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Question (1):
Prove that if a line be drawn Parallel to the base of aD , the portion intercepted by the
sides is bisected by the median to the base. [3 Mark]
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Answer:
Given:
In D ABC, PQ||BC. AD is the median of BC and meets BC in D.
To prove:
PR = RQ
R is midpoint of PQ.
Proof:
Since PQ||BC,
D APR and D ABD are equiangular.
D APR ~ DABD (AAA)
 ....(1)
Similarly, D ARQ ~ D ADC
 ....(2)
 (From (1) and (2) both are equal to )
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Question (2):
In  [3 Mark]
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Answer:
Given:
or
Proof:
In D ADC and D BDA,
 (given)
 (SAS similarity criterion)
 The two triangles are equiangular.
 ....(1)
Now in D ABC,
 (Angle sum property of a triangle)
 (From (1))
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Question (3):
ABC is a triangle in which AB = AC and D is a point on the side AC such that BC2 = AC
x CD. Prove that BD = BC. [2 Mark]
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Answer:
Given:
AB = AC and BC2 = AC x CD
i.e., 
To prove:
BD = BC
In D BCA and D DCB,
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| That we name triangle ABC as BCA to keep up the correspondence with D DCB. The
corresponding angles at the vertices C in both triangles are equal.
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Proof:
 (given)
 (common angle)
 (SAS similarity criterion)
 ....(1)
But AB = AC (given)
 BC = BD (From |
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Question (4):
In the given figure, D ABC ~ D DEF. If AB = 2DE and area of D ABC = 56 sq.cm, then find the area of D DEF. [2 Mark]
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Answer:
We know that,
if D ABC ~ D DEF, then
 [The ratio of areas of two similar  is equal to the ratio of the squares of
their corresponding sides]
\ Area of D DEF = 14 sq.cm
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Question (5):
In the given figure, DE ll BC and AD:DB = 5:4. find  [2 Mark]
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Answer:
 (By AA similarity criterion)
 ....(1)
In D DFE and D CFB,
 (Alternate angle)
 (Vertically opposite angle)
 (AA Similarity)
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Question (6):
D, E and F are the mid-point of the sides BC, CA and AB respectively of a D ABC.
Determine the ratio of the areas of D DEF and D ABC. [3 Mar />
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Answer:
By mid-point theorem
DE||BC
 ....(1)
 AFDE is a parallelogram. (Since opposite sides are parallel)
 (Opposite angles of a parallelogram are equal0
Similarly 
 are equiangular and hence they are similar.
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Question (7):
ABC is a triangle and DE is a line segment meeting AB in D and AC in E such that
DE||BC and divides D ABC into two parts of equal area. Find
 [3 Mark]
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Answer:
Given:
Area (D ADE) = Area (trapezium DECB) ... (1)
Area (D ABC) = Area (D ADE) + Area (trap DECB)
= Area (D ADE) + Area (D ADE)
 Area (D ABC) = 2(Area (D ADE)) ... (2)
 (By AA similarity criterion)
But  (From (2))
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Question (8):
In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects
BC at O, then prove that [3 Mark]
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Answer:
Construction:
Draw AE BC and DF BC
Proof:
 ....(1)
Now in D AOE and D DOF, we have
 (Vertically opposite angles)
 (each angle = 90o)
 (AA similarity)
 ....(2)
From (1) and (2), we get
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| If two triangles have a common vertex and their bases are along the same straight line, the ratio between their areas is equal to the ratio between the lengths of their bases.
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The triangles ABC, ABD, ADC, AEC have a common vertex A and their bases are along the same
line BC.
Then, we have
i. 
ii. 
iii. 
The proof of (i) is shown below
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Question (9):
In the figure, ABCD is a parallelogram. P is a point on BC such that BP:PC = 1:2. DP
produced meets AB produced at Q. If area of D CPQ = 20 cm2, find [3 Mark]
i. Area of D BPQ
ii. Area of D CDP
iii. Area of parallelogram ABCD.
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Answer:
D CPQ and D BPQ have the common vertex Q, bases on the same line, therefore
 ... (1)
ii.  (AAA Similarity)
 (From 1)
= 4 x 10
= 40cm2
iii. 
