, by mid-point theorem,
VR = ½ PU and
R is midpoint of QU i.e. QR = RU  ——————–(i)
Also, in ΔPQU, R is the midpoint of the QU and RT || PQ (PQRS is parallelogram), by mid-point theorem,
TR = ½ PQ and
T is midpoint of PU
i.e. PT = TU
⇒ TU = ½ PU.

5. Find the ration of (AB + CD)/EG if EG || AB and AB || DC and E and G are the midpoints.

a) 1
b) 1/2
c) 2
d) 3/2
View Answer

Answer: c
Explanation: In ΔABD, EF || AB and E is the midpoint of the AD, by mid-point theorem,
F is the midpoint of BD
and EF = ½ AB  —————— (i)
Also, in ΔBDC, G and F are the midpoints of the BC and BD and FG || DC, by mid-point theorem,
FG = ½ CD  —————— (ii)
Adding equation i and ii, EF + FG = ½ AB + ½ CD
⇒ EG = ½ (AB + CD).
advertisement

6. Find the perimeter of ΔABC, if perimeter of ΔPQR is 36cm and A, B and C are midpoints.

a) 9cm
b) 18cm
c) 20cm
d) 36cm
View Answer

Answer: b
Explanation: In ΔPQR, A and B are the midpoints of the PQ and PR, by mid-point theorem,
AB || QR and AB = ½ QR  —————— (i)
Similarly, A and C are the midpoints of the PQ and QR, by mid-point theorem,
AC || PR and AC = ½ PR  —————— (i)
and B and C are the midpoints of the PR and QR, by mid-point theorem,
BC || PQ and BC = ½ PQ  —————— (i)
Now, perimeter of ΔABC = AB + BC + AC
⇒ perimeter of ΔABC = ½ QR + ½ PQ + ½ PR  (from equation i, ii and iii)
⇒ perimeter of ΔABC = ½ (QR + PQ + PR)
⇒ perimeter of ΔABC = ½ (perimeter of ΔPQR) = ½ x 36 = 18cm.

7. Find the ratio of the angles D : E : F of ΔDEF formed by joining the midpoints of the sides of ΔABC.

a) 4 : 2 : 3
b) 4 : 3 : 2
c) 2 : 3 : 5
d) 5 : 3 : 2
View Answer

Answer: a
Explanation: Since D, E and F are the midpoints of the sides AB, BC and AC of ΔABC, by mid-point theorem, DF || BC, DE || AC and FE || AB
As EF || AB and DF || BC, ⇒ EF || BD and DF || BE ⇒ DFEB is a parallelogram
⇒ ∠B = ∠DFE = 60°  ————— (i) (Opposite angles of parallelogram are equal)
Similarly, As EF || AB and DE || AC, ⇒ EF || AD and DE || AF ⇒ AFED is a parallelogram
⇒ ∠A = ∠DEF = 40° ——————– (ii) (Opposite angles of parallelogram are equal)
and as DE || AC and DF || BC, ⇒ DE || FC and DF || EC ⇒ DECF is a parallelogram
⇒ ∠C = ∠EDF = 80°  ——————— (iii) (Opposite angles of parallelogram are equal)
Now, ∠D: ∠E: ∠F = 80 : 40 : 60  (from equation i,ii and iii)
⇒ ∠D: ∠E: ∠F = 4 : 2 : 3.
advertisement

8. If ABCR and PQRS are rectangles and B is the midpoint of PR, find the ratio of AC/PR.

a) 1/2
b) 2
c) 1
d) 5/2
View Answer

Answer: a
Explanation: In ΔPSR, B is the midpoint of PR and AB || SP, by mid-point theorem,
A is midpoint of SR i.e. AS = AR  —————— (i)
Similarly, in ΔSRQ, A is the midpoint of SR and AC || SR, by mid-point theorem,
AC = ½ SQ and C is midpoint of RQ i.e. RC = CQ  —————— (i)
Now, AC = ½ SQ
⇒ AC = ½ PR  (Diagonals of rectangle are equal)
⇒ AC/PR = 1/2.

9. From the diagram below, if D and E are mid points of the lines AB and AC respectively and if DE ║ BC, then what is the relation between DE and BC?

a) DE = BC
b) 3DE = BC
c) 2DE = BC
d) There is no definite relation between DE and BC
View Answer

Answer: c
Explanation: According to mid-point theorem, if D and E are mid points of the lines AB and AC respectively and if DE ║ BC, then 2DE = BC.
advertisement

10. From the diagram given below, if P, Q and R are mid points of sides AB, BC and AC respectively, then ΔRQP is congruent to ΔBDE, ΔDAF and ΔEFC.

a) True
b) False
View Answer

Answer: a
Explanation: P, Q and R are mid points of sides AB, BC and AC respectively (given).
Therefore, according to theorem 8.9, PQ ║ AC, QR ║ AB and PR ║ BC.
Hence, APQR, PBQR and PQCR are all parallelograms.
Now, PQ is the diagonal of the parallelogram PBQR. Therefore, according to theorem 8.1,
ΔPQR is congruent to ΔQPB.
Similarly, QR and RP are the diagonals of parallelograms PQCR and APQR respectively.
Therefore, ΔPQR is congruent to ΔCRQ and
ΔPQR is congruent to ΔRAP.

Sanfoundry Global Education & Learning Series – Mathematics – Class 9.

To practice all areas of Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

advertisement
advertisement
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter