Frank Solutions for Class 9 Maths Chapter 15 Mid-point and Intercept Theorems act as an important study tool to prepare for Class 9 annual examinations. These solutions are prepared in accordance with the current ICSE Board syllabus, keeping in mind the understanding capacity of students. Practising these solutions on a daily basis will help them to achieve desired scores as well as enhance their problem-solving skills. Frank Solutions are provided here in PDF for students to practise offline without any time constraints and thus encourage concept-based learning.
Chapter 15 has problems related to the midpoints of a triangle. Mid-point theorem states that if any two sides of a triangle midpoints are connected, then the resulting line segment is parallel to the third side of the triangle. By referring to these solutions, students can cross-check their answers and clear doubts while solving textbook problems. Access Frank Solutions for Class 9 Maths Chapter 15 Mid-point and Intercept Theorems from the link below.
Frank Solutions for Class 9 Maths Chapter 15 Mid-point and Intercept Theorems Download PDF
Access Frank Solutions for Class 9 Maths Chapter 15 Mid-point and Intercept Theorems
1. In â–³ABC, D is the mid-point of AB, and E is the mid-point of BC
Calculate:
(i) DE, if AC = 8.6 cm
(ii) ∠DEB, if ∠ACB = 720
Solution:
In â–³ABC,
D and E are the mid-points of AB and BC, respectively
Hence, by mid-point theorem DE || AC and DE = (1/2) AC
(i) DE = (1/2) AC = (1/2) x 8.6
We get,
= 4.3 cm
(ii) ∠DEB = ∠C = 720 (Corresponding angles are equal, since DE || AC)
2. In â–³ABC, AB = 12 cm and AC = 9 cm. If M is the mid-point of AB and a straight line through M parallel to AC cuts BC in N, what is the length of MN?
Solution:
MN || AC and M is the mid-point of AB
Hence, N is the mid-point of BC
Therefore, MN = (1/2) AC
= (9/2) cm
We get,
= 4.5 cm
3. (a) In â–³ABC, D, E, and F are the mid-points of BC, CA and AB, respectively. Find FE, if BC = 14 cm
(b) In â–³ABC, D, E, and F are the mid-points of BC, CA and AB, respectively. Find DE, if AB = 8 cm
(c) In △ABC, D, E, and F are the mid-points of BC, CA and AB, respectively. Fine ∠FDB if ∠ACB = 1150
Solution:
(a)
F is the mid-point of AB, and E is the mid-point of AC
Hence,
FE = (1/2) BC ………… (Mid-point Theorem)
= (1/2) x 14
We get,
= 7 cm
(b)
In â–³ABC,
D is the mid-point of BC, and E is the mid-point of AC
Hence,
DE = (1/2) AB …….. (Mid-point Theorem)
= (1/2) x 8
We get,
= 4 cm
(c)
In â–³ABC,
FD || AC
Hence,
∠FDB = ∠ACB = 1150 …….(Corresponding angles are equal)
4. In parallelogram PQRS, L is the mid-point of side SR, and SN is drawn parallel to LQ, which meets RQ produced at N and cuts side PQ at M. Prove that M is the mid-point of PQ
Solution:
In â–³NSR,
MQ = (1/2) SR
But L is the mid-point of SR and SR = PQ (Sides of parallelogram)
So, it can be written as,
MQ = (1/2) PQ
MQ = PM = LS= LR
Hence, M is the mid-point of PQ
5. In â–³ABC, BE and CF are medians. P is a point on BE produced such that BE = EP, and Q is a point on CF produced such that CF = FQ. Prove that:
(a) QAP is a straight line
(b) A is the mid-point of PQ
Solution:
Since BE and CF are the medians,
F is the mid-point of AB, and E is the mid-point of AC
We know that the line joining the mid-points of any two sides is parallel and half of the third side
We have,
In â–³ACQ,
EF|| AQ and EF = (1/2) AQ ………(1)
In â–³ABP,
EF || AP and EF = (1/2) AP ……….(2)
(a) From (1) and (2)
We get,
AP || AQ (both are parallel to EF)
As AP and AQ are parallel and have a common point, A
This is possible only if QAP is a straight line
Thus, proved
(b) From (1) and (2),
EF = (1/2) AQ and EF = (1/2) AP
⇒ (1/2) AQ = (1/2) AP
⇒ AQ = AP
Therefore, A is the mid-point of QP
6. Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus
Solution:
In the given rectangle ABCD,
Join AC and BD
In â–³ABC, P and Q are the mid-points of AB and BC, respectively
PQ = (1/2) AC …….. (1) and PQ ||AC
In â–³BDC, R and Q are the mid-points of CD and BC, respectively
QR = (1/2) BD ……..(2) and QR || BD
But AC and BD are diagonals of the rectangle
From equations (1) and (2)
PQ = QR
Similarly,
QR = RS and RS = SP
And,
RS || AC and SP || BD
Therefore, PQ = QR = RS = SP
Hence, PQRS is a rhombus
7. D, E and F are the mid-points of the sides AB, BC and CA of an isosceles â–³ABC in which AB = BC. Prove that â–³DEF is also isosceles.
Solution:
E and F are mid-points of BC and AC
Hence, EF = (1/2) AB …….(1)
D and F are the mid-points of AB and AC
Hence, DF = (1/2) BC ……..(2)
But AB = BC
From (1) and (2)
EF = DF
Thus, â–³DEF is an isosceles triangle
8. The diagonals of a quadrilateral intersect each other at right angle. Prove that the figure obtained by joining the mid-points of the adjacent sides of the quadrilateral is a rectangle.
Solution:
P and Q are the mid-points of AB and BC
Hence,
PQ || AC and PQ = (1/2) AC ……. (i)
S and R are the mid-points of AD and DC
Hence,
SR || AC and SR = (1/2) AC …….(ii)
From (i) and (ii),
PQ || SR and PQ = SR
Therefore, PQRS is a parallelogram
Further, AC and BC intersect at right angles
∴ SP || BD and BD ⊥ AC
∴ SP ⊥ AC
⇒ SP ⊥ SR
⇒ ∠RSP = 900
∴ ∠RSP = ∠SRQ = ∠RQP = ∠SPQ = 900
Hence, PQRS is a rectangle
9. In a right-angled triangle ABC. ∠ABC = 900, and D is the mid-point of AC. Prove that BD = (1/2) AC
Solution:
Draw line segment DE ||CB, which meets AB at point E
Now,
DE || CB and AB are the transversal,
∴∠AED = ∠ABC …. (corresponding angles)
∠ABC = 900 (given)
⇒ ∠AED = 900
Also,
Since D is the mid-point of AC and DE || CB,
DE bisects side AB,
Hence,
AE = BE …..(i)
In â–³AED and â–³BED,
∠AED = ∠BED …..(Each 900)
AE = BE ……[From (i)]
DE = DE ….(Common)
Therefore, △AED ≅ △BED ….(By SAS)
⇒ AD = BD …..(C.P.C.T.C)
⇒ BD = (1/2) AC
Hence, proved
10. In a parallelogram ABCD, E and F are the mid-points of the sides AB and CD, respectively. The line segments AF and BF meet the line segments DE and CE at points G and H, respectively. Prove that:
(a) △GEA ≅ △GFD
(b) △HEB ≅ △HFC
(c) EGFH is a parallelogram
Solution:
Since ABCD is a parallelogram,
AB = CD and AD = BC
Now,
E and F are the mid-points of AB and CD, respectively,
Hence,
AE = EB = DF = FC ……….(1)
(a) In â–³GEA and â–³GFD,
AE = DF …… [From (1)]
∠AGE = ∠DGF ……… (vertically opposite angles)
∠GAE = ∠GFD …….. (Alternate interior angles)
Therefore,
△GEA ≅ △GFD
(b) In â–³HEB and â–³HFC,
BE = FC …..[From (1)]
∠EHB = ∠FHC (vertically opposite angles)
∠HBE = ∠HFC (Alternate interior angles)
Therefore,
△HEB ≅ △HFC
(c) In quadrilateral AECF,
AE = CF …..[From (1)]
AE || CF ……(since AB || DC)
Hence,
AECF is a parallelogram
⇒ EC || AF or EH || GF …… (i)
In quadrilateral BFDE,
BE = DF ….[From (1)]
BE || DF ….(since AB || DC)
⇒ BEDF is a parallelogram
⇒ BF || ED or HF || EG ……(ii)
From equations (i) and (ii),
We get,
EGFH is a parallelogram
11. In â–³ABC, the medians BE and CD are produced to the points P and Q, respectively, such that BE = EP and CD = DQ. Prove that:
(a) Q, A and P are collinear
(b) A is the mid-point of PQ
Solution:
In â–³BDC and â–³ADQ,
CD = DQ … (given)
∠BDC = ∠ADQ ……(vertically opposite angles)
BD = AD ……(D is the mid-point of AB)
Therefore,
△BDC ≅ △ADQ
⇒ ∠DBC = ∠DAQ (cpct)…….(i)
And, BC = AQ (cpct) …….(ii)
Similarly,
We can prove
△CEB ≅ △ AEP
⇒ ∠ECB = ∠EAP (cpct)……..(iii)
And, BC = AP (cpct)……..(iv)
(a) In â–³ABC,
∠ABC + ∠ACB + ∠BAC = 1800
⇒ ∠DBC + ∠ECB + ∠BAC = 1800
⇒ ∠DAQ + ∠EAP + ∠BAC = 1800 [From (i) and (iii)]
⇒ Q, A, P are collinear
(b) From (ii) and (iv),
AQ = AP
Therefore, A is the mid-point of PQ
12. In â–³ABC, D and E are two points on the side AB such that AD = DE = EB. Through D and E, lines are drawn parallel to BC, which meet the side AC at points F and G, respectively. Through F and G, lines are drawn parallel to AB, which meets the side BC at points M and N, respectively. Prove that BM = MN = NC
Solution:
In â–³AEG,
D is the mid-point of AE and DF || EG
Hence,
F is the mid-point of AG
AF = FG ……….. (1)
In â–³ABC,
DF || EG|| BC
DE = BE
Hence,
GF = GC …………. (2)
From (1) and (2), we get,
AF = FG = GC
Similarly, since GN|| FM|| AB
Thus, BM = MN = NC (proved)
13. In the given figure, the lines l, m and n are parallel to each other. D is the mid-point of CE. Find
(a) BC
(b) EF
(c) CG
(d) BD
Solution:
According to the equal intercept theorem,
Since CD = DE
AB = BC ………(i)
EF = GF ……….(ii)
(a) BC = AB = 6 cm ….. [From (i)]
(b) EG = EF + FG
EG = 2EF …… [From (ii)]
9 = 2EF
EF = (9/2)
EF = 4.5 cm
(c) CG = 2DF
CG = 2 x 4.2
CG = 8.4 cm
(d) AE = 2BD
BD = (1/2) AE
BD = (1/2) x 12
We get,
BD = 6 cm
14. The diagonals AC and BD of a quadrilateral ABCD intersect at right angles. Prove that the quadrilateral formed by joining the mid-points of quadrilateral ABCD is a rectangle.
Solution:
The figure is as shown below
Let ABCD be a quadrilateral where P, Q, R, and S are the mid-points of sides AB, BC, CD, and DA. Diagonals AC and BD intersect at point ‘O’.
We need to prove that PQRS is a rectangle
Proof:
In â–³ABC and â–³ADC,
2PQ = AC and PQ || AC …….. (1)
2RS = AC and RS || AC …….. (2)
From (1) and (2)
We get,
PQ = RS and PQ || RS
Similarly, we can show that
PS = RQ and PS || RQ
Hence,
PQRS is a parallelogram
PQ || AC
Therefore, ∠AOD = ∠PXO = 900 …. [Corresponding angles]
Again BD || RQ
Therefore, ∠PXO = ∠RQX = 900 …[Corresponding angles]
Similarly,
∠QRS = ∠RSP = ∠SPQ = 900
Hence,
PQRS is a rectangle
15. In â–³ABC, D and E are the midpoints of the sides AB and AC, respectively. F is any point on the side BC. If DE intersects AF at P, show that DP = PE.
Solution:
Note: The given question is incomplete
According to the question given, F could be any point on BC, as shown below
So, this makes it impossible to prove DP = DE
Since P too would shift as F shift because P too would be any point on DE as F is
Note: If we are given F to be the mid-point of BC, the result can be proved
Here,
D and E are the mid-points of AB and AC, respectively
DE || BC and DE = (1/2) BC
But F is the mid-point of BC
BF = FC = (1/2) BC = DE
Since D is the mid-point of AB, and DP || BF
Since P is the mid-point of AF and E is the mid-point of AC,
PE = (1/2) FC
Also,
D and P are the mid-points of AB and AF, respectively
⇒ DP = (1/2) BF = (1/2) FC = PE …….. (since BF = FC)
⇒ DP = PE
Hence, proved.
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