Question 1
Question: 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 />
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.
Question 2
Question: 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]
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))
Question 3
Question: 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]
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
| Note: |
| 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. |
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
Note: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.Question 4
Question: 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.
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)
Question 5
Question: In a triangle, AD^ BC, prove that AB2- BD2= AC2- CD2. [2 Mark]
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.
Question 6
Question: In an equilateral triangle with side 'a', prove that the length of the altitude is 
and its area is 
[3 Mark]
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 
Question 7
Question: 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]
Answer: In D ABD, by Pythagoras Theorem
....(1)
In D BEC,
....(2)
Adding (1) and (2), we have
Question 8
Question: In D ABC, if AD is the median, show that 
Answer: 
Draw AE ^ BC.
In 
.... (1) (By Pythagoras Theorem)
In D AEC, AC2 = AE2 + EC2 . . . (2)
Adding (1) and (2), we have
Question 9
Question: 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. 
Answer: Proof of (i)
(AA Similarity)
Proof of (ii)
(AA Similarity)
....(1)
In D CDB,
(From (1))
Question 10
Question: 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
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)