= 40 + 20 = 60cm2
Area of (parallelogram ABCD) = 2 x Area (D QDC) = 120 cm2 ( If a parallelogram and a triangle are on the same base and between the same parallels; area of the parallelogram is twice the area of the triangle)
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Question (10):
In a triangle, AD^ BC, prove that AB2- BD2= AC2- CD2. [2 Mark]
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Answer:
Given:
In D ADB,
 (Pythagoras Theorem)
 ... (1)
In D ADC,
 (Pythagoras theorem)
 ...(2)
L.H.S of (1) and (2) are equal.
\ R.H.S of (1) and (2) must be equal.
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Question (11):
In an equilateral triangle with side 'a', prove that the length of the altitude is  and
its area is  [3 Mark]
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Answer:
Let the altitude from A to BC be AD = h.
 BD = DC
Given AB = BC = CA = a
D ADB is right angles triangle.
Area of 
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Question (12):
In the given figure, ABC is a right triangle, right-angled at B. AD and CE are the two
medians drawn from A and C respectively. If AC = 5cm and AD = cm, find the length of CE.
[3 Mark]
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Answer:
In D ABD, by Pythagoras Theorem
 ....(1)
In D BEC,
 ....(2)
Adding (1) and (2), we have
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Question (13):
In D ABC, if AD is the median, show that
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Answer:
Draw AE ^ BC.
In  .... (1) (By Pythagoras Theorem)
In D AEC, AC2 = AE2 + EC2 . . . (2)
Adding (1) and (2), we have
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Question (14):
ABC is a right triangle, right angled at C. If P is the length of the perpendicular drawn
from C to AB and a, b, c have the usual meaning, then prove that
[3 Mark]
i. 
ii. 
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Answer:
Proof of (i)
 (AA Similarity)
Proof of (ii)
 (AA Similarity)
 ....(1)
In D CDB,
 (From (1))
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Question (15):
In a right triangle ABC, right angled at C, P and Q are the points on the sides CA and
CB respectively which divides these sides in the ratio 1:2. Prove that [5 Mark]
i. 9AQ2 = 9AC2 + 4BC2
ii. 9BP2 = 9BC2 + 4AC2
iii. 9(AQ2 + BP2) = 13AB2
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Answer:
P divides CA in the ratio 1:2.
 ....(1)
Similarly CQ ....(2)
Proof of (i)
 (Applying Pythagoras theorem to right-angled  ACQ)
 (Multiplying by 9 throughout)
 (From 2)
 .... (3)
Proof of (ii)
Similarly in 
 (From (1))
 .... (4)
Proof of (iii)
Adding (3) and (4),
= 13 AB2 (from D ABC)
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Question (16):
ABC is an equilateral triangle P is a point on BC such that BP:PC = 2:1, prove that 9AP2 = 7AB2. [5 Mark]
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Answer:
Draw AD ^ BC
Let PC = x, then BP = 2x
BC = BP + PC = 3x
D is the mid-point of BC. 
In 
 .... (1)
Now in 
 (From (1))
 
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Question (17):
In an isosceles triangles ABC, with AB = AC, BD is perpendicular from B to the side
AC. Prove that BD2 - CD2 = 2CD.AD. [2 Mark]
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Answer:
From D ADB,
 |
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Question (18):
In a quadrilateral ABCD, and AD2 = AB2 + BC2 + CD2,
Prove that [3 Mark]
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Answer:
Given:
To prove:
Proof:
AD2 = AB2 + BC2 + CD2 (given)
But AC2 = AB2 + BC2 ( ABC is right angled; By Pythagoras Theorem)
 AD2 = AC2 + CD2
 (By converse of Pythagoras theorem) |
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Question (19):
Prove that in any triangle the sum of the squares on any two sides is equal to twice
the square on half the third side together with twice the square of the median.
[5 Mark]
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Answer:
Given:
In D ABC, AD is the median.
To prove:
AB2 + AC2 = 2BD2 + 2AD2
Construction:
Draw AM ^ BC.
Proof:
In D ABM,
 ....(1)
In D AMC,
Adding (1) and (2) , we get
 (DC = BD) |
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Question (20):
In the figure, D and E trisects BC, prove that
8AE2 = 3AC2 + 5AD2. >[5 Mark]
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Answer:
Given:
D and E trisects BC. Let BD = DE = EC = x
To prove:
8AE2 = 3AC2 + 5AD2
Proof:
BC = 3x, EB = 2x
In 
 ....(1) (Pythagoras theorem)
Similarly in D ABE,
 ....(2)
In D ABC,
 ... (3)
Now 
= 8AE2 (From (2))
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